Group Theory - James Milne

Group TheoryJ.S. MilnerfS 3DD 1 2 32 3 1 1 2 31 3 2Version 3.10September 24, 2010

The first version of these notes was written for a first-year graduate algebra course. As inmost such courses, the notes concentrated on abstract groups and, in particular, on finitegroups. However, it is not as abstract groups that most mathematicians encounter groups,but rather as algebraic groups, topological groups, or Lie groups, and it is not just the groupsthemselves that are of interest, but also their linear representations. It is my intention (oneday) to expand the notes to take account of this, and to produce a volume that, while stillmodest in size (c200 pages), will provide a more comprehensive introduction to grouptheory for beginning graduate students in mathematics, physics, and related fields.BibTeX information@misc{milneGT,author={Milne, James S.},title={Group Theory (v3.10)},year={2010},note={Available at www.jmilne.org/math/},pages={131}}Please send comments and corrections to me at the address on my website www.jmilne.org/math/.v2.01 (August 21, 1996). First version on the web; 57 pages.v2.11 (August 29,2003). Fixed many minor errors; numbering unchanged; 85 pages.v3.00 (September 1, 2007). Revised and expanded; 121 pages.v3.01 (May 17, 2008). Minor fixes and changes; 124 pages.v3.02 (September 21, 2009). Minor fixes; changed TeX styles; 127 pages.v3.10 (September 24, 2010). Many minor improvements; 131 pages.The multiplication table of S 3 on the front page was produced by Group Explorer.Copyright c1996, 2002, 2003, 2007, 2008, 2010 J.S. Milne.Single paper copies for noncommercial personal use may be made without explicit permissionfrom the copyright holder.

ContentsContents 31 Basic Definitions and Results 7Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Multiplication tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Groups of small order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Normal subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18Kernels and quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Theorems concerning homomorphisms . . . . . . . . . . . . . . . . . . . . . . . 21Direct products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Commutative groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25The order of ab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 Free Groups and Presentations; Coxeter Groups 31Free monoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31Free groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Generators and relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Finitely presented groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Coxeter groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403 Automorphisms and Extensions 41Automorphisms of groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41Characteristic subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43Semidirect products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44Extensions of groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48The Hölder program. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514 Groups Acting on Sets 53Definition and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53Permutation groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60The Todd-Coxeter algorithm. . . . . . . . . . . . . . . 66Primitive actions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 695 The Sylow Theorems; Applications 733

The Sylow theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73Alternative approach to the Sylow theorems . . . . . . . . . . . . . . . . . . . . 77Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806 Subnormal Series; Solvable and Nilpotent Groups 81Subnormal Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81Solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83Nilpotent groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87Groups with operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90Krull-Schmidt theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 937 Representations of Finite Groups 95Matrix representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95Roots of 1 in fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96Linear representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96Maschke’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97The group algebra; semisimplicity . . . . . . . . . . . . . . . . . . . . . . . . . 99Semisimple modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100Simple F -algebras and their modules . . . . . . . . . . . . . . . . . . . . . . . . 101Semisimple F -algebras and their modules . . . . . . . . . . . . . . . . . . . . . 105The representations of G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107The characters of G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108The character table of a group . . . . . . . . . . . . . . . . . . . . . . . . . . . 111Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111A Additional Exercises 113B Solutions to the Exercises 117C Two-Hour Examination 125Bibliography 127Index 1294

NOTATIONS.We use the standard (Bourbaki) notations: N D f0;1;2;:::g; Z is the ring of integers; Qis the field of rational numbers; R is the field of real numbers; C is the field of complexnumbers; F q is a finite field with q elements where q is a power of a prime number. Inparticular, F p D Z=pZ for p a prime number.For integers m and n, mjn means that m divides n, i.e., n 2 mZ. Throughout the notes,p is a prime number, i.e., p D 2;3;5;7;11;:::;1000000007;:::.Given an equivalence relation, Œ denotes the equivalence class containing . Theempty set is denoted by ;. The cardinality of a set S is denoted by jSj (so jSj is the numberof elements in S when S is finite). Let I and A be sets; a family of elements of A indexedby I , denoted .a i / i2I , is a function i 7! a i WI ! A. 1Rings are required to have an identity element 1, and homomorphisms of rings arerequired to take 1 to 1. An element a of a ring is a unit if it has an inverse (element b suchthat ab D 1 D ba). The identity element of a ring is required to act as 1 on a module overthe ring.X Y X is a subset of Y (not necessarily proper);D Y X is defined to be Y , or equals Y by definition;X Y X is isomorphic to Y ;X ' Y X and Y are canonically isomorphic (or there is a given or unique isomorphism);X defPREREQUISITESAn undergraduate “abstract algebra” course.COMPUTER ALGEBRA PROGRAMSGAP is an open source computer algebra program, emphasizing computational group theory.To get started with GAP, I recommend going to Alexander Hulpke’s page http://www.math.colostate.edu/~hulpke/CGT/education.html where you will find versionsof GAP for both Windows and Macs and a guide “Abstract Algebra in GAP”. TheSage page http://www.sagemath.org/ provides a front end for GAP and other programs.I also recommend N. Carter’s “Group Explorer” http://groupexplorer.sourceforge.net for exploring the structure of groups of small order. Earlier versions of these notes(v3.02) described how to use Maple for computations in group theory.ACKNOWLEDGEMENTSI thank the following for providing corrections and comments for earlier versions of thesenotes: Dustin Clausen, Benoît Claudon, Keith Conrad, Demetres Christofides, Adam Glesser,Sylvan Jacques, Martin Klazar, Mark Meckes, Victor Petrov, Efthymios Sofos, Dave Simpson,Robert Thompson, Michiel Vermeulen.Also, I have benefited from the posts to mathoverflow by Richard Borcherds, RobinChapman, Steve Dalton, Leonid Positselski, Noah Snyder, Richard Stanley, Qiaochu Yuan,and others (a reference mo9990 means http://mathoverflow.net/questions/9990/).1 A family should be distinguished from a set. For example, if f is the function Z ! Z=3Z sending aninteger to its equivalence class, then ff .i/ j i 2 Zg is a set with three elements whereas .f .i// i2Z is familywith an infinite index set.5

The theory of groups of finite order may be said to date from the time of Cauchy. Tohim are due the first attempts at classification with a view to forming a theory from anumber of isolated facts. Galois introduced into the theory the exceedingly importantidea of a [normal] sub-group, and the corresponding division of groups into simpleand composite. Moreover, by shewing that to every equation of finite degree therecorresponds a group of finite order on which all the properties of the equation depend,Galois indicated how far reaching the applications of the theory might be, and therebycontributed greatly, if indirectly, to its subsequent developement.Many additions were made, mainly by French mathematicians, during the middlepart of the [nineteenth] century. The first connected exposition of the theory was givenin the third edition of M. Serret’s “Cours d’Algèbre Supérieure,” which was publishedin 1866. This was followed in 1870 by M. Jordan’s “Traité des substitutions et deséquations algébriques.” The greater part of M. Jordan’s treatise is devoted to a developementof the ideas of Galois and to their application to the theory of equations.No considerable progress in the theory, as apart from its applications, was madetill the appearance in 1872 of Herr Sylow’s memoir “Théorèmes sur les groupes desubstitutions” in the fifth volume of the Mathematische Annalen. Since the date of thismemoir, but more especially in recent years, the theory has advanced continuously.W. Burnside, Theory of Groups of Finite Order, 1897.Galois introduced the concept of a normal subgroup in 1832, and Camille Jordan in thepreface to his Traité. . . in 1870 flagged Galois’ distinction between groupes simplesand groupes composées as the most important dichotomy in the theory of permutationgroups. Moreover, in the Traité, Jordan began building a database of finite simplegroups — the alternating groups of degree at least 5 and most of the classical projectivelinear groups over fields of prime cardinality. Finally, in 1872, Ludwig Sylowpublished his famous theorems on subgroups of prime power order.R. Solomon, Bull. Amer. Math. Soc., 2001.Why are the finite simple groups classifiable?It is unlikely that there is any easy reason why a classification is possible, unless someonecomes up with a completely new way to classify groups. One problem, at leastwith the current methods of classification via centralizers of involutions, is that everysimple group has to be tested to see if it leads to new simple groups containing it inthe centralizer of an involution. For example, when the baby monster was discovered,it had a double cover, which was a potential centralizer of an involution in a largersimple group, which turned out to be the monster. The monster happens to have nodouble cover so the process stopped there, but without checking every finite simplegroup there seems no obvious reason why one cannot have an infinite chain of largerand larger sporadic groups, each of which has a double cover that is a centralizer ofan involution in the next one. Because of this problem (among others), it was unclearuntil quite late in the classification whether there would be a finite or infinite numberof sporadic groups.Richard Borcherds, mo38161.

CHAPTER 1Basic Definitions and ResultsThe axioms for a group are short and natural. . . .Yet somehow hidden behind these axioms is themonster simple group, a huge and extraordinarymathematical object, which appears to rely on numerousbizarre coincidences to exist. The axiomsfor groups give no obvious hint that anything likethis exists.Richard Borcherds, in Mathematicians 2009.Definitions and examplesDEFINITION 1.1 A group is a set G together with a binary operationsatisfying the following conditions:G1: (associativity) for all a;b;c 2 G,.a;b/ 7! a bWG G ! G.a b/ c D a .b c/IG2: (existence of a neutral element) there exists an element e 2 G such thata e D a D e a (1)for all a 2 G;G3: (existence of inverses) for each a 2 G, there exists an a 0 2 G such thata a 0 D e D a 0 a:We usually abbreviate .G;/ to G. Also, we usually write ab for a b and 1 for e; alternatively,we write a C b for a b and 0 for e. In the first case, the group is said to bemultiplicative, and in the second, it is said to be additive.1.2 In the following, a;b;::: are elements of a group G.7

8 1. BASIC DEFINITIONS AND RESULTS(a) An element e satisfying (1) is called a neutral element. If e 0 is a second such element,then e 0 D e e 0 D e. In fact, e is the unique element of G such that e e D e (applyG3).(b) If b a D e and a c D e, thenb D b e D b .a c/ D .b a/ c D e c D c:Hence the element a 0 in (G3) is uniquely determined by a. We call it the inverse ofa, and denote it a 1 (or the negative of a, and denote it a).(c) Note that (G1) shows that the product of any ordered triple a 1 , a 2 , a 3 of elements ofG is unambiguously defined: whether we form a 1 a 2 first and then .a 1 a 2 /a 3 , or a 2 a 3first and then a 1 .a 2 a 3 /, the result is the same. In fact, (G1) implies that the productof any ordered n-tuple a 1 , a 2 ,. . . , a n of elements of G is unambiguously defined. Weprove this by induction on n. In one multiplication, we might end up with.a 1 a i /.a iC1 a n / (2)as the final product, whereas in another we might end up with.a 1 a j /.a j C1 a n /: (3)Note that the expression within each pair of parentheses is well defined because ofthe induction hypotheses. Thus, if i D j , (2) equals (3). If i ¤ j , we may supposei < j . Then.a 1 a i /.a iC1 a n / D .a 1 a i / .a iC1 a j /.a j C1 a n / .a 1 a j /.a j C1 a n / D .a 1 a i /.a iC1 a j / .a j C1 a n /and the expressions on the right are equal because of (G1).(d) The inverse of a 1 a 2 a n is an 1an 11 a 1 1 , i.e., the inverse of a product is theproduct of the inverses in the reverse order.(e) (G3) implies that the cancellation laws hold in groups,ab D ac H) b D c;ba D ca H) b D c(multiply on left or right by a 1 ). Conversely, if G is finite, then the cancellation lawsimply (G3): the map x 7! axWG ! G is injective, and hence (by counting) bijective;in particular, e is in the image, and so a has a right inverse; similarly, it has a leftinverse, and the argument in (b) above shows that the two inverses are equal.Two groups .G;/ and .G 0 ; 0 / are isomorphic if there exists a one-to-one correspondencea $ a 0 , G $ G 0 , such that .a b/ 0 D a 0 0 b 0 for all a;b 2 G.The order jGj of a group G is its cardinality. A finite group whose order is a power ofa prime p is called a p-group.For an element a of a group G, define8 0 .n copies of a/e n D 0a 1 a 1 a 1 n < 0 (jnj copies of a 1 )

Definitions and examples 9The usual rules hold:It follows from (4) that the seta m a n D a mCn ; .a m / n D a mn ; all m;n 2 Z: (4)fn 2 Z j a n D egis an ideal in Z, and so equals mZ for some integer m 0. When m D 0, a n ¤ e unlessn D 0, and a is said to have infinite order. When m ¤ 0, it is the smallest integer m > 0such that a m D e, and a is said to have finite order m. In this case, a 1 D a m 1 , anda n D e ” mjn:EXAMPLES1.3 Let C 1 be the group .Z;C/, and, for an integer m 1, let C m be the group .Z=mZ;C/.1.4 Permutation groups. Let S be a set and let Sym.S/ be the set of bijections ˛WS ! S.We define the product of two elements of Sym.S/ to be their composite:For any ˛;ˇ; 2 Sym.S/ and s 2 S,˛ˇ D ˛ ı ˇ:..˛ ı ˇ/ ı /.s/ D .˛ ı ˇ/..s// D ˛.ˇ..s/// D .˛ ı .ˇ ı //.s/; (5)and so associativity holds. The identity map s 7! s is an identity element for Sym.S/,and inverses exist because we required the elements of Sym.S/ to be bijections. ThereforeSym.S/ is a group, called the group of symmetries of S. For example, the permutationgroup on n letters S n is defined to be the group of symmetries of the set f1;:::;ng — it hasorder nŠ.1.5 When G and H are groups, we can construct a new group G H , called the (direct)product of G and H . As a set, it is the cartesian product of G and H , and multiplication isdefined by.g;h/.g 0 ;h 0 / D .gg 0 ;hh 0 /:1.6 A group G is commutative (or abelian) 1 ifab D ba; all a;b 2 G:In a commutative group, the product of any finite (not necessarily ordered) family S of elementsis well defined, for example, the empty product is e. Usually, we write commutativegroups additively. With this notation, Equation (4) becomes:ma C na D .m C n/a;m.na/ D mna:When G is commutative,m.a C b/ D ma C mb for m 2 Z and a;b 2 G,1 “Abelian group” is more common than “commutative group”, but I prefer to use descriptive names wherepossible.

10 1. BASIC DEFINITIONS AND RESULTSand so the map.m;a/ 7! maWZ G ! Gmakes A into a Z-module. In a commutative group G, the elements of finite order form asubgroup G tors of G, called the torsion subgroup.1.7 Let F be a field. The n n matrices with coefficients in F and nonzero determinantform a group GL n .F / called the general linear group of degree n. For a finite dimensionalF -vector space V , the F -linear automorphisms of V form a group GL.V / called thegeneral linear group of V . Note that if V has dimension n, then the choice of a basis determinesan isomorphism GL.V / ! GL n .F / sending an automorphism to its matrix withrespect to the basis.1.8 Let V be a finite dimensional vector space over a field F . A bilinear form on V is amapping WV V ! F that is linear in each variable. An automorphism of such a is anisomorphism ˛WV ! V such that.˛v;˛w/ D .v;w/ for all v;w 2 V: (6)The automorphisms of form a group Aut./. Let e 1 ;:::;e n be a basis for V , and letP D ..e i ;e j // 1i;j nbe the matrix of . The choice of the basis identifies Aut./ with the group of invertiblematrices A such that 2 A T P A D P . (7)When is symmetric, i.e.,.v;w/ D .w;v/ all v;w 2 V;and nondegenerate, Aut./ is called the orthogonal group of .When is skew-symmetric, i.e.,.v;w/ D .w;v/ all v;w 2 V;and nondegenerate, Aut./ is called the symplectic group of . In this case, there exists abasis for V for which the matrix of is 0 ImJ 2m D; 2m D n;I m 02 When we use the basis to identify V with F n , the pairing becomesa 1! 0 10 1b 1:::b 1:::: ;@A 7! .a:1 ;:::;a n / P @ A:a n b n b nIf A is the matrix of ˛ with respect to the basis, then ˛ corresponds to the map(6) becomes the statement that0 10 1b 1:::b 1:::.a 1 ;:::;a n / A T P A @ A D .a 1 ;:::;a n / P @ A for allb n b na 1!: :a na 1!: :a n0;@7! A1b 1:::b na 1!: :a nA 2 F n :On examining this statement on the standard basis vectors for F n , we see that it is equivalent to (7).:Therefore,

Multiplication tables 11and the group of invertible matrices A such thatis called the symplectic group Sp 2m .A T J 2m A D J 2mREMARK 1.9 A set S together with a binary operation .a;b/ 7! a bWS S ! S is called amagma. When the binary operation is associative, .S;/ is called a semigroup. The productQ AdefD a 1 a nof any sequence A D .a i / 1in of elements in a semigroup S is well-defined (see 1.2(c)),and for any pair A and B of such sequences,. Q A/. Q B/ D Q .A t B/. (8)Let ; be the empty sequence, i.e., the sequence of elements in S indexed by the empty set.What should Q ; be? Clearly, we should have. Q ;/. Q A/ D Q .; t A/ D Q A D Q .A t ;/ D . Q A/. Q ;/:In other words, Q ; should be a neutral element. A semigroup with a neutral elementis called a monoid. In a monoid, the product of any finite (possibly empty) sequence ofelements is well-defined, and (8) holds.ASIDE 1.10 (a) The group conditions (G2,G3) can be replaced by the following weaker conditions(existence of a left neutral element and left inverses): (G2 0 ) there exists an e such that e a D a forall a; (G3 0 ) for each a 2 G, there exists an a 0 2 G such that a 0 a D e. To see that these imply (G2)and (G3), let a 2 G, and apply (G3 0 ) to find a 0 and a 00 such that a 0 a D e and a 00 a 0 D e. Thenwhence (G3), anda a 0 D e .a a 0 / D .a 00 a 0 / .a a 0 / D a 00 .a 0 a/ a 0 D a 00 a 0 D e;a D e a D .a a 0 / a D a .a 0 a/ D a e;whence (G2).(b) A group can be defined to be a set G with a binary operation satisfying the followingconditions: (g1) is associative; (g2) G is nonempty; (g3) for each a 2 G, there exists an a 0 2 Gsuch that a 0 a is neutral. As there is at most one neutral element in a set with an associative binaryoperation, these conditions obviously imply those in (a). They are minimal in the sense that thereexist sets with a binary operation satisfying any two of them but not the third. For example, .N;C/satisfies (g1) and (g2) but not (g3); the empty set satisfies (g1) and (g3) but not (g2); the set of 2 2matrices with coefficents in a field and with A B D AB BA satisfies (g2) and (g3) but not (g1).Multiplication tablesA binary operation on a finite set can be described by its multiplication table:e a b c :::e ee ea eb ec :::a ae a 2 ab ac :::b be ba b 2 bc :::c ce ca cb c 2 ::::::::

12 1. BASIC DEFINITIONS AND RESULTSThe element e is an identity element if and only if the first row and column of the tablesimply repeat the elements. Inverses exist if and only if each element occurs exactly oncein each row and in each column (see 1.2e). If there are n elements, then verifying theassociativity law requires checking n 3 equalities.For the multiplication table of S 3 , see the front page. Note that each colour occursexactly once in each row and and each column.This suggests an algorithm for finding all groups of a given finite order n, namely, listall possible multiplication tables and check the axioms. Except for very small n, this is notpractical! The table has n 2 positions, and if we allow each position to hold any of the nelements, then that gives a total of n n2 possible tables very few of which define groups. Forexample, there are 8 64 D 6277101735386680763835789423207666416102355444464034512896 binary operations on a set with 8 elements, but only five isomorphism classesof groups of order 8 (see 4.21).SubgroupsPROPOSITION 1.11 Let S be a nonempty subset of a group G. IfS1: a;b 2 S H) ab 2 S;andS2: a 2 S H) a 1 2 S;then the binary operation on G makes S into a group.PROOF. (S1) implies that the binary operation on G defines a binary operation S S ! Son S, which is automatically associative. By assumption S contains at least one element a,its inverse a 1 , and the product e D aa 1 . Finally (S2) shows that the inverses of elementsin S lie in S.✷A nonempty subset S satisfying (S1) and (S2) is called a subgroup of G. When S isfinite, condition (S1) implies (S2): let a 2 S; then fa;a 2 ;:::g S, and so a has finite order,say a n D e; now a 1 D a n 1 2 S. The example .N;C/ .Z;C/ shows that (S1) does notimply (S2) when S is infinite.EXAMPLE 1.12 The centre of a group G is the subsetIt is a subgroup of G.Z.G/ D fg 2 G j gx D xg for all x 2 Gg:PROPOSITION 1.13 An intersection of subgroups of G is a subgroup of G:PROOF. It is nonempty because it contains e, and (S1) and (S2) obviously hold.✷REMARK 1.14 It is generally true that an intersection of subobjects of an algebraic objectis a subobject. For example, an intersection of subrings of a ring is a subring, an intersectionof submodules of a module is a submodule, and so on.PROPOSITION 1.15 For any subset X of a group G, there is a smallest subgroup of G containingX. It consists of all finite products of elements of X and their inverses (repetitionsallowed).

Subgroups 13PROOF. The intersection S of all subgroups of G containing X is again a subgroup containingX, and it is evidently the smallest such group. Clearly S contains with X, all finiteproducts of elements of X and their inverses. But the set of such products satisfies (S1) and(S2) and hence is a subgroup containing X. It therefore equals S.✷The subgroup S given by the proposition is denoted hXi, and is called the subgroupgenerated by X. For example, h;i D feg. If every element of X has finite order, forexample, if G is finite, then the set of all finite products of elements of X is already a groupand so equals hXi.We say that X generates G if G D hXi, i.e., if every element of G can be written as afinite product of elements from X and their inverses. Note that the order of an element a ofa group is the order of the subgroup hai it generates.EXAMPLES1.16 The cyclic groups. A group is said to be cyclic if it is generated by a single element,i.e., if G D hri for some r 2 G. If r has finite order n, thenG D fe;r;r 2 ;:::;r n 1 g C n ; r i $ i mod n;and G can be thought of as the group of rotational symmetries about the centre of a regularpolygon with n-sides. If r has infinite order, thenG D f:::;r i ;:::;r 1 ;e;r;:::;r i ;:::g C 1 ; r i $ i:Thus, up to isomorphism, there is exactly one cyclic group of order n for each n 1. Infuture, we shall loosely use C n to denote any cyclic group of order n (not necessarily Z=nZor Z).1.17 The dihedral groups D n . 3 For n 3, D n is the group of symmetries of a regularpolygon with n-sides. 4 Number the vertices 1;:::;n in the counterclockwise direction. Letr be the rotation through 2=n about the centre of polygon (so i 7! i C 1 mod n/, and lets be the reflection in the line (= rotation about the line) through the vertex 1 and the centreof the polygon (so i 7! n C 2 i mod n). For example, the picturess1r2 3 1 $ 1s D2 $ 3r D 1 ! 2 ! 3 ! 12r1s48

14 1. BASIC DEFINITIONS AND RESULTSillustrate the groups D 3 and D 4 . In the general caseThese equalites imply thatr n D eI s 2 D eI srs D r 1 (so sr D r n 1 s/:D n D fe;r;:::;r n 1 ;s;rs;:::;r n 1 sg;and it is clear from the geometry that the elements of the set are distinct, and so jD n j D 2n.Let t be the reflection in the line through the midpoint of the side joining the vertices1 and 2 and the centre of the polygon (so i 7! n C 3 i mod n/. Then r D ts. HenceD n D hs;ti ands 2 D e; t 2 D e; .ts/ n D e D .st/ n :We define D 1 to be C 2 D f1;rg and D 2 to be C 2 C 2 D f1;r;s;rsg. The group D 2is also called the Klein Viergruppe or, more simply, the 4-group. Note that D 3 is the fullgroup of permutations of f1;2;3g. It is the smallest noncommutative group. p 1.18 The quaternion group Q: Let a D p 0 1and b D 0 11 0 1 0. Thena 4 D e; a 2 D b 2 ; bab 1 D a 3 (so ba D a 3 b).The subgroup of GL 2 .C/ generated by a and b isQ D fe;a;a 2 ;a 3 ;b;ab;a 2 b;a 3 bg:The group Q can also be described as the subset f˙1;˙i;˙j;˙kg of the quaternion algebraH. Recall thatH D R1 ˚ Ri ˚ Rj ˚ Rkwith the multiplication determined byi 2 D 1 D j 2 ; ij D k D j i:The map i 7! a, j 7! b extends uniquely to a homomorphism H ! M 2 .C/ of R-algebras,which maps the group hi;j i isomorphically onto ha;bi.1.19 Recall that S n is the permutation group on f1;2;:::;ng. A transposition is a permutationthat interchanges two elements and leaves all other elements unchanged. It is notdifficult to see that S n is generated by transpositions (see (4.26) below for a more precisestatement).Groups of small order[For] n D 6, there are three groups, a group C 6 ,and two groups C 2 C 3 and S 3 .Cayley, American J. Math. 1 (1878), p. 51.For each prime p, there is only one group of order p, namely C p (see 1.28 below). Inthe following table, c C n D t means that there are c commutative groups and n noncommutativegroups (up to isomorphism, of course).

Homomorphisms 15jGj c C n D t Groups Ref.4 2 C 0 D 2 C 4 , C 2 C 2 4.186 1 C 1 D 2 C 6 ; S 3 4.238 3 C 2 D 5 C 8 , C 2 C 4 , C 2 C 2 C 2 ; Q, D 4 4.219 2 C 0 D 2 C 9 , C 3 C 3 4.1810 1 C 1 D 2 C 10 ; D 5 5.1412 2 C 3 D 5 C 12 , C 2 C 6 ; C 2 S 3 , A 4 , C 4 ⋊ C 3 5.1614 1 C 1 D 2 C 14 ; D 7 5.1415 1 C 0 D 1 C 15 5.1416 5 C 9 D 14 See Wild 200518 2 C 3 D 5 C 18 , C 3 C 6 ; D 9 ; S 3 C 3 , .C 3 C 3 / ⋊ C 220 2 C 3 D 5 C 20 ,C 2 C 10 ;D 10 ,C 5 ⋊ C 4 ,ha;b ja 4 D b 5 D 1; ba D ab 2 i21 1 C 1 D 2 C 21 ; ha;b j a 3 D b 7 D 1, ba D ab 2 i22 1 C 1 D 2 C 22 ; D 11 5.1424 3 C 12D15 See opensourcemath.org/gap/small groups.htmlHere ha;b j a 4 D b 5 D 1; ba D ab 2 i is the group with generators a and b and relationsa 4 D b 5 D 1 and ba D ab 2 (see Chapter 2).Roughly speaking, the more high powers of primes divide n, the more groups of ordern there should be. In fact, if f .n/ is the number of isomorphism classes of groups of ordern, thenf .n/ n . 227 Co.1//e.n/2where e.n/ is the largest exponent of a prime dividing n and o.1/ ! 0 as e.n/ ! 1 (seePyber 1993).By 2001, a complete irredundant list of groups of order 2000 had been found — upto isomorphism, there are exactly 49,910,529,484 (Besche et al. 2001).HomomorphismsDEFINITION 1.20 A homomorphism from a group G to a second G 0 is a map ˛WG ! G 0such that ˛.ab/ D ˛.a/˛.b/ for all a;b 2 G. An isomorphism is a bijective homomorphism.For example, the determinant map detWGL n .F / ! F is a homomorphism.1.21 Let ˛ be a homomorphism. For any elements a 1 ;:::;a m of G,˛.a 1 a m / D ˛.a 1 .a 2 a m //D ˛.a 1 /˛.a 2 a m /D ˛.a 1 /˛.a m /,

16 1. BASIC DEFINITIONS AND RESULTSand so homomorphisms preserve all products. In particular, for m 1,˛.a m / D ˛.a/ m : (9)Moreover ˛.e/ D ˛.ee/ D ˛.e/˛.e/, and so ˛.e/ D e (apply 1.2a). Alsoaa 1 D e D a 1 a H) ˛.a/˛.a 1 / D e D ˛.a 1 /˛.a/;and so ˛.a 1 / D ˛.a/ 1 . It follows that (9) holds for all m 2 Z, and so a homomorphismof commutative groups is also a homomorphism of Z-modules.As we noted above, each row of the multiplication table of a group is a permutation ofthe elements of the group. As Cayley pointed out, this allows one to realize the group as agroup of permutations.THEOREM 1.22 (CAYLEY) There is a canonical injective homomorphism˛WG ! Sym.G/:PROOF. For a 2 G, define a L WG ! G to be the map x 7! ax (left multiplication by a).For x 2 G,.a L ı b L /.x/ D a L .b L .x// D a L .bx/ D abx D .ab/ L .x/;and so .ab/ L D a L ı b L . As e L D id, this implies thata L ı .a 1 / L D id D .a 1 / L ı a L ;and so a L is a bijection, i.e., a L 2 Sym.G/. Hence a 7! a L is a homomorphism G !Sym.G/, and it is injective because of the cancellation law.✷COROLLARY 1.23 A finite group of order n can be realized as a subgroup of S n .PROOF. List the elements of the group as a 1 ;:::;a n .✷Unfortunately, unless n is small, S n is too large to be manageable. We shall see later(4.22) that G can often be embedded in a permutation group of much smaller order than nŠ.CosetsFor a subset S of a group G and an element a of G, we letaS D fas j s 2 SgSa D fsa j s 2 Sg:Because of the associativity law, a.bS/ D .ab/S , and so we can denote this set unambiguouslyby abS:When H is a subgroup of G, the sets of the form aH are called the left cosets of Hin G, and the sets of the form Ha are called the right cosets of H in G. Because e 2 H ,aH D H if and only if a 2 H .

Cosets 17EXAMPLE 1.24 Let G D .R 2 ;C/, and let H be a subspace of dimension 1 (line throughthe origin). Then the cosets (left or right) of H are the lines a C H parallel to H .PROPOSITION 1.25 Let H be a subgroup of a group G.(a) An element a of G lies in a left coset C of H if and only if C D aH:(b) Two left cosets are either disjoint or equal.(c) aH D bH if and only if a 1 b 2 H:(d) Any two left cosets have the same number of elements (possibly infinite).PROOF. (a) Certainly a 2 aH . Conversely, if a lies in the left coset bH , then a D bh forsome h, and soaH D bhH D bH:(b) If C and C 0 are not disjoint, then they have a common element a, and C D aH andC 0 D aH by (a).(c) If a 1 b 2 H , then H D a 1 bH , and so aH D aa 1 bH D bH . Conversely, ifaH D bH , then H D a 1 bH , and so a 1 b 2 H .(d) The map .ba 1 / L Wah 7! bh is a bijection aH ! bH:✷The index .G W H / of H in G is defined to be the number of left cosets of H in G. 5 Forexample, .G W 1/ is the order of G.As the left cosets of H in G cover G, (1.25b) shows that they form a partition G. Inother words, the condition “a and b lie in the same left coset” is an equivalence relation onG.THEOREM 1.26 (LAGRANGE) If G is finite, then.G W 1/ D .G W H /.H W 1/:In particular, the order of every subgroup of a finite group divides the order of the group.PROOF. The left cosets of H in G form a partition of G, there are .G W H / of them, andeach left coset has .H W 1/ elements.✷COROLLARY 1.27 The order of each element of a finite group divides the order of thegroup.PROOF. Apply Lagrange’s theorem to H D hgi, recalling that .H W 1/ D order.g/.✷EXAMPLE 1.28 If G has order p, a prime, then every element of G has order 1 or p. Butonly e has order 1, and so G is generated by any element a ¤ e. In particular, G is cyclicand so G C p . This shows, for example, that, up to isomorphism, there is only one groupof order 1;000;000;007 (because this number is prime). In fact there are only two groupsof order 1;000;000;014;000;000;049 (see 4.18).5 More formally, .G W H / is the cardinality of the set faH j a 2 Gg.

18 1. BASIC DEFINITIONS AND RESULTS1.29 For a subset S of G, let S 1 D fg 1 j g 2 Sg. Then .aH / 1 is the right coset Ha 1 ,and .Ha/ 1 D a 1 H . Therefore S 7! S 1 defines a one-to-one correspondence betweenthe set of left cosets and the set of right cosets under which aH $ Ha 1 . Hence .G W H /is also the number of right cosets of H in G: But, in general, a left coset will not be a rightcoset (see 1.34 below).1.30 Lagrange’s theorem has a partial converse: if a prime p divides m D .G W 1/, then Ghas an element of order p (Cauchy’s theorem 4.13); if a prime power p n divides m, then Ghas a subgroup of order p n (Sylow’s theorem 5.2). However, note that the 4-group C 2 C 2has order 4, but has no element of order 4, and A 4 has order 12, but has no subgroup oforder 6 (see Exercise 4-14).More generally, we have the following result.PROPOSITION 1.31 For any subgroups H K of G,.G W K/ D .G W H /.H W K/(meaning either both are infinite or both are finite and equal).PROOF. Write G D F i2I g iH (disjoint union), and H D F j 2J h j K (disjoint union). Onmultiplying the second equality by g i , we find that g i H D F j 2J g ih j K (disjoint union),and so G D F i;j 2I J g ih j K (disjoint union). This shows that.G W K/ D jI jjJ j D .G W H /.H W K/:✷Normal subgroupsWhen S and T are two subsets of a group G, we letST D fst j s 2 S, t 2 T g:Because of the associativity law, R.ST / D .RS/T , and so we can denote this set unambiguouslyas RST .A subgroup N of G is normal, denoted N G G, if gNg 1 D N for all g 2 G.REMARK 1.32 To show that N is normal, it suffices to check that gNg 1 N for all g,because multiplying this inclusion on the left and right with g 1 and g respectively givesthe inclusion N g 1 Ng, and rewriting this with g 1 for g gives that N gNg 1 for allg. However, the next example shows that there can exist a subgroup N of a group G andan element g of G such that gNg 1 N but gNg 1 ¤ N .EXAMPLE 1.33 Let G D GL 2 .Q/, and let H D ˚ 1 n0 1ˇˇn 2 Z . Then H is a subgroup ofG; in fact H ' Z. Let g D 5 0 1 0 . Then 1 ng g 1 5 0 1 n 510 1 5nDD :0 1 0 1 0 1 0 1 0 1Hence gHg 1 H (and g 1 Hg 6 H ).

Normal subgroups 19PROPOSITION 1.34 A subgroup N of G is normal if and only if every left coset of N inG is also a right coset, in which case, gN D Ng for all g 2 G:PROOF. Clearly,gNg 1 D N ” gN D Ng:Thus, if N is normal, then every left coset is a right coset (in fact, gN D Ng). Conversely,if the left coset gN is also a right coset, then it must be the right coset Ng by (1.25a). HencegN D Ng, and so gNg 1 D N .✷1.35 The proposition says that, in order for N to be normal, we must have that for allg 2 G and n 2 N , there exists an n 0 2 N such that gn D n 0 g (equivalently, for all g 2 Gand n 2 N , there exists an n 0 such that ng D gn 0 ). In other words, to say that N is normalamounts to saying that an element of G can be moved past an element of N at the cost ofreplacing the element of N by another element of N .EXAMPLE 1.36 (a) Every subgroup of index two is normal. Indeed, let g 2 G H . ThenG D H t gH (disjoint union). Hence gH is the complement of H in G. Similarly, Hg isthe complement of H in G, and so gH D Hg:(b) Consider the dihedral groupD n D fe;r;:::;r n 1 ;s;:::;r n 1 sg:Then C n D fe;r;:::;r n 1 g has index 2, and hence is normal. For n 3 the subgroup fe;sgis not normal because r 1 sr D r n 2 s … fe;sg.(c) Every subgroup of a commutative group is normal (obviously), but the converseis false: the quaternion group Q is not commutative, but every subgroup is normal (seeExercise 1-1).A group G is said to be simple if it has no normal subgroups other than G and feg. Sucha group can still have lots of nonnormal subgroups — in fact, the Sylow theorems (Chapter5) imply that every finite group has nontrivial subgroups unless it is cyclic of prime order.PROPOSITION 1.37 If H and N are subgroups of G and N is normal, then HN is asubgroup of G. If H is also normal, then HN is a normal subgroup of G.PROOF. The set HN is nonempty, and.h 1 n 1 /.h 2 n 2 / 1.35 D h 1 h 2 n 0 1 n 2 2 HN;and so it is closed under multiplication. Since.hn/ 1 D n 1 h 1 1.35D h 1 n 0 2 HNit is also closed under the formation of inverses, and so HN is a subgroup. If both H andN are normal, thengHNg 1 D gHg 1 gNg 1 D HNfor all g 2 G.✷

20 1. BASIC DEFINITIONS AND RESULTSAn intersection of normal subgroups of a group is again a normal subgroup (cf. 1.14).Therefore, we can define the normal subgroup generated by a subset X of a group G tobe the intersection of the normal subgroups containing X. Its description in terms of Xis a little complicated. We say that a subset X of a group G is normal or closed underconjugation if gXg 1 X for all g 2 G.LEMMA 1.38 If X is normal, then the subgroup hXi generated by it is normal.PROOF. The map “conjugation by g”, a 7! gag 1 , is a homomorphism G ! G. If a 2 hXi,say, a D x 1 x m with each x i or its inverse in X, thengag 1 D .gx 1 g 1 /.gx m g 1 /.As X is closed under conjugation, each gx i g 1 or its inverse lies in X, and so ghXig 1 hXi.✷LEMMA 1.39 For any subset X of G, the subset S g2G gXg 1 is normal, and it is thesmallest normal set containing X.PROOF. Obvious.✷On combining these lemmas, we obtain the following proposition.PROPOSITION 1.40 The normal subgroup generated by a subset X of G is h S g2G gXg 1 i.Kernels and quotientsThe kernel of a homomorphism ˛WG ! G 0 isKer.˛/ D fg 2 Gj ˛.g/ D eg:If ˛ is injective, then Ker.˛/ D feg. Conversely, if Ker.˛/ D feg, then ˛ is injective, because˛.g/ D ˛.g 0 / H) ˛.g 1 g 0 / D e H) g 1 g 0 D e H) g D g 0 .PROPOSITION 1.41 The kernel of a homomorphism is a normal subgroup.PROOF. It is obviously a subgroup, and if a 2 Ker.˛/, so that ˛.a/ D e, and g 2 G, then˛.gag 1 / D ˛.g/˛.a/˛.g/ 1 D ˛.g/˛.g/ 1 D e:Hence gag 1 2 Ker.˛/.✷For example, the kernel of the homomorphism detWGL n .F / ! F is the group of nnmatrices with determinant 1 — this group SL n .F / is called the special linear group ofdegree n.PROPOSITION 1.42 Every normal subgroup occurs as the kernel of a homomorphism.More precisely, if N is a normal subgroup of G, then there is a unique group structureon the set G=N of cosets of N in G for which the natural map a 7! ŒaWG ! G=N is ahomomorphism.

Theorems concerning homomorphisms 21PROOF. Write the cosets as left cosets, and define .aN /.bN / D .ab/N . We have to check(a) that this is well-defined, and (b) that it gives a group structure on the set of cosets. Itwill then be obvious that the map g 7! gN is a homomorphism with kernel N .(a). Let aN D a 0 N and bN D b 0 N ; we have to show that abN D a 0 b 0 N . ButabN D a.bN / D a.b 0 N / 1.34 D aN b 0 D a 0 N b 0 1.34D a 0 b 0 N:(b). The product is certainly associative, the coset N is an identity element, and a 1 Nis an inverse for aN .✷The group G=N is called the 6 quotient of G by N .Propositions 1.41 and 1.42 show that the normal subgroups are exactly the kernels ofhomomorphisms.PROPOSITION 1.43 The map a 7! aN WG ! G=N hasthe following universal property: for any homomorphism˛WG ! G 0 of groups such that ˛.N / D feg, there existsa unique homomorphism G=N ! G 0 making the diagramat right commute:Ga 7! aN˛G=NG 0 :PROOF. Note that for n 2 N , ˛.gn/ D ˛.g/˛.n/ D ˛.g/, and so ˛ is constant on each leftcoset gN of N in G. It therefore defines a mapN˛WG=N ! G 0 ;N˛.gN / D ˛.g/;and N˛ is a homomorphism becauseN˛..gN / .g 0 N // D N˛.gg 0 N / D ˛.gg 0 / D ˛.g/˛.g 0 / D N˛.gN / N˛.g 0 N /.The uniqueness of N˛ follows from the surjectivity of G ! G=N .✷EXAMPLE 1.44 (a) Consider the subgroup mZ of Z. The quotient group Z=mZ is a cyclicgroup of order m.(b) Let L be a line through the origin in R 2 . Then R 2 =L is isomorphic to R (because itis a one-dimensional vector space over R).(c) For n 2, the quotient D n =hri D fNe; Nsg (cyclic group of order 2).Theorems concerning homomorphismsThe theorems in this subsection are sometimes called the isomorphism theorems (first, second,. . . , or first, third, . . . , or . . . ).FACTORIZATION OF HOMOMORPHISMSRecall that the image of a map ˛WS ! T is ˛.S/ D f˛.s/ j s 2 Sg.6 Some authors say “factor” instead of “quotient”, but this can be confused with “direct factor”.

22 1. BASIC DEFINITIONS AND RESULTSTHEOREM 1.45 (HOMOMORPHISM THEOREM) For any homomorphism ˛WG ! G 0 ofgroups, the kernel N of ˛ is a normal subgroup of G, the image I of ˛ is a subgroupof G 0 , and ˛ factors in a natural way into the composite of a surjection, an isomorphism,and an injection:G˛! G0x??surjective?injectiveyg7!gNG=N gN 7!˛.g/ !isomorphism I:PROOF. We have already seen (1.41) that the kernel is a normal subgroup of G. If b D ˛.a/and b 0 D ˛.a 0 /, then bb 0 D ˛.aa 0 / and b 1 D ˛.a 1 /, and so I D def˛.G/ is a subgroup of G 0 .The universal property of quotients (1.43) shows that the map x 7! ˛.x/WG ! I definesa homomorphism N˛WG=N ! I with N˛.gN / D ˛.g/. The homomorphism N˛ is certainlysurjective, and if N˛.gN / D e, then g 2 Ker.˛/ D N , and so N˛ has trivial kernel. Thisimplies that it is injective (p. 20).✷THE ISOMORPHISM THEOREMTHEOREM 1.46 (ISOMORPHISM THEOREM) Let H be a subgroup of G and N a normalsubgroup of G. Then HN is a subgroup of G, H \ N is a normal subgroup of H , and themaph.H \ N / 7! hN WH=H \ N ! HN=Nis an isomorphism.PROOF. We have already seen (1.37) that HN is a subgroup. Consider the mapH ! G=N;h 7! hN:This is a homomorphism, and its kernel is H \ N , which is therefore normal in H . Accordingto Theorem 1.45, the map induces an isomorphism H=H \ N ! I where I is itsimage. But I is the set of cosets of the form hN with h 2 H , i.e., I D HN=N .✷It is not necessary to assume that N be normal in G as long as hN h 1 D N for allh 2 H (i.e., H is contained in the normalizer of N — see later). Then H \ N is stillnormal in H , but it need not be a normal subgroup of G.THE CORRESPONDENCE THEOREMThe next theorem shows that if NG is a quotient group of G, then the lattice of subgroups inNG captures the structure of the lattice of subgroups of G lying over the kernel of G ! NG.THEOREM 1.47 (CORRESPONDENCE THEOREM) Let ˛WG ↠ NG be a surjective homomorphism,and let N D Ker.˛/. Then there is a one-to-one correspondencefsubgroups of G containing N g 1W1$ fsubgroups of NGgunder which a subgroup H of G containing N corresponds to HND ˛.H / and a subgroupHNof NG corresponds to H D ˛ 1. H N /. Moreover, if H $ HNand H 0 $ HN0 , then

Direct products 23(a)(b)HN HN0 ” H H 0 , in which case . HN0 W H N / D .H 0 W H /;HNis normal in NG if and only if H is normal in G, in which case, ˛ induces anisomorphismG=H ! ' NG= H N :PROOF. If HNis a subgroup of NG, then ˛ 1. H N / is easily seen to be a subgroup of G containingN , and if H is a subgroup of G, then ˛.H / is a subgroup of NG (see 1.45). Clearly,˛ 1˛.H / D HN , which equals H if and only if H N , and ˛˛ 1. H N / D H N . Therefore,the two operations give the required bijection. The remaining statements are easily verified.For example, a decomposition H 0 D F i2I a iH of H 0 into a disjoint union of left cosets ofH gives a similar decomposition HN0 D F i2I a NiH of HN0 .✷COROLLARY 1.48 Let N be a normal subgroup of G; then there is a one-to-one correspondencebetween the set of subgroups of G containing N and the set of subgroups ofG=N , H $ H=N . Moreover H is normal in G if and only if H=N is normal in G=N , inwhich case the homomorphism g 7! gN WG ! G=N induces an isomorphismG=H ' ! .G=N /=.H=N /.PROOF. This is the special case of the theorem in which ˛ is g 7! gN WG ! G=N .✷EXAMPLE 1.49 Let G D D 4 and let N be its subgroup hr 2 i. Recall (1.17) that srs 1 D r 3 ,and so sr 2 s 1 D r 3 2 D r 2 . Therefore N is normal. The groups G and G=N have thefollowing lattices of subgroups:D 4D 4 =hr 2 ihr 2 ;si hri hr 2 ;rsi hNsi hNri hNr Nsihsi hr 2 si hr 2 i hrsi hr 3 si 11Direct productsLet G be a group, and let H 1 ;:::;H k be subgroups of G. We say that G is a direct productof the subgroups H i if the map.h 1 ;h 2 ;:::;h k / 7! h 1 h 2 h k W H 1 H 2 H k ! Gis an isomorphism of groups. This means that each element g of G can be written uniquelyin the form g D h 1 h 2 h k , h i 2 H i , and that if g D h 1 h 2 h k and g 0 D h 0 1 h0 2 h0 k , thengg 0 D .h 1 h 0 1 /.h 2h 0 2 /.h kh 0 k /:

24 1. BASIC DEFINITIONS AND RESULTSThe following propositions give criteria for a group to be a direct product of subgroups.PROPOSITION 1.50A group G is a direct product of subgroups H 1 , H 2 if and only if(a) G D H 1 H 2 ,(b) H 1 \ H 2 D feg, and(c) every element of H 1 commutes with every element of H 2 .PROOF. If G is the direct product of H 1 and H 2 , then certainly (a) and (c) hold, and (b)holds because, for any g 2 H 1 \H 2 , the element .g;g 1 / maps to e under .h 1 ;h 2 / 7! h 1 h 2and so equals .e;e/.Conversely, (c) implies that .h 1 ;h 2 / 7! h 1 h 2 is a homomorphism, and (b) implies thatit is injective:h 1 h 2 D e H) h 1 D h 12 2 H 1 \ H 2 D feg:Finally, (a) implies that it is surjective.PROPOSITION 1.51 A group G is a direct product of subgroups H 1 , H 2 if and only if(a) G D H 1 H 2 ,(b) H 1 \ H 2 D feg, and(c) H 1 and H 2 are both normal in G.PROOF. Certainly, these conditions are implied by those in the previous proposition, and soit remains to show that they imply that each element h 1 of H 1 commutes with each elementh 2 of H 2 . Two elements h 1 ;h 2 of a group commute if and only if their commutatoris e. ButŒh 1 ;h 2 defD .h 1 h 2 /.h 2 h 1 / 1.h 1 h 2 /.h 2 h 1 / 1 D h 1 h 2 h1 1 h 2 1 D .h1 h 2 h1 1/ h 21h 1 h 2 h 1 ;1 h 21which is in H 2 because H 2 is normal, and is in H 1 because H 1 is normal. Therefore (b)implies h 1 and h 2 commute.✷PROPOSITION 1.52 A group G is a direct product of subgroups H 1 ;H 2 ;:::;H k if andonly if(a) G D H 1 H 2 H k ;(b) for each j , H j \ .H 1 H j 1 H j C1 H k / D feg, and(c) each of H 1 ;H 2 ;:::;H k is normal in G,PROOF. The necessity of the conditions being obvious, we shall prove only the sufficiency.For k D 2, we have just done this, and so we argue by induction on k. An inductionargument using (1.37) shows that H 1 H k 1 is a normal subgroup of G. The conditions(a,b,c) hold for the subgroups H 1 ;:::;H k 1 of H 1 H k 1 , and so the induction hypothesisshows that.h 1 ;h 2 ;:::;h k 1 / 7! h 1 h 2 h k 1 WH 1 H 2 H k 1 ! H 1 H 2 H k 1✷

Commutative groups 25is an isomorphism. The pair H 1 H k1 , H k satisfies the hypotheses of (1.51), and so.h;h k / 7! hh k W.H 1 H k1 / H k ! Gis also an isomorphism. The composite of these isomorphismsH 1 H k 1 H k.h 1 ;:::;h k /7!.h 1 h k 1 ;h k /! H 1 H k 1 H k.h;h k /7!hh k! Gsends .h 1 ;h 2 ;:::;h k / to h 1 h 2 h k :✷Commutative groupsThe classification of finitely generated commutative groups is most naturally studied as partof the theory of modules over a principal ideal domain, but, for the sake of completeness, Iinclude an elementary exposition here.Let M be a commutative group, written additively. The subgroup hx 1 ;:::;x k i of Mgenerated by the elements x 1 ;:::;x k consists of the sums P m i x i , m i 2 Z. A subsetfx 1 ;:::;x k g of M is a basis for M if it generates M andm 1 x 1 C C m k x k D 0;m i 2 Z H) m i x i D 0 for every iIthenM D hx 1 i ˚ ˚ hx k i:LEMMA 1.53 Suppose that M is generated by fx 1 ;:::;x k g and let c 1 ;:::;c k be integerssuch that gcd.c 1 ;:::;c k / D 1. Then there exist generators y 1 ;:::;y k for M such that y 1 Dc 1 x 1 C C c k x k .PROOF. If c i < 0, we change the signs of both c i and x i . This allows us to assume that allc i 2 N. We argue by induction on s D c 1 C C c k . The lemma certainly holds if s D 1,and so we assume s > 1. Then, at least two c i are nonzero, say, c 1 c 2 > 0. Now˘ fx 1 ;x 2 C x 1 ;x 3 ;:::;x k g generates M ,˘ gcd.c 1 c 2 ;c 2 ;c 3 ;:::;c k / D 1, and˘ .c 1 c 2 / C c 2 C C c k < s,and so, by induction, there exist generators y 1 ;:::;y k for M such thaty 1 D .c 1 c 2 /x 1 C c 2 .x 1 C x 2 / C c 3 x 3 C C c k x kD c 1 x 1 C C c k x k .✷THEOREM 1.54 Every finitely generated commutative group M has a basis; hence it is afinite direct sum of cyclic groups.PROOF. 7 We argue by induction on the number of generators of M . If M can be generatedby one element, the statement is trivial, and so we may assume that it requires at least k > 17 John Stillwell tells me that, for finite commutative groups, this is similar to the first proof of the theorem,given by Kronecker in 1870.

26 1. BASIC DEFINITIONS AND RESULTSgenerators. Among the generating sets fx 1 ;:::;x k g for M with k elements there is one forwhich the order of x 1 is the smallest possible. We shall show that M is then the direct sumof hx 1 i and hx 2 ;:::;x k i. This will complete the proof, because the induction hypothesisprovides us with a basis for the second group, which together with x 1 forms a basis for M .If M is not the direct sum of hx 1 i and hx 2 ;:::;x k i, then there exists a relationm 1 x 1 C m 2 x 2 C C m k x k D 0 (10)with m 1 x 1 ¤ 0; we may suppose that m 1 2 N and m 1 < order.x 1 /. Let d D gcd.m 1 ;:::;m k / >0, and let c i D m i =d. According to the lemma, there exists a generating set y 1 ;:::;y k suchthat y 1 D c 1 x 1 C C c k x k . Butdy 1 D m 1 x 1 C m 2 x 2 C C m k x k D 0and d m 1 < order.x 1 /, and so this is a contradiction.✷COROLLARY 1.55 A finite commutative group is cyclic if, for each n > 0, it contains atmost n elements of order dividing n.PROOF. After the Theorem 1.54, we may suppose that G D C n1 C nr with n i 2 N.If n divides n i and n j with i ¤ j , then G has more than n elements of order dividing n.Therefore, the hypothesis implies that the n i are relatively prime. Let a i generate the ithfactor. Then .a 1 ;:::;a r / has order n 1 n r , and so generates G.✷EXAMPLE 1.56 Let F be a field. The elements of order dividing n in F are the roots ofthe polynomial X n 1. Because unique factorization holds in F ŒX, there are at most n ofthese, and so the corollary shows that every finite subgroup of F is cyclic.THEOREM 1.57 A nonzero finitely generated commutative group M can be expressedfor certain integers n 1 ;:::;n s 2 and r 0. Moreover,M C n1 C ns C r 1 (11)(a) r is uniquely determined by M ;(b) the n i can be chosen so that n 1 2 and n 1 jn 2 ;:::;n s 1 jn s , and then they are uniquelydetermined by M ;(c) the n i can be chosen to be powers of prime numbers, and then they are uniquelydetermined by M .The number r is called the rank of M . By r being uniquely determined by M , we meanthat in any two decompositions of M of the form (11), the number of copies of C 1 will bethe same (and similarly for the n i in (b) and (c)). The integers n 1 ;:::;n s in (b) are calledthe invariant factors of M . Statement (c) says that M can be expressedM C pe 11 C pett C r 1 , e i 1, (12)for certain prime powers p e ii(repetitions of primes allowed), and that the integers p e 11 ;:::;pe ttare uniquely determined by M ; they are called the elementary divisors of M .

Commutative groups 27PROOF. The first assertion is a restatement of Theorem 1.54.(a) For a prime p not dividing any of the d i ,M=pM .C 1 =pC 1 / r .Z=pZ/ r ;and so r is the dimension of M=pM as an F p -vector space.(b,c) If gcd.m;n/ D 1, then C m C n contains an element of order mn, and soC m C n C mn : (13)Use (13) to decompose the C ni into products of cyclic groups of prime power order. Oncethis has been achieved, (13) can be used to combine factors to achieve a decomposition asin (b); for example, C ns D Q C e p i where the product is over the distinct primes among theip i and e i is the highest exponent for the prime p i .In proving the uniqueness statements in (b) and (c), we can replace M with its torsionsubgroup (and so assume r D 0). A prime p will occur as one of the primes p i in (12) ifand only M has an element of order p, in which case p will occur exact a times where p ais the number of elements of order dividing p. Similarly, p 2 will divide some p e iiin (12)if and only if M has an element of order p 2 , in which case it will divide exactly b of thep e iiwhere p a b p 2b is the number of elements in M of order dividing p 2 . Continuing inthis fashion, we find that the elementary divisors of M can be read off from knowing thenumbers of elements of M of each prime power order.The uniqueness of the invariant factors can be derived from that of the elementary divisors,or it can be proved directly: n s is the smallest integer > 0 such that n s M D 0; n s 1is the smallest integer > 0 such that n s 1 M is cyclic; n s 2 is the smallest integer such thatn s 2 can be expressed as a product of two cyclic groups, and so on. ✷SUMMARY 1.58 Each finite commutative group is isomorphic to exactly one of the groupsC n1 C nr ; n 1 jn 2 ;:::;n r 1 jn r :The order of this group is n 1 n r . For example, each commutative group of order 90 isisomorphic to exactly one of C 90 or C 3 C 30 — to see this, note that the largest invariantfactor must be a factor of 90 divisible by all the prime factors of 90.THE LINEAR CHARACTERS OF A COMMUTATIVE GROUPLet .C/ D fz 2 C j jzj D 1g. This is an infinite group. For any integer n, the set n .C/ ofelements of order dividing n is cyclic of order n; in fact, n .C/ D fe 2im=n j 0 m n 1g D f1;;:::; n 1 gwhere D e 2i=n is a primitive nth root of 1.A linear character (or just character) of a group G is a homomorphism G ! .C/.The homomorphism a 7! 1 is called the trivial (or principal) character.EXAMPLE 1.59 The quadratic residue modulo p of an integer a not divisible by p isdefined by a 1 if a is a square in Z=pZDp 1 otherwise.

28 1. BASIC DEFINITIONS AND RESULTSClearly, this depends only on a modulo p, and if neither a nor b is divisible by p, thenabpD a bp p(because .Z=pZ/ is cyclic). Therefore Œa 7! apW.Z=pZ/ ! f˙1g D 2 .C/ is a character of .Z=pZ/ .The set of characters of a group G becomes a group G _ under the addition,. C 0 /.g/ D .g/.g 0 /;called the dual group of G. For example, the dual group Z _ of Z is isomorphic to .C/ bythe map 7! .1/.THEOREM 1.60 Let G be a finite commutative group.(a) The dual of G _ is isomorphic to G.(b) The map G ! G __ sending an element a of G to the character 7! .a/ of G _ isan isomorphism.In other words, G G _ and G ' G __ .PROOF. The statements are obvious for cyclic groups, and .G H / _ ' G _ H _ .ASIDE 1.61 The statement that the natural map G ! G __ is an isomorphism is a special caseof the Pontryagin theorem. For infinite groups, it is necessary to consider groups together with atopology. For example, as we observed above, Z _ ' .C/. Each m 2 Z does define a character 7! m W.C/ ! .C/, but there are many homomorphisms .C/ ! .C/ not of this form, andso the dual of .C/ is larger than Z. However, these are the only continuous homomorphisms. Ingeneral, let G be a commutative group endowed with a locally compact topology for which thegroup operations are continuous; then the group G _ of continuous characters G ! .C/ has anatural topology for which it is locally compact, and the Pontryagin duality theorem says that thenatural map G ! G __ is an isomorphism.THEOREM 1.62 (ORTHOGONALITY RELATIONS) Let G be a finite commutative group.For any characters and of G,X .a/ .a 1 jGj if D/ Da2G 0 otherwise.In particular,Xa2G .a/ D jGj if is trivial0 otherwise.PROOF. If D , then .a/ .a 1 / D 1, and so the sum is jGj. Otherwise there exists ab 2 G such that .b/ ¤ .b/. As a runs over G, so also does ab, and soXa2G .a/ .a 1 / D X a2G .ab/ ..ab/ 1 / D .b/ .b/ 1 X a2G .a/ .a 1 /:Because .b/ .b/ 1 ¤ 1, this implies that P a2G .a/ .a 1 / D 0.COROLLARY 1.63 For any a 2 G,X2G _ .a/ D jGj if a D e0 otherwise.✷✷PROOF. Apply the theorem to G _ , noting that .G _ / _ ' G.✷

The order of ab 29The order of abLet a and b be elements of a group G. If a has order m and b has order n, what can we sayabout the order of ab? The next theorem shows that we can say nothing at all.THEOREM 1.64 For any integers m;n;r > 1, there exists a finite group G with elements aand b such that a has order m, b has order n, and ab has order r.PROOF. We shall show that, for a suitable prime power q, there exist elements a and b ofSL 2 .F q / such that a, b, and ab have orders 2m, 2n, and 2r respectively. As I is theunique element of order 2 in SL 2 .F q /, the images of a, b, ab in SL 2 .F q /=f˙I g will thenhave orders m, n, and r as required.Let p be a prime number not dividing 2mnr. Then p is a unit in the finite ringZ=2mnrZ, and so some power of it, q say, is 1 in the ring. This means that 2mnr dividesq 1. As the group F q has order q 1 and is cyclic (see 1.56), there exist elementsu, v, and w of F q having orders 2m, 2n, and 2r respectively. Let u 1v 0a D0 u 1 and b Dt v 1 (elements of SL 2 .F q /);where t has been chosen so thatuv C t C u 1 v 1 D w C w 1 :The characteristic polynomial of a is .X u/.X u 1 /, and so a is similar to diag.u;u 1 /.Therefore a has order 2m. Similarly b has order 2n. The matrix uv C t v1ab Du 1 t u 1 v 1 ;has characteristic polynomialX 2 .uv C t C u 1 v 1 /X C 1 D .X w/.X w 1 /,and so ab is similar to diag.w;w 1 /. Therefore ab has order 2r.✷Exercises1-1 Show that the quaternion group has only one element of order 2, and that it commuteswith all elements of Q. Deduce that Q is not isomorphic to D 4 , and that every subgroup ofQ is normal.1-2 Consider the elementsa D 0 11 0b D 0 11 1in GL 2 .Z/. Show that a 4 D 1 and b 3 D 1, but that ab has infinite order, and hence that thegroup ha;bi is infinite.1-3 Show that every finite group of even order contains an element of order 2.

30 1. BASIC DEFINITIONS AND RESULTS1-4 Let n D n 1 C C n r be a partition of the positive integer n. Use Lagrange’s theoremto show that nŠ is divisible by Q riD1 n iŠ.1-5 Let N be a normal subgroup of G of index n. Show that if g 2 G, then g n 2 N . Givean example to show that this may be false when the subgroup is not normal.1-6 A group G is said to have finite exponent if there exists an m > 0 such that a m D efor every a in G; the smallest such m is then called the exponent of G.(a) Show that every group of exponent 2 is commutative.(b) Show that, for an odd prime p, the group of matrices8019< 1 a b=@ 0 1 cA:0 0 1 ˇ a;b;c 2 F p;has exponent p, but is not commutative.1-7 Show that any set of representatives for the conjugacy classes in a finite group generatesthe group.1-8 Two subgroups H and H 0 of a group G are said to be commensurable if H \ H 0 isof finite index in both H and H 0 . Show that commensurability is an equivalence relationon the subgroups of G.1-9 Show that a nonempty finite set with an associative binary operation satisfying thecancellation laws is a group.

CHAPTER 2Free Groups and Presentations;Coxeter GroupsIt is frequently useful to describe a group by giving a set of generators for the group anda set of relations for the generators from which every other relation in the group can bededuced. For example, D n can be described as the group with generators r;s and relationsr n D e; s 2 D e; srsr D e:In this chapter, we make precise what this means. First we need to define the free group on aset X of generators — this is a group generated by X and with no relations except for thoseimplied by the group axioms. Because inverses cause problems, we first do this for monoids.Recall that a monoid is a set S with an associative binary operation having an identityelement e. A homomorphism ˛WS ! S 0 of monoids is a map such that ˛.ab/ D ˛.a/˛.b/for all a;b 2 S and ˛.e/ D e — unlike the case of groups, the second condition is notautomatic. A homomorphism of monoids preserves all finite products.Free monoidsLet X D fa;b;c;:::g be a (possibly infinite) set of symbols. A word is a finite sequence ofsymbols from X in which repetition is allowed. For example,aa; aabac; bare distinct words. Two words can be multiplied by juxtaposition, for example,aaaa aabac D aaaaaabac:This defines on the set of all words an associative binary operation. The empty sequenceis allowed, and we denote it by 1. (In the unfortunate case that the symbol 1 is already anelement of X, we denote it by a different symbol.) Then 1 serves as an identity element.Write SX for the set of words together with this binary operation. Then SX is a monoid,called the free monoid on X.31

32 2. FREE GROUPS AND PRESENTATIONS; COXETER GROUPSWhen we identify an element a of X with the word a, X becomesa subset of SX and generates it (i.e., no proper submonoid of SXcontains X). Moreover, the map X ! SX has the following universalproperty: for any map of sets ˛WX ! S from X to a monoidS, there exists a unique homomorphism SX ! S making the diagramat right commute:Xa 7! a˛SXS:Free groupsWe want to construct a group FX containing X and having the same universal property asSX with “monoid” replaced by “group”. Define X 0 to be the set consisting of the symbolsin X and also one additional symbol, denoted a 1 , for each a 2 X; thusX 0 D fa;a 1 ;b;b 1 ;:::g:Let W 0 be the set of words using symbols from X 0 . This becomes a monoid under juxtaposition,but it is not a group because a 1 is not yet the inverse of a, and we can’t cancel outthe obvious terms in words of the following form:aa 1 or a 1 a A word is said to be reduced if it contains no pairs of the form aa 1 or a 1 a. Starting witha word w, we can perform a finite sequence of cancellations to arrive at a reduced word(possibly empty), which will be called the reduced form w 0 of w. There may be manydifferent ways of performing the cancellations, for example,cabb 1 a 1 c 1 ca ! caa 1 c 1 ca ! cc 1 ca ! ca;cabb 1 a 1 c 1 ca ! cabb 1 a 1 a ! cabb 1 ! ca:We have underlined the pair we are cancelling. Note that the middle a 1 is cancelled withdifferent a’s, and that different terms survive in the two cases (the ca at the right in the firstcancellation, and the ca at left in the second). Nevertheless we ended up with the sameanswer, and the next result says that this always happens.PROPOSITION 2.1 There is only one reduced form of a word.PROOF. We use induction on the length of the word w. If w is reduced, there is nothing toprove. Otherwise a pair of the form a 0 a0 1 or a0 1a0 occurs — assume the first, since theargument is the same in both cases.Observe that any two reduced forms of w obtained by a sequence of cancellations inwhich a 0 a0 1 is cancelled first are equal, because the induction hypothesis can be appliedto the (shorter) word obtained by cancelling a 0 a0 1.Next observe that any two reduced forms of w obtained by a sequence of cancellationsin which a 0 a0 1 is cancelled at some point are equal, because the result of such a sequenceof cancellations will not be affected if a 0 a0 1 is cancelled first.Finally, consider a reduced form w 0 obtained by a sequence in which no cancellationcancels a 0 a0 1 directly. Since a 0 a0 1 does not remain in w 0 , at least one of a 0 or a0 1 must

Free groups 33be cancelled at some point. If the pair itself is not cancelled, then the first cancellationinvolving the pair must look like 6 a 10 6 a 0a 10 or a 0 6 a 10 6 a 0 where our original pair is underlined. But the word obtained after this cancellation is thesame as if our original pair were cancelled, and so we may cancel the original pair instead.Thus we are back in the case just proved.✷We say two words w;w 0 are equivalent, denoted w w 0 , if they have the same reducedform. This is an equivalence relation (obviously).PROPOSITION 2.2 Products of equivalent words are equivalent, i.e.,w w 0 ; v v 0 H) wv w 0 v 0 :PROOF. Let w 0 and v 0 be the reduced forms of w and of v. To obtain the reduced formof wv, we can first cancel as much as possible in w and v separately, to obtain w 0 v 0 andthen continue cancelling. Thus the reduced form of wv is the reduced form of w 0 v 0 . Asimilar statement holds for w 0 v 0 , but (by assumption) the reduced forms of w and v equalthe reduced forms of w 0 and v 0 , and so we obtain the same result in the two cases. ✷Let FX be the set of equivalence classes of words. Proposition 2.2 shows that thebinary operation on W 0 defines a binary operation on FX, which obviously makes it into amonoid. It also has inverses, because.ab gh/ h 1 g 1 b 1 a 1 1:Thus FX is a group, called the free group on X. To summarize: the elements of FX arerepresented by words in X 0 ; two words represent the same element of FX if and only ifthey have the same reduced forms; multiplication is defined by juxtaposition; the emptyword represents 1; inverses are obtained in the obvious way. Alternatively, each element ofFX is represented by a unique reduced word; multiplication is defined by juxtaposition andpassage to the reduced form.When we identify a 2 X with the equivalence class of the (reduced) word a, then Xbecomes identified with a subset of FX — clearly it generates FX. The next propositionis a precise statement of the fact that there are no relations among the elements of X whenregarded as elements of FX except those imposed by the group axioms.PROPOSITION 2.3 For any map of sets ˛WX ! G from X to a group G, there exists aunique homomorphism FX ! G making the following diagram commute:Xa 7! aFX˛G:

34 2. FREE GROUPS AND PRESENTATIONS; COXETER GROUPSPROOF. Consider a map ˛WX ! G. We extend it to a map of sets X 0 ! G by setting˛.a 1 / D ˛.a/ 1 . Because G is, in particular, a monoid, ˛ extends to a homomorphism ofmonoids SX 0 ! G. This map will send equivalent words to the same element of G, and sowill factor through FX D SX 0 =. The resulting map FX ! G is a group homomorphism.It is unique because we know it on a set of generators for FX.✷REMARK 2.4 The universal property of the map WX ! FX, x 7! x, characterizes it: if 0 WX ! F 0 is a second map with the same universal property, then there is a unique isomorphism˛WFX ! F 0 such that ˛ ı D 0 ,FXX˛ 0F 0 :We recall the proof: by the universality of , there exists a unique homomorphism ˛WFX !F 0 such that ˛ ı D 0 ; by the universality of 0 , there exists a unique homomorphismˇWF 0 ! FX such that ˇ ı 0 D ; now .ˇ ı ˛/ ı D , but by the universality of , id F Xis the unique homomorphism FX ! FX such that id F X ı D , and so ˇ ı ˛ D id F X ;similarly, ˛ ı ˇ D id F 0, and so ˛ and ˇ are inverse isomorphisms.COROLLARY 2.5 Every group is a quotient of a free group.PROOF. Choose a set X of generators for G (e.g., X D G), and let F be the free groupgenerated by X. According to (2.3), the map a 7! aW X ! G extends to a homomorphismF ! G, and the image, being a subgroup containing X, must equal G:✷The free group on the set X D fag is simply the infinite cyclic group C 1 generated bya, but the free group on a set consisting of two elements is already very complicated.I now discuss, without proof, some important results on free groups.THEOREM 2.6 (NIELSEN-SCHREIER) 1 Subgroups of free groups are free.The best proof uses topology, and in particular covering spaces—see Serre 1980 orRotman 1995, Theorem 11.44.Two free groups FX and F Y are isomorphic if and only if X and Y have the samecardinality. Thus we can define the rank of a free group G to be the cardinality of any freegenerating set (subset X of G for which the homomorphism FX ! G given by (2.3) isan isomorphism). Let H be a finitely generated subgroup of a free group G. Then thereis an algorithm for constructing from any finite set of generators for H a free finite set ofgenerators. If G has finite rank n and .G W H / D i < 1, then H is free of rankni i C 1:In particular, H may have rank greater than that of F (or even infinite rank 2 ). For proofs,see Rotman 1995, Chapter 11, and Hall 1959, Chapter 7.1 Nielsen (1921) proved this for finitely generated subgroups, and in fact gave an algorithm for decidingwhether a word lies in the subgroup; Schreier (1927) proved the general case.2 For example, the commutator subgroup of the free group on two generators has infinite rank.

Generators and relations 35Generators and relationsConsider a set X and a set R of words made up of symbols in X 0 . Each element of Rrepresents an element of the free group FX, and the quotient G of FX by the normalsubgroup generated by these elements (1.40) is said to have X as generators and R asrelations (or as a set of defining relations). One also says that .X;R/ is a presentation forG, and denotes G by hX j Ri.EXAMPLE 2.7 (a) The dihedral group D n has generators r;s and defining relationsr n ;s 2 ;srsr:(See 2.9 below for a proof.)(b) The generalized quaternion group Q n , n 3, has generators a;b and relations 3a 2n 1 D 1;a 2n 2 D b 2 ;bab 1 D a 1 :For n D 3 this is the group Q of (1.18). In general, it has order 2 n (for more on it, seeExercise 2-5).(c) Two elements a and b in a group commute if and only if their commutator Œa;b Ddefaba 1 b 1 is 1. The free abelian group on generators a 1 ;:::;a n has generators a 1 ;a 2 ;:::;a nand relationsŒa i ;a j ; i ¤ j:(d) Let G D hs;t j s 3 t;t 3 ;s 4 i. Then G D f1g becauses D ss 3 t D s 4 t D t1 D s 3 tt3 D s 3 ss 3 D s:For the remaining examples, see Massey 1967, which contains a good account of theinterplay between group theory and topology. For example, for many types of topologicalspaces, there is an algorithm for obtaining a presentation for the fundamental group.(e) The fundamental group of the open disk with one point removed is the free groupon where is any loop around the point (ibid. II 5.1).(f) The fundamental group of the sphere with r points removed has generators 1 ;:::; r( i is a loop around the ith point) and a single relation 1 r D 1:(g) The fundamental group of a compact Riemann surface of genus g has 2g generatorsu 1 ;v 1 ;:::;u g ;v g and a single relation(ibid. IV Exercise 5.7).u 1 v 1 u 11 v 11 u gv g u 1g v 1g D 13 Strictly speaking, I should say the relations a 2n 1 , a 2n 2 b 2 , bab 1 a.

Finitely presented groups 37Finitely presented groupsA group is said to be finitely presented if it admits a presentation .X;R/ with both X andR finite.EXAMPLE 2.11 Consider a finite group G. Let X D G, and let R be the set of wordsfabc 1 j ab D c in Gg:I claim that .X;R/ is a presentation of G, and so G is finitely presented. Let G 0 D hX jRi. The extension of a 7! aWX ! G to FX sends each element of R to 1, and thereforedefines a homomorphism G 0 ! G, which is obviously surjective. But every element of G 0is represented by an element of X, and so jG 0 j jGj. Therefore the homomorphism isbijective.Although it is easy to define a group by a finite presentation, calculating the propertiesof the group can be very difficult — note that we are defining the group, which may be quitesmall, as the quotient of a huge free group by a huge subgroup. I list some negative results.THE WORD PROBLEMLet G be the group defined by a finite presentation .X;R/. The word problem for G askswhether there exists an algorithm (decision procedure) for deciding whether a word on X 0represents 1 in G. The answer is negative: Novikov and Boone showed that there existfinitely presented groups G for which no such algorithm exists. Of course, there do existother groups for which there is an algorithm.The same ideas lead to the following result: there does not exist an algorithm thatwill determine for an arbitrary finite presentation whether or not the corresponding groupis trivial, finite, abelian, solvable, nilpotent, simple, torsion, torsion-free, free, or has asolvable word problem.See Rotman 1995, Chapter 12, for proofs of these statements.THE BURNSIDE PROBLEMRecall that a group is said to have exponent e if g e D 1 for all g 2 G and e is the smallestnatural number with this property. It is easy to write down examples of infinite groupsgenerated by a finite number of elements of finite order (see Exercise 1-2 or Example 2.10),but does there exist such a group with finite exponent? (Burnside problem). In 1968, Adjanand Novikov showed the answer is yes: there do exist infinite finitely generated groups offinite exponent.THE RESTRICTED BURNSIDE PROBLEMThe Burnside group of exponent e on r generators B.r;e/ is the quotient of the free groupon r generators by the subgroup generated by all eth powers. The Burnside problem askedwhether B.r;e/ is finite, and it is known to be infinite except some small values of r ande. The restricted Burnside problem asks whether B.r;e/ has only finitely many finite quotients;equivalently, it asks whether there is one finite quotient of B.r;e/ having all otherfinite quotients as quotients. The classification of the finite simple groups (see p. 50) showedthat in order prove that B.r;e/ always has only finitely many finite quotients, it suffices to

38 2. FREE GROUPS AND PRESENTATIONS; COXETER GROUPSprove it for e equal to a prime power. This was shown by Efim Zelmanov in 1989 afterearlier work of Kostrikin. See Feit 1995.TODD-COXETER ALGORITHMThere are some quite innocuous looking finite presentations that are known to define quitesmall groups, but for which this is very difficult to prove. The standard approach to thesequestions is to use the Todd-Coxeter algorithm (see Chapter 4 below).We shall develop various methods for recognizing groups from their presentations (seealso the exercises).Coxeter groupsA Coxeter system is a pair .G;S/ consisting of a group G and a set of generators S for Gsubject only to relations of the form .st/ m.s;t/ D 1, where8

Coxeter groups 39A reflection is a linear map sWV ! V sending some nonzero vector ˛ to ˛ and fixing thepoints of the hyperplane H˛ orthogonal to ˛. We write s˛ for the reflection defined by ˛; itis given by the formula2.v;˛/s˛v D v.˛;˛/ ˛;because this is certainly correct for v D ˛ and for v 2 H˛, and hence (by linearity) on thewhole of V D h˛i˚H˛. A finite reflection group is a finite group generated by reflections.For such a group G, it is possible to choose a set S of generating reflections for which.G;S/ is a Coxeter system (Humphreys 1990, 1.9). Thus, the finite reflection groups are allCoxeter groups (in fact, they are precisely the finite Coxeter groups, ibid., 6.4).2.15 Let S n act on R n by permuting the coordinates,.a 1 ;:::;a n / D .a .1/ ;:::;a .n/ /:The transposition .ij / interchanging i and j , sends the vector˛ D .0;:::;0; i 1;0;:::;0; j 1;0;:::/to its negative, and leaves the points of the hyperplaneH˛ D .a 1 ;:::; ia i ;:::; j a i ;:::;a n /fixed. Therefore, .ij / is a reflection. As S n is generated by the transpositions, this showsthat it is a finite reflection group (hence also a Coxeter group).THE STRUCTURE OF COXETER GROUPSTHEOREM 2.16 Let .G;S/ be the the Coxeter system defined by a map mWS S ! N [f1g satisfying (14).(a) The natural map S ! G is injective.(b) Each s 2 S has order 2 in G.(c) For each s ¤ t in S, st has order m.s;t/ in G.PROOF. Note that the order of s is 1 or 2, and the order of st divides m.s;t/, and so thetheorem says that the elements of S remain distinct in G and that each s and each st hasthe largest possible order.If S has only a single element, then G ' C 2 (see 2.12), and so the statements are obvious.Otherwise, let s and t be distinct elements of S, and let G 0 D hs;t j s 2 ;t 2 ;.st/ m.s;t/ i.The map S ! G 0 sending s to s, t to t, and all other elements of S to 1 extends to a homomorphismG ! G 0 . We know that s and t are distinct elements of order 2 in G 0 and that sthas order m.s;t/ in G 0 (see 2.13), and it follows that the same is true in G.✷REMARK 2.17 Let V be the R-vector space with basis a family .e s / s2S indexed by S.The standard proof of Theorem 2.16 defines a “geometry” on V for which there exist “reflections” s , s 2 S, such that s t has order m.s;t/. According to (2.8), the map s 7! sextends to homomorphism of group G ! GL.V /. This proves the theorem, and it realizesG as a group of automorphisms of a “geometry”. See Humphreys 1990, Chapter 5, or v3.02of these notes.

40 2. FREE GROUPS AND PRESENTATIONS; COXETER GROUPSExercises2-1 Let D n D ha;bja n ;b 2 ;ababi be the nth dihedral group. If n is odd, prove that D 2n ha n i ha 2 ;bi, and hence that D 2n C 2 D n .2-2 Prove that the group with generators a 1 ;:::;a n and relations Œa i ;a j D 1, i ¤ j , isthe free abelian group on a 1 ;:::;a n . [Hint: Use universal properties.]2-3 Let a and b be elements of an arbitrary free group F . Prove:(a) If a n D b n with n > 1, then a D b.(b) If a m b n D b n a m with mn 6D 0, then ab D ba.(c) If the equation x n D a has a solution x for every n, then a D 1.2-4 Let F n denote the free group on n generators. Prove:(a) If n < m, then F n is isomorphic to both a subgroup and a quotient group of F m .(b) Prove that F 1 F 1 is not a free group.(c) Prove that the centre Z.F n / D 1 provided n > 1.2-5 Prove that Q n (see 2.7b) has a unique subgroup of order 2, which is Z.Q n /. Provethat Q n =Z.Q n / is isomorphic to D 2 n 1.2-6 (a) Prove that ha;b j a 2 ;b 2 ;.ab/ n i ' D n (cf. 2.9).(b) Prove that G D ha;b j a 2 ;ababi is an infinite group. (This is usually known as theinfinite dihedral group.)2-7 Let G D ha;b;c j a 3 ;b 3 ;c 4 ;acac 1 ;aba 1 bc 1 b 1 i. Prove that G is the trivialgroup f1g. [Hint: Expand .aba 1 / 3 D .bcb 1 / 3 .]2-8 Let F be the free group on the set fx;yg and let G D C 2 , with generator a ¤ 1. Let ˛be the homomorphism F ! G such that ˛.x/ D a D ˛.y/. Find a minimal generating setfor the kernel of ˛. Is the kernel a free group?2-9 Let G D hs;t j t 1 s 3 t D s 5 i. Prove that the elementg D s 1 t1 s 1 tst1 stis in the kernel of every map from G to a finite group.Coxeter came to Cambridge and gave a lecture [in which he stated a] problem for which he gaveproofs for selected examples, and he asked for a unified proof. I left the lecture room thinking.As I was walking through Cambridge, suddenly the idea hit me, but it hit me while I was inthe middle of the road. When the idea hit me I stopped and a large truck ran into me. . . . So Ipretended that Coxeter had calculated the difficulty of this problem so precisely that he knewthat I would get the solution just in the middle of the road. . . . Ever since, I’ve called that theorem“the murder weapon”. One consequence of it is that in a group if a 2 D b 3 D c 5 D .abc/ 1 ,then c 610 D 1.John Conway, Math. Intelligencer 23 (2001), no. 2, pp. 8–9.

CHAPTER 3Automorphisms and ExtensionsAutomorphisms of groupsAn automorphism of a group G is an isomorphism of the group with itself. The set Aut.G/of automorphisms of G becomes a group under composition: the composite of two automorphismsis again an automorphism; composition of maps is always associative (see (5),p. 9); the identity map g 7! g is an identity element; an automorphism is a bijection, andtherefore has an inverse, which is again an automorphism.For g 2 G, the map i g “conjugation by g”,x 7! gxg 1 W G ! Gis an automorphism of G. An automorphism of this form is called an inner automorphism,and the remaining automorphisms are said to be outer.Note that.gh/x.gh/ 1 D g.hxh 1 /g 1 , i.e., i gh .x/ D .i g ı i h /.x/;and so the map g 7! i g WG ! Aut.G/ is a homomorphism. Its image is denoted by Inn.G/.Its kernel is the centre of G,Z.G/ D fg 2 G j gx D xg all x 2 Gg;and so we obtain from (1.45) an isomorphismG=Z.G/ ! Inn.G/:In fact, Inn.G/ is a normal subgroup of Aut.G/: for g 2 G and ˛ 2 Aut.G/,.˛ ı i g ı ˛ 1/.x/ D ˛.g ˛ 1.x/ g 1 / D ˛.g/ x ˛.g/ 1 D i˛.g/ .x/:EXAMPLE 3.1 (a) Let G D F n p . The automorphisms of G as a commutative group are justthe automorphisms of G as a vector space over F p ; thus Aut.G/ D GL n .F p /. Because Gis commutative, all nontrivial automorphisms of G are outer.(b) As a particular case of (a), we see thatAut.C 2 C 2 / D GL 2 .F 2 /:41

42 3. AUTOMORPHISMS AND EXTENSIONS(c) Since the centre of the quaternion group Q is ha 2 i, we have thatIn fact, Aut.Q/ S 4 . See Exercise 3-4.Inn.Q/ ' Q=ha 2 i C 2 C 2 :ASIDE 3.2 The inner automorphisms of a group are the only automorphisms that extend to everyovergroup (Schupp 1987).COMPLETE GROUPSDEFINITION 3.3 A group G is complete if the map g 7! i g WG ! Aut.G/ is an isomorphism.Thus, a group G is complete if and only if (a) the centre Z.G/ of G is trivial, and (b)every automorphism of G is inner.EXAMPLE 3.4 (a) For n ¤ 2;6, S n is complete. The group S 2 is commutative and hencefails (a); Aut.S 6 /=Inn.S 6 / C 2 and hence S 6 fails (b). See Rotman 1995, Theorems 7.5,7.10.(b) If G is a simple noncommutative group, then Aut.G/ is complete. See Rotman1995, Theorem 7.14.According to Exercise 3-3, GL 2 .F 2 / S 3 , and so the nonisomorphic groups C 2 C 2and S 3 have isomorphic automorphism groups.AUTOMORPHISMS OF CYCLIC GROUPSLet G be a cyclic group of order n, say G D hai. Let m be an integer 1. The smallestnmultiple of m divisible by n is m gcd.m;n/ . Therefore, am nhas order , and so thegcd.m;n/generators of G are exactly the elements a m with gcd.m;n/ D 1. An automorphism ˛ of Gmust send a to another generator of G, and so ˛.a/ D a m for some m relatively prime to n.The map ˛ 7! m defines an isomorphismwhereAut.C n / ! .Z=nZ/ .Z=nZ/ D funits in the ring Z=nZg D fm C nZ j gcd.m;n/ D 1g:This isomorphism is independent of the choice of a generator a for G: if ˛.a/ D a m , thenfor any other element b D a i of G,˛.b/ D ˛.a i / D ˛.a/ i D a mi D .a i / m D .b/ m :It remains to determine .Z=nZ/ . If n D p r 11 pr ssproduct of powers of distinct primes, thenis the factorization of n into aZ=nZ ' Z=p r 11 Z Z=pr ss Z; m mod n $ .m mod pr 1;:::/by the Chinese remainder theorem. This is an isomorphism of rings, and so.Z=nZ/ ' .Z=p r 11 Z/ .Z=p r ss Z/ :

Characteristic subgroups 43It remains to consider the case n D p r , p prime.Suppose first that p is odd. The set f0;1;:::;p r 1g is a complete set of representativesfor Z=p r Z, and 1 p of these elements are divisible by p. Hence .Z=pr Z/ has order p rp rp D pr 1 .p 1/. The homomorphism.Z=p r Z/ ! .Z=pZ/ is surjective with kernel of order p r 1 , and we know that .Z=pZ/ is cyclic. Let a 2.Z=p r Z/ map to a generator of .Z=pZ/ . Then a pr .p 1/ D 1 and a pr again maps to agenerator of .Z=pZ/ . Therefore .Z=p r Z/ contains an element D defa pr of order p 1.Using the binomial theorem, one finds that 1 C p has order p r 1 in .Z=p r Z/ . Therefore.Z=p r Z/ is cyclic with generator .1 C p/ (cf. (13), p. 27), and every element can bewritten uniquely in the form i .1 C p/ j ; 0 i < p 1; 0 j < p r 1 :On the other hand,is not cyclic..Z=8Z/ D fN1; N3; N5; N7g D hN3; N5i C 2 C 2SUMMARY 3.5 (a) For a cyclic group of G of order n, Aut.G/ ' .Z=nZ/ : The automorphismof G corresponding to Œm 2 .Z=nZ/ sends an element a of G to a m .(b) If n D p r 11 pr ss is the factorization of n into a product of powers of distinct primesp i , then.Z=nZ/ ' .Z=p r 11 Z/ .Z=p r ss Z/ ; m mod n $ .m mod p r 1;:::/:(c) For a prime p,8ˆ< C .p 1/p r 1 p odd,.Z=p r Z/ C 2 pˆ:r D 2 2C 2 C 2 r 2 p D 2, r > 2:Characteristic subgroupsDEFINITION 3.6 A characteristic subgroup of a group G is a subgroup H such that˛.H / D H for all automorphisms ˛ of G.The same argument as in (1.32) shows that it suffices to check that ˛.H / H for all ˛ 2Aut.G/. Thus, a subgroup H of G is normal if it is stable under all inner automorphisms ofG, and it is characteristic if it stable under all automorphisms. In particular, a characteristicsubgroup is normal.REMARK 3.7 (a) Consider a group G and a normal subgroup N . An inner automorphismof G restricts to an automorphism of N , which may be outer (for an example, see 3.16below). Thus a normal subgroup of N need not be a normal subgroup of G. However, acharacteristic subgroup of N will be a normal subgroup of G. Also a characteristic subgroupof a characteristic subgroup is a characteristic subgroup.

44 3. AUTOMORPHISMS AND EXTENSIONS(b) The centre Z.G/ of G is a characteristic subgroup, becausezg D gz all g 2 G H) ˛.z/˛.g/ D ˛.g/˛.z/ all g 2 G;and as g runs over G, ˛.g/ also runs over G. Expect subgroups with a general grouptheoreticdefinition to be characteristic.(c) If H is the only subgroup of G of order m, then it must be characteristic, because˛.H / is again a subgroup of G of order m.(d) Every subgroup of a commutative group is normal but not necessarily characteristic.For example, every subspace of dimension 1 in F 2 p is subgroup of F2 p , but it is notcharacteristic because it is not stable under Aut.F 2 p / D GL 2.F p /.Semidirect productsLet N be a normal subgroup of G. Each element g of G defines an automorphism of N ,n 7! gng 1 , and this defines a homomorphismWG ! Aut.N /;g 7! i g jN:If there exists a subgroup Q of G such that G ! G=N maps Q isomorphically onto G=N ,then I claim that we can reconstruct G from N , Q, and the restriction of to Q. Indeed,an element g of G can be written uniquely in the formg D nq; n 2 N; q 2 Q;— q must be the unique element of Q mapping to gN 2 G=N , and n must be gq 1 . Thus,we have a one-to-one correspondence of setsIf g D nq and g 0 D n 0 q 0 , thenG 1-1 ! N Q:gg 0 D .nq/ n 0 q 0 D n.qn 0 q 1 /qq 0 D n .q/.n 0 / qq 0 :DEFINITION 3.8 A group G is a semidirect product of its subgroups N and Q if N isnormal and the homomorphism G ! G=N induces an isomorphism Q ! G=N .Equivalently, G is a semidirect product of subgroup N and Q ifN G G; NQ D G; N \ Q D f1g: (15)Note that Q need not be a normal subgroup of G. When G is the semidirect product ofsubgroups N and Q, we write G D N ⋊ Q (or N ⋊ Q where WQ ! Aut.N / gives theaction of Q on N by inner automorphisms).EXAMPLE 3.9 (a) In D n , n 2, let C n D hri and C 2 D hsi; thenwhere .s/.r i / D r i (see 1.17).D n D hri ⋊ hsi D C n ⋊ C 2

Semidirect products 45(b) The alternating subgroup A n is a normal subgroup of S n (because it has index 2),and C 2 D f.12/g maps isomorphically onto S n =A n . Therefore S n D A n ⋊ C 2 .(c) The quaternion group can not be written as a semidirect product in any nontrivialfashion (see Exercise 3-1).(d) A cyclic group of order p 2 , p prime, is not a semidirect product (because it has onlyone subgroup of order p).(e) Let G D GL n .F /. Let B be the subgroup of upper triangular matrices in G, T thesubgroup of diagonal matrices in G, and U the subgroup of upper triangular matrices withall their diagonal coefficients equal to 1. Thus, when n D 2, B D0 ; T D 00 1; U D0 1.Then, U is a normal subgroup of B, U T D B, and U \ T D f1g. Therefore,B D U ⋊ T .Note that, when n 2, the action of T on U is not trivial, for example, a 0 1 c a10 1 ac=b0 b 0 1 0 b 1 D;0 1and so B is not the direct product of T and U .We have seen that, from a semidirect product G D N ⋊ Q, we obtain a triple.N;Q;WQ ! Aut.N //;and that the triple determines G. We now prove that every triple .N;Q;/ consisting of twogroups N and Q and a homomorphism WQ ! Aut.N / arises from a semidirect product.As a set, let G D N Q, and define.n;q/.n 0 ;q 0 / D .n .q/.n 0 /;qq 0 /:PROPOSITION 3.10 The composition law above makes G into a group, in fact, the semidirectproduct of N and Q:PROOF. Write q n for .q/.n/, so that the composition law becomesThen.n;q/.n 0 ;q 0 / D .n q n 0 ;qq 0 /...n;q/;.n 0 ;q 0 //.n 00 ;q 00 / D .n q n 0 qq0 n 00 ;qq 0 q 00 / D .n;q/..n 0 ;q 0 /.n 00 ;q 00 //and so the associative law holds. Because .1/ D 1 and .q/.1/ D 1,and so .1;1/ is an identity element. Next.1;1/.n;q/ D .n;q/ D .n;q/.1;1/,.n;q/. q 1 n 1 ;q 1 / D .1;1/ D . q 1 n 1 ;q 1 /.n;q/;and so . q 1 n 1 ;q 1 / is an inverse for .n;q/. Thus G is a group, and it is obvious thatN G G, NQ D G, and N \ Q D f1g, and so G D N ⋊ Q. Moreover, when N and Q areregarded as subgroups of G, the action of Q on N is that given by .✷

46 3. AUTOMORPHISMS AND EXTENSIONSEXAMPLES3.11 A group of order 12. Let be the (unique) nontrivial homomorphismC 4 ! Aut.C 3 / ' C 2 ;namely, that sending a generator of C 4 to the map a 7! a 2 . Then G defD C 3 ⋊ C 4 is anoncommutative group of order 12, not isomorphic to A 4 . If we denote the generators ofC 3 and C 4 by a and b, then a and b generate G, and have the defining relations3.12 Direct products. The bijection of setsa 3 D 1; b 4 D 1; bab 1 D a 2 :.n;q/ 7! .n;q/WN Q ! N ⋊ Qis an isomorphism of groups if and only if is the trivial homomorphism Q ! Aut.N /,i.e., .q/.n/ D n for all q 2 Q, n 2 N .3.13 Groups of order 6. Both S 3 and C 6 are semidirect products of C 3 by C 2 — theycorrespond to the two distinct homomorphisms C 2 ! C 2 ' Aut.C 3 /.3.14 Groups of order p 3 (element of order p 2 ). Let N D hai be cyclic of order p 2 ,and let Q D hbi be cyclic of order p, where p is an odd prime. Then AutN C p 1 C p(see 3.5), and C p is generated by ˛Wa 7! a 1Cp (note that ˛2.a/ D a 1C2p ;:::). DefineQ ! AutN by b 7! ˛. The group G D defN ⋊ Q has generators a;b and defining relationsa p2 D 1; b p D 1; bab 1 D a 1Cp :It is a noncommutative group of order p 3 , and possesses an element of order p 2 .3.15 Groups of order p 3 (no element of order p 2 ). Let N D ha;bi be the product oftwo cyclic groups hai and hbi of order p, and let Q D hci be a cyclic group of order p.Define WQ ! Aut.N / to be the homomorphism such that.c i /.a/ D ab i ; .c i /.b/ D b.(If we regard N as the additive group N D F 2 p with a and b the standard basis elements, then 1 0.c i / is the automorphism of N defined by the matrix .) The group G D defN ⋊i 1 Qis a group of order p 3 , with generators a;b;c and defining relationsa p D b p D c p D 1; ab D cac 1 ; Œb;a D 1 D Œb;c:Because b ¤ 1, the middle equality shows that the group is not commutative. When p isodd, all elements except 1 have order p. When p D 2, G D 4 , which does have an elementof order 2 2 : Note that this shows that a group can have quite different representations as asemidirect product:.3.9a)D 4 C 4 ⋊ C 2 .C 2 C 2 / ⋊ C 2 :

Semidirect products 47For an odd prime p, a noncommutative group of order p 3 is isomorphic to the group in(3.14) if it has an element of order p 2 and to the group in (3.15) if it doesn’t (see Exercise4-3). In particular, up to isomorphism, there are exactly two noncommutative groups oforder p 3 .3.16 Making outer automorphisms inner. Let ˛ be an automorphism, possibly outer,of a group N . We can realize N as a normal subgroup of a group G in such a way that˛ becomes the restriction to N of an inner automorphism of G. To see this, let WC 1 !Aut.N / be the homomorphism sending a generator a of C 1 to ˛ 2 Aut.N /, and let G DN ⋊ C 1 . The element g D .1;a/ of G has the property that g.n;1/g 1 D .˛.n/;1/ for alln 2 N .CRITERIA FOR SEMIDIRECT PRODUCTS TO BE ISOMORPHICIt will be useful to have criteria for when two triples .N;Q;/ and .N;Q; 0 / determineisomorphic groups.LEMMA 3.17 If there exists an ˛ 2 Aut.N / such that 0 .q/ D ˛ ı .q/ ı ˛ 1; all q 2 Q;then the map.n;q/ 7! .˛.n/;q/WN ⋊ Q ! N ⋊ 0 Qis an isomorphism.PROOF. For .n;q/ 2 N ⋊ Q, let .n;q/ D .˛.n/;q/. Then.n;q/ .n 0 ;q 0 / D .˛.n/;q/ .˛.n 0 /;q 0 /D .˛.n/ 0 .q/.˛.n 0 //;qq 0 /D .˛.n/ .˛ ı .q/ ı ˛ 1/.˛.n 0 //;qq 0 /D .˛.n/ ˛..q/.n 0 //;qq 0 /;and..n;q/ .n 0 ;q 0 // D .n .q/.n 0 /;qq 0 /D .˛.n/ ˛ .q/.n 0 / ;qq 0 /:Therefore is a homomorphism. The map.n;q/ 7! .˛ 1.n/;q/WN ⋊ 0 Q ! N ⋊ Qis also a homomorphism, and it is inverse to , and so both are isomorphisms.✷LEMMA 3.18 If D 0 ı ˛ with ˛ 2 Aut.Q/, then the mapis an isomorphism..n;q/ 7! .n;˛.q//WN ⋊ Q N ⋊ 0 Q

48 3. AUTOMORPHISMS AND EXTENSIONSPROOF. Routine verification.✷LEMMA 3.19 If Q is cyclic and the subgroup .Q/ of Aut.N / is conjugate to 0 .Q/, thenN ⋊ Q N ⋊ 0 Q:PROOF. Let a generate Q. By assumption, there exists an a 0 2 Q and an ˛ 2 Aut.N / suchthat 0 .a 0 / D ˛ .a/ ˛ 1:The element 0 .a 0 / generates 0 .Q/, and so we can choose a 0 to generate Q, say a 0 D a iwith i relatively prime to the order of Q. Now the map .n;q/ 7! .˛.n/;q i / is an isomorphismN ⋊ Q ! N ⋊ 0 Q.✷SUMMARY 3.20 Let G be a group with subgroups H 1 and H 2 such that G D H 1 H 2 andH 1 \ H 2 D feg, so that each element g of G can be written uniquely as g D h 1 h 2 withh 1 2 H 1 and h 2 2 H 2 .(a) If H 1 and H 2 are both normal, then G is the direct product of H 1 and H 2 , G DH 1 H 2 (1.51).(b) If H 1 is normal in G, then G is the semidirect product of H 1 and H 2 , G D H 1 ⋊ H 2((15), p. 44).(c) If neither H 1 nor H 2 is normal, then G is the Zappa-Szép (or knit) product of H 1and H 2 (see http://en.wikipedia.org/wiki/Zappa-Szep_product).Extensions of groupsA sequence of groups and homomorphisms1 ! N ! G ! Q ! 1 (16)is exact if is injective, is surjective, and Ker./ D Im./. Thus .N / is a normal subgroupof G (isomorphic by to N / and G=.N / ! Q. We often identify N with the'subgroup .N / of G and Q with the quotient G=N:An exact sequence (16) is also called an extension of Q by N . 1 An extension is centralif .N / Z.G/. For example, a semidirect product N ⋊ Q gives rise to an extension ofQ by N ,1 ! N ! N ⋊ Q ! Q ! 1;which is central if and only if is the trivial homomorphism.Two extensions of Q by N are said to be isomorphic if there exists a commutativediagram1 ! N ! G ! Q ! 1??y1 ! N ! G 0 ! Q ! 1:1 This is Bourbaki’s terminology (Algèbre, I 6); some authors call (16) an extension of N by Q.

Extensions of groups 49An extension of Q by N ,1 ! N ! G ! Q ! 1;is said to be split if it is isomorphic to the extension defined by a semidirect product N ⋊ Q.Equivalent conditions:(a) there exists a subgroup Q 0 G such that induces an isomorphism Q 0 ! Q; or(b) there exists a homomorphism sWQ ! G such that ı s D id:In general, an extension will not split. For example, the extensions1 ! N ! Q ! Q=N ! 1 (17)with N any subgroup of order 4 in the quaternion group Q, and1 ! C p ! C p 2 ! C p ! 1do not split. We give two criteria for an extension to split.THEOREM 3.21 (SCHUR-ZASSENHAUS) An extension of finite groups of relatively primeorder is split.PROOF. Rotman 1995, 7.41.✷PROPOSITION 3.22 An extension (17) splits if N is complete. In fact, G is then directproduct of N with the centralizer of N in G,C G .N / defD fg 2 G j gn D ng all n 2 N g.PROOF. Let Q D C G .N /. We shall check that N and Q satisfy the conditions of Proposition1.51.Observe first that, for any g 2 G, n 7! gng 1 WN ! N is an automorphism of N , and(because N is complete), it must be the inner automorphism defined by an element of N ;thusgng 1 D n 1 all n 2 N .This equation shows that 1 g 2 Q, and hence g D . 1 g/ 2 NQ. Since g was arbitrary,we have shown that G D NQ.Next note that every element of N \ Q is in the centre of N , which (because N iscomplete) is trivial; hence N \ Q D 1.Finally, for any element g D nq 2 G,gQg 1 D n.qQq 1 /n 1 D nQn 1 D Q(recall that every element of N commutes with every element of Q). Therefore Q is normalin G.✷An extension1 ! N ! G ! Q ! 1gives rise to a homomorphism 0 WG ! Aut.N /, namely, 0 .g/.n/ D gng 1 :

50 3. AUTOMORPHISMS AND EXTENSIONSLet Qq 2 G map to q in Q; then the image of 0 . Qq/ in Aut.N /=Inn.N / depends only on q;therefore we get a homomorphism W Q ! Out.N / defD Aut.N /=Inn.N /:This map depends only on the isomorphism class of the extension, and we write Ext 1 .Q;N / for the set of isomorphism classes of extensions with a given : These sets have been extensivelystudied.When Q and N are commutative and is trivial, the group G is also commutative, andthere is a commutative group structure on the set Ext 1 .Q;N /. Moreover, endomorphismsof Q and N act as endomorphisms on Ext 1 .Q;N /. In particular, multiplication by m onQ or N induces multiplication by m on Ext 1 .Q;N /. Thus, if Q and N are killed by mand n respectively, then Ext 1 .Q;N / is killed by m and by n, and hence by gcd.m;n/. Thisproves the Schur-Zassenhaus theorem in this case.The Hölder program.It would be of the greatest interest if it were possibleto give an overview of the entire collection of finitesimple groups.Otto Hölder, Math. Ann., 1892Recall that a group G is simple if it contains no normal subgroup except 1 and G. Inother words, a group is simple if it can’t be realized as an extension of smaller groups.Every finite group can be obtained by taking repeated extensions of simple groups. Thusthe simple finite groups can be regarded as the basic building blocks for all finite groups.The problem of classifying all simple groups falls into two parts:A. Classify all finite simple groups;B. Classify all extensions of finite groups.A. THE CLASSIFICATION OF FINITE SIMPLE GROUPSThere is a complete list of finite simple groups. They are(a) the cyclic groups of prime order,(b) the alternating groups A n for n 5 (see the next chapter),(c) certain infinite families of matrix groups, and(d) the 26 “sporadic groups”.By far the largest class is (c), but the 26 sporadic groups are of more interest thantheir small number might suggest. Some have even speculated that the largest of them, theFischer-Griess monster, is built into the fabric of the universe.As an example of a matrix group, considerSL m .F q / defD fm m matrices A with entries in F q such that detA D 1g:

Exercises 51Here q D p n , p prime, and F q is “the”0field with1q elements. This group is not simple if 0 00 0q ¤ 2, because the scalar matricesB@: ::CA , m D 1, are in the centre for any m0 0 dividing q 1, but these are the only matrices in the centre, and the groupsPSL m .F q / defD SL n .F q /=fcentregare simple when m 3 (Rotman 1995, 8.23) and when m D 2 and q > 3 (ibid. 8.13). Otherfinite simple groups can be obtained from the groups in (1.8). The smallest noncommutativegroup is A 5 , and the second smallest is PSL 3 .F 2 /, which as order 168 (see Exercise 4-7).B THE CLASSIFICATION OF ALL EXTENSIONS OF FINITE GROUPSMuch is known about the extensions of finite groups, for example, about the extensions ofone simple group by another. However, as Solomon writes (2001, p. 347):. . . the classification of all finite groups is completely infeasible. Neverthelessexperience shows that most of the finite groups which occur in “nature”. . . are “close” either to simple groups or to groups such as dihedral groups,Heisenberg groups, etc., which arise naturally in the study of simple groups.As we noted earlier, by the year 2001, a complete irredundant list of finite groups wasavailable only for those up to an order of about 2000, and the number of groups on the listis overwhelming.NOTES The dream of classifying the finite simple groups goes back at least to Hölder 1892. Howevera clear strategy for accomplishing this did not begin to emerge until the 1950s, when work ofBrauer and others suggested that the key was to study the centralizers of elements of order 2 (theinvolution centralizers). For example, Brauer and Fowler (1955) showed that, for any finite groupH , the determination of the finite simple groups with an involution centralizer isomorphic to H is afinite problem. Later work showed that the problem is even tractable, and so the strategy became: (a)list the groups H that are candidates for being an involution centralizer in some finite simple group,and (b) for each H in (a) list the finite simple groups for which H occurs as an involution centralizer.Of course, this approach applies only to the finite simple groups containing an element of order 2,but an old conjecture said that, except for the cyclic groups of prime order, every finite simple grouphas even order and hence contains an element of order 2 by Cauchy’s theorem (4.13). With theproof of this conjecture by Feit and Thompson (1963), the effort to complete the classification of thefinite simple groups began in earnest. A complete classification was announced in 1982, but thereremained sceptics, because the proof depended on thousands of pages of rarely read journal articles,and, in fact, in reworking the proof, gaps were discovered. However, these have been closed, andwith the publication of Aschbacher and Smith 2004 it has become generally accepted that the proofof the classification is indeed complete.For a popular account of the history of the classification, see the book Ronan 2006, and for amore technical account, see the expository article Solomon 2001.Exercises3-1 Let G be the quaternion group (1.18). Prove that G can’t be written as a semidirectproduct in any nontrivial fashion.

52 3. AUTOMORPHISMS AND EXTENSIONS3-2 Let G be a group of order mn where m and n have no common factor. If G containsexactly one subgroup M of order m and exactly one subgroup N of order n, prove that Gis the direct product of M and N .3-3 Prove that GL 2 .F 2 / S 3 .3-4 Let G be the quaternion group (1.18). Prove that Aut.G/ S 4 .0a 01b3-5 Let G be the set of all matrices in GL 3 .R/ of the form @ 0 a c A, ad ¤ 0. Check0 0 dthat G is a subgroup of GL 3 .R/, and prove that it is a semidirect product of R 2 (additivegroup) by R R . Is it a direct product of these two groups?3-6 Find the automorphism groups of C 1 and S 3 .3-7 Let G D N ⋊ Q where N and Q are finite groups, and let g D nq be an element ofG with n 2 N and q 2 Q. Denote the order of an element x by o.x/:(a) Show that o.g/ D k o.q/ for some divisor k of jN j.(b) When Q acts trivially on N , show that o.g/ D lcm.o.n/;o.q//:(c) Let G D S 5 D A 5 ⋊ Q with Q D h.1;2/i. Let n D .1;4;3;2;5/ and let q D .1;2/.Show that o.g/ D 6, o.n/ D 5, and o.q/ D 2.(d) Suppose that G D .C p / p ⋊ Q where Q is cyclic of order p and that, for somegenerator q of Q,q.a 1 ;:::;a n /q 1 D .a n ;a 1 ;:::;a n 1 /:Show inductively that, for i p,..1;0;:::;0/;q/ i D ..1;:::;1;0;:::;0/;q i /(i copies of 1). Deduce that ..1;0;:::;0/;q/ has order p 2 (hence o.g/ D o.n/ o.q/ in thiscase).(e) Suppose that G D N ⋊ Q where N is commutative, Q is cyclic of order 2, and thegenerator q of Q acts on N by sending each element to its inverse. Show that .n;1/ hasorder 2 no matter what n is (in particular, o.g/ is independent of o.n/).3-8 Let G be the semidirect G D N ⋊ Q of its subgroups N and Q, and let(centralizer of Q in N ). Show thatC N .Q/ D fn 2 N j nq D qn for all q 2 QgZ.G/ D fn q j n 2 C N .Q/, q 2 Z.Q/, nn 0 n 1 D q 1 n 0 q for all n 0 2 N g:Let be the homomorphism Q ! Aut.N / giving the action of Q on N (by conjugation).Show that if N is commutative, thenand if N and Q are commutative, thenZ.G/ D fn q j n 2 C N .Q/, q 2 Z.Q/ \ Ker./g;Z.G/ D fn q j n 2 C N .Q/, q 2 Ker./g:

CHAPTER 4Groups Acting on SetsDefinition and examplesDEFINITION 4.1 Let X be a set and let G be a group. A left action of G on X is a mapping.g;x/ 7! gxWG X ! X such that(a) 1x D x, for all x 2 XI(b) .g 1 g 2 /x D g 1 .g 2 x/, all g 1 , g 2 2 G, x 2 X:A set together with a (left) action of G is called a (left) G-set. An action is trivial if gx D xfor all g 2 G.The conditions imply that, for each g 2 G, left translation by g,g L WX ! X;x 7! gx;has .g 1 / L as an inverse, and therefore g L is a bijection, i.e., g L 2 Sym.X/. Axiom (b)now says thatg 7! g L WG ! Sym.X/ (18)is a homomorphism. Thus, from a left action of G on X, we obtain a homomorphismG ! Sym.X/I conversely, every such homomorphism defines an action of G on X. Theaction is said to be faithful (or effective) if the homomorphism (18) is injective, i.e., ifgx D x for all x 2 X H) g D 1:EXAMPLE 4.2 (a) Every subgroup of the symmetric group S n acts faithfully on f1;2;:::;ng.(b) Every subgroup H of a group G acts faithfully on G by left translation,H G ! G;.h;x/ 7! hx:(c) Let H be a subgroup of G. The group G acts on the set of left cosets of H ,G G=H ! G=H;.g;C / 7! gC:The action is faithful if, for example, H ¤ G and G is simple.(d) Every group G acts on itself by conjugation,G G ! G;.g;x/ 7! g x defD gxg 1 :53

54 4. GROUPS ACTING ON SETSFor any normal subgroup N , G acts on N and G=N by conjugation.(e) For any group G, Aut.G/ acts on G:(f) The group of rigid motions of R n is the group of bijections R n ! R n preservinglengths. It acts on R n on the left.A right action X G ! G is defined similarly. To turn a right action into a left action,set g x D xg 1 . For example, there is a natural right action of G on the set of rightcosets of a subgroup H in G, namely, .C;g/ 7! Cg, which can be turned into a left action.g;C / 7! Cg 1 .A map of G-sets (alternatively, a G-map or a G-equivariant map) is a map 'WX ! Ysuch that'.gx/ D g'.x/; all g 2 G; x 2 X:An isomorphism of G-sets is a bijective G-map; its inverse is then also a G-map.ORBITSLet G act on X.A subset S X is said to be stable under the action of G ifg 2 G; x 2 S H) gx 2 S:The action of G on X then induces an action of G on S.Write x G y if y D gx, some g 2 G. This relation is reflexive because x D 1x,symmetric becausey D gx H) x D g 1 y(multiply by g 1 on the left and use the axioms), and transitive becausey D gx;z D g 0 y H) z D g 0 .gx/ D .g 0 g/x:It is therefore an equivalence relation. The equivalence classes are called G-orbits. Thusthe G-orbits partition X. Write GnX for the set of orbits.By definition, the G-orbit containing x 0 isGx 0 D fgx 0 j g 2 Gg:It is the smallest G-stable subset of X containing x 0 .EXAMPLE 4.3 (a) Suppose G acts on X, and let ˛ 2 G be an element of order n. Then theorbits of h˛i are the sets of the formfx 0 ;˛x 0 ;:::;˛n 1 x 0 g:(These elements need not be distinct, and so the set may contain fewer than n elements.)(b) The orbits for a subgroup H of G acting on G by left multiplication are the rightcosets of H in G. We write H nG for the set of right cosets. Similarly, the orbits for Hacting by right multiplication are the left cosets, and we write G=H for the set of left cosets.Note that the group law on G will not induce a group law on G=H unless H is normal.(c) For a group G acting on itself by conjugation, the orbits are called conjugacyclasses: for x 2 G, the conjugacy class of x is the setfgxg 1 j g 2 Gg

Definition and examples 55of conjugates of x. The conjugacy class of x 0 always contains x 0 , and it consists only of x 0if and only if x 0 is in the centre of G. In linear algebra the conjugacy classes in G D GL n .k/are called similarity classes, and the theory of rational canonical forms provides a set ofrepresentatives for the conjugacy classes: two matrices are similar (conjugate) if and onlyif they have the same rational canonical form.Note that a subset of X is stable if and only if it is a union of orbits. For example, asubgroup H of G is normal if and only if it is a union of conjugacy classes.The action of G on X is said to be transitive, and G is said to act transitively on X,if there is only one orbit, i.e., for any two elements x and y of X, there exists a g 2 Gsuch that gx D y. The set X is then called a homogeneous G-set. For example, S n actstransitively on f1;2;:::ng. For any subgroup H of a group G, G acts transitively on G=H ,but the action of G on itself is never transitive if G ¤ 1 because f1g is always a conjugacyclass.The action of G on X is doubly transitive if for any two pairs .x 1 ;x 2 /, .y 1 ;y 2 / ofelements of X with x 1 ¤ x 2 and y 1 ¤ y 2 , there exists a (single) g 2 G such that gx 1 D y 1and gx 2 D y 2 . Define k-fold transitivity for k 3 similarly.STABILIZERSLet G act on X. The stabilizer (or isotropy group) of an element x 2 X isStab.x/ D fg 2 G j gx D xg:It is a subgroup, but it need not be a normal subgroup (see the next lemma). The action isfree if Stab.x/ D feg for all x.LEMMA 4.4 For any g 2 G and x 2 X,PROOF. Certainly, if g 0 x D x, thenStab.gx/ D g Stab.x/ g 1 :.gg 0 g 1 /gx D gg 0 x D gx D y;and so g Stab.x/ g 1 Stab.gx/. Conversely, if g 0 .gx/ D gx, then.g 1 g 0 g/x D g 1 g 0 .gx/ D g 1 y D x;and so g 1 g 0 g 2 Stab.x/, i.e., g 0 2 g Stab.x/ g 1 .✷Clearly\Stab.x/ D Ker.G ! Sym.X//;x2Xwhich is a normal subgroup of G. The action is faithful if and only if T Stab.x/ D f1g.EXAMPLE 4.5 (a) Let G act on itself by conjugation. ThenStab.x/ D fg 2 G j gx D xgg:

56 4. GROUPS ACTING ON SETSThis group is called the centralizer C G .x/ of x in G. It consists of all elements of G thatcommute with, i.e., centralize, x. The intersection\C G .x/ D fg 2 G j gx D xg for all x 2 Ggx2Gis the centre of G.(b) Let G act on G=H by left multiplication. Then Stab.H / D H , and the stabilizer ofgH is gHg 1 :(c) Let G be the group of rigid motions of R n (4.2f). The stabilizer of the origin is theorthogonal group O n for the standard positive definite form on R n (Artin 1991, Chap. 4,5.16). Let T ' .R n ;C/ be the subgroup of G of translations of R n , i.e., maps of the formv 7! v C v 0 some v 0 2 R n . Then T is a normal subgroup of G and G ' T ⋊ O (cf. Artin1991, Chap. 5, 2).For a subset S of X, we define the stabilizer of S to beStab.S/ D fg 2 G j gS D Sg:Then Stab.S/ is a subgroup of G, and the same argument as in the proof of (4.4) shows thatStab.gS/ D g Stab.S/ g 1 :EXAMPLE 4.6 Let G act on G by conjugation, and let H be a subgroup of G. The stabilizerof H is called the normalizer N G .H / of H in G:N G .H / D fg 2 G j gHg 1 D H g:Clearly N G .H / is the largest subgroup of G containing H as a normal subgroup.It is possible for gS S but g 2 Stab.S/ (see 1.33).TRANSITIVE ACTIONSPROPOSITION 4.7 If G acts transitively on X, then for any x 0 2 X, the mapg Stab.x 0 / 7! gx 0 WG=Stab.x 0 / ! Xis an isomorphism of G-sets.PROOF. It is well-defined because, if h 2 Stab.x 0 /, then ghx 0 D gx 0 . It is injective becausegx 0 D g 0 x 0 H) g 1 g 0 x 0 D x 0 H) g;g 0 lie in the same left coset of Stab.x 0 /:It is surjective because G acts transitively. Finally, it is obviously G-equivariant.✷Thus every homogeneous G-set X is isomorphic to G=H for some subgroup H of G,but such a realization of X is not canonical: it depends on the choice of x 0 2 X: To saythis another way, the G-set G=H has a preferred point, namely, the coset H ; to give ahomogeneous G-set X together with a preferred point is essentially the same as to give asubgroup of G.

Definition and examples 57COROLLARY 4.8 Let G act on X, and let O D Gx 0 be the orbit containing x 0 . Then thecardinality of O isjOj D .G W Stab.x 0 //: (19)For example, the number of conjugates gHg 1 of a subgroup H of G is .GWN G .H //.PROOF. The action of G on O is transitive, and so g 7! gx 0 defines a bijection G=Stab.x 0 / !Gx 0 .✷The equation (19) is frequently useful for computing jOj.PROPOSITION 4.9 Let x 0 2 X. If G acts transitively on X, thenKer.G ! Sym.X//is the largest normal subgroup contained in Stab.x 0 /.PROOF. WhenKer.G ! Sym.X// D \x2XStab.x/ D \ g2GStab.gx 0 / (4.4)D \ g Stab.x 0 / g 1 :Hence, the proposition is a consequence of the following lemma.✷LEMMA 4.10 For any subgroup H of a group G, T g2G gHg 1 is the largest normalsubgroup contained in H .PROOF. Note that N 0defD T g2G gHg 1 , being an intersection of subgroups, is itself a subgroup.It is normal becauseg 1 N 0 g 11 D \ g2G.g 1 g/N 0 .g 1 g/ 1 D N 0— for the second equality, we used that, as g runs over the elements of G, so also doesg 1 g. Thus N 0 is a normal subgroup of G contained in eHe 1 D H . If N is a second suchgroup, thenN D gNg 1 gHg 1for all g 2 G, and soN \ g2GgHg 1 D N 0 :✷

58 4. GROUPS ACTING ON SETSTHE CLASS EQUATIONWhen X is finite, it is a disjoint union of a finite number of orbits:X Dm[iD1O i(disjoint union):Hence:PROPOSITION 4.11 The number of elements in X isjXj DmXjO i j DiD1mX.G W Stab.x i //; x i in O i : (20)iD1When G acts on itself by conjugation, this formula becomes:PROPOSITION 4.12 (CLASS EQUATION)jGj D X .G W C G .x// (21).x runs over a set of representatives for the conjugacy classes), orjGj D jZ.G/j C X .G W C G .y// (22).y runs over set of representatives for the conjugacy classes containing more than one element).THEOREM 4.13 (CAUCHY) If the prime p divides jGj, then G contains an element oforder p.PROOF. We use induction on jGj. If for some y not in the centre of G, p does not divide.G W C G .y//, then pjC G .y/ and we can apply induction to find an element of order pin C G .y/. Thus we may suppose that p divides all of the terms .G W C G .y// in the classequation (second form), and so also divides Z.G/. But Z.G/ is commutative, and it followsfrom the structure theorem 1 of such groups that Z.G/ will contain an element of order p. ✷COROLLARY 4.14 A finite group G is a p-group if and only if every element has order apower of p.PROOF. If jGj is a power of p, then Lagrange’s theorem (1.26) shows that the order ofevery element is a power of p. The converse follows from Cauchy’s theorem.✷COROLLARY 4.15Every group of order 2p, p an odd prime, is cyclic or dihedral.1 Here is a direct proof that the theorem holds for an abelian group Z. We use induction on the order of Z.It suffices to show that Z contains an element whose order is divisible by p; because then some power of theelement will have order exactly p. Let g ¤ 1 be an element of Z. If p doesn’t divide the order of g, then itdivides the order of Z=hgi, in which case there exists (by induction) an element of G whose order in Z=hgi isdivisible by p. But the order of such an element must itself be divisible by p.

Definition and examples 59PROOF. From Cauchy’s theorem, we know that such a G contains elements s and r of orders2 and p respectively. Let H D hri. Then H is of index 2, and so is normal. Obviouslys … H , and so G D H [ Hs WG D f1;r;:::;r p 1 ;s;rs;:::;r p 1 sg:As H is normal, srs 1 D r i , some i. Because s 2 D 1, r D s 2 rs 2 D s.srs 1 /s 1 D r i 2 ,and so i 2 1 mod p. Because Z=pZ is a field, its only elements with square 1 are ˙1, andso i 1 or 1 mod p. In the first case, the group is commutative (any group generated bya set of commuting elements is obviously commutative); in the second srs 1 D r 1 andwe have the dihedral group (2.9).✷p-GROUPSTHEOREM 4.16 Every nontrivial finite p-group has nontrivial centre.PROOF. By assumption, .G W 1/ is a power of p, and so .G W C G .y// is power of p (¤ p 0 )for all y not in the centre of G. As p divides every term in the class equation (22) except(perhaps) jZ.G/j, it must divide jZ.G/j also.✷COROLLARY 4.17 A group of order p n has normal subgroups of order p m for all m n.PROOF. We use induction on n. The centre of G contains an element g of order p, and soN D hgi is a normal subgroup of G of order p. Now the induction hypothesis allows us toassume the result for G=N; and the correspondence theorem (1.47) then gives it to us forG: ✷PROPOSITION 4.18 Every group of order p 2 is commutative, and hence is isomorphic toC p C p or C p 2.PROOF. We know that the centre Z is nontrivial, and that G=Z therefore has order 1 or p.In either case it is cyclic, and the next result implies that G is commutative.✷LEMMA 4.19 Suppose G contains a subgroup H in its centre (hence H is normal) suchthat G=H is cyclic. Then G is commutative.PROOF. Let a be an element of G whose image in G=H generates it. Then every elementof G can be written g D a i h with h 2 H , i 2 Z. Nowa i h a i 0 h 0 D a i a i 0 hh 0 because H Z.G/D a i 0 a i h 0 hD a i 0 h 0 a i h:✷REMARK 4.20 The above proof shows that if H Z.G/ and G contains a set of representativesfor G=H whose elements commute, then G is commutative.

60 4. GROUPS ACTING ON SETSFor p odd, it is now not difficult to show that any noncommutative group of order p 3is isomorphic to exactly one of the groups constructed in (3.14, 3.15) (Exercise 4-3). Thus,up to isomorphism, there are exactly two noncommutative groups of order p 3 .EXAMPLE 4.21 Let G be a noncommutative group of order 8. Then G must contain anelement a of order 4 (see Exercise 1-6). If G contains an element b of order 2 not inhai, then G ' hai ⋊ hbi where is the unique isomorphism Z=2Z ! .Z=4Z/ , and soG D 4 . If not, any element b of G not in hai must have order 4, and a 2 D b 2 . Nowbab 1 is an element of order 4 in hai. It can’t equal a, because otherwise G would becommutative, and so bab 1 D a 3 . Therefore G is the quaternion group (1.18, 2.7b).ACTION ON THE LEFT COSETSThe action of G on the set of left cosets G=H of H in G is a very useful tool in the studyof groups. We illustrate this with some examples.Let X D G=H . Recall that, for any g 2 G,and the kernel ofStab.gH / D g Stab.H /g 1 D gHg 1G ! Sym.X/is the largest normal subgroup T g2G gHg 1 of G contained in H .REMARK 4.22 (a) Let H be a subgroup of G not containing a normal subgroup of G otherthan 1. Then G ! Sym.G=H / is injective, and we have realized G as a subgroup of asymmetric group of order much smaller than .G W 1/Š. For example, if G is simple, then theSylow theorems (see Chapter 5) show that G has many proper subgroups H ¤ 1 (unless Gis cyclic), but (by definition) it has no such normal subgroup.(b) If .G W 1/ does not divide .G W H /Š, thenG ! Sym.G=H /can’t be injective (Lagrange’s theorem, 1.26), and we can conclude that H contains a normalsubgroup ¤ 1 of G. For example, if G has order 99, then it will have a subgroup N oforder 11 (Cauchy’s theorem, 4.13), and the subgroup must be normal. In fact, G D N Q.EXAMPLE 4.23 Corollary 4.15 shows that every group G of order 6 is either cyclic ordihedral. Here we present a slightly different argument. According to Cauchy’s theorem(4.13), G must contain an element r of order 3 and an element s of order 2. MoreoverN D defhri must be normal because 6 doesn’t divide 2Š (or simply because it has index 2).Let H D hsi. Either (a) H is normal in G, or (b) H is not normal in G. In the firstcase, rsr 1 D s, i.e., rs D sr, and so G ' hri hsi C 2 C 3 . In the second case,G ! Sym.G=H / is injective, hence surjective, and so G S 3 D 3 .Permutation groupsConsider Sym.X/ where X has n elements. Since (up to isomorphism) a symmetry groupSym.X/ depends only on the number of elements in X, we may take X D f1;2;:::;ng,

Permutation groups 61and so work with S n . The symbol1 2 3 4 5 6 72 5 7 4 3 1 6denotes the permutation sending 1 7! 2,2 7! 5, 3 7! 7, etc..Consider a permutation 1 2 3 ::: n D:.1/ .2/ .3/ ::: .n/The pairs .i;j / with i < j and .i/ > .j / are called the inversions of , and is saidto be even or odd according as the number its inversions is even or odd.. The signature,sign./, of is C1 or 1 according as is even or odd. For example, sign./ D 1 if isa transposition.REMARK 4.24 To compute the signature of , connect (by a line) each element i in the toprow to the element i in the bottom row, and count the number of times that the lines cross: is even or odd according as this number is even or odd. For example,1 2 3 4 53 5 1 4 2is even (6 intersections). This works, because there is one crossing for each inversion.For a permutation , consider the productsYV D .j i/ D .2 1/.3 1/.n 1/1i

62 4. GROUPS ACTING ON SETSLet p be the element of P defined byp.z 1 ;:::;z n / DThe same argument as above shows thatY1i

Permutation groups 63be the decomposition of f1;:::;ng into a disjoint union of orbits for hi, and let j be thecycle associated (as above) with O j . Then D 1 mis a decomposition of into a product of disjoint cycles. For the uniqueness, note thata decomposition D 1 m into a product of disjoint cycles must correspond to a decompositionof f1;:::;ng into orbits (ignoring cycles of length 1 and orbits with only oneelement). We can drop cycles of length one, change the order of the cycles, and change howwe write each cycle (by choosing different initial elements), but that’s all because the orbitsare intrinsically attached to :✷For example, 1 2 3 4 5 6 7 8D .15/.27634/.8/: (24)5 7 4 2 1 3 6 8It has order lcm.2;5/ D 10.COROLLARY 4.27 Each permutation can be written as a product of transpositions; thenumber of transpositions in such a product is even or odd according as is even or odd.PROOF. The cycle.i 1 i 2 :::i r / D .i 1 i 2 /.i r 2 i r 1 /.i r 1 i r /;and so the first statement follows from the proposition. Because sign is a homomorphism,and the signature of a transposition is 1, sign./ D . 1/ #transpositions . ✷Note that the formula in the proof shows that the signature of a cycle of length r is. 1/ r 1 , that is, an r-cycle is even or odd according as r is odd or even.It is possible to define a permutation to be even or odd according as it is a product of aneven or odd number of transpositions, but then one has to go through an argument as aboveto show that this is a well-defined notion.The corollary says that S n is generated by transpositions. For A n there is the followingresult.COROLLARY 4.28 The alternating group A n is generated by cycles of length three.PROOF. Any 2 A n is the product (possibly empty) of an even number of transpositions, D t 1 t1 0 t mtm 0 , but the product of two transpositions can always be written as a productof 3-cycles:8ˆ< .ij /.jl/ D .ijl/ case j D k;.ij /.kl/ D .ij /.jk/.jk/.kl/ D .ijk/.jkl/ case i;j;k;l distinct,ˆ:1 case .ij / D .kl/:✷Recall that two elements a and b of a group G are said to be conjugate a b if thereexists an element g 2 G such that b D gag 1 , and that conjugacy is an equivalence relation.For a group G, it is useful to determine the conjugacy classes in G.

64 4. GROUPS ACTING ON SETSEXAMPLE 4.29 In S n , the conjugate of a cycle is given by:g.i 1 :::i k /g 1 D .g.i 1 /:::g.i k //:Hence g.i 1 :::i r /.l 1 :::l u /g 1 D .g.i 1 /:::g.i r //.g.l 1 /:::g.l u // (even if the cyclesare not disjoint, because conjugation is a homomorphism). In other words, to obtain gg 1 ,replace each element in each cycle of by its image under g:We shall now determine the conjugacy classes in S n . By a partition of n, we mean asequence of integers n 1 ;:::;n k such that1 n 1 n 2 n k n andn 1 C n 2 C C n k D n:For example, there are exactly 5 partitions of 4, namely,4 D 1 C 1 C 1 C 1; 4 D 1 C 1 C 2; 4 D 1 C 3; 4 D 2 C 2; 4 D 4;and 1;121;505 partitions of 61. Note that a partitionf1;2;:::;ng D O 1 [ ::: [ O k(disjoint union)of f1;2;:::;ng determines a partition of n,n D n 1 C n 2 C ::: C n k ; n i D jO i j;provided the numbering has been chosen so that jO i j jO iC1 j. Since the orbits of anelement of S n form a partition of f1;:::;ng, we can attach to each such a partition ofn. For example, the partition of 8 attached to .15/.27634/.8/ is 1;2;5 and the partitionattached of n attached to D .i 1 :::i n1 /.l 1 :::l nk /; (disjoint cycles) 1 < n i n iC1 ;is 1;1;:::;1;n 1 ;:::;n k .n P n i ones/:PROPOSITION 4.30 Two elements and of S n are conjugate if and only if they definethe same partitions of n.PROOF. H) W We saw in (4.29) that conjugating an element preserves the type of its disjointcycle decomposition.(H W Since and define the same partitions of n, their decompositions into productsof disjoint cycles have the same type: D .i 1 :::i r /.j 1 :::j s /:::.l 1 :::l u /;If we define g to bethen D .i1 0 :::i0 r /.j 1 0 :::j s 0 /:::.l0 1 :::l0 u /: i1 i r j 1 j s l 1 l ui1 0 ir 0 j1 0 js 0 l1 0 lu0 ;gg 1 D :✷

Permutation groups 65EXAMPLE 4.31 .ijk/ D . 1234:::ijk4::: /.123/.1234::: ijk4::: / 1 :REMARK 4.32 For 1 < k n, there areneeded so that we don’t countn.n 1/.n kC1/kdistinct k-cycles in S n . The 1 k is.i 1 i 2 :::i k / D .i k i 1 :::i k 1 / D :::k times. Similarly, it is possible to compute the number of elements in any conjugacyclass in S n , but a little care is needed when the partition of n has several terms equal. Forexample, the number of permutations in S 4 of type .ab/.cd/ is12 4 32 2 12D 3:The 1 2 is needed so that we don’t count .ab/.cd/ D .cd/.ab/ twice. For S 4 we have thefollowing table:Partition Element No. in Conj. Class Parity1 C 1 C 1 C 1 1 1 even1 C 1 C 2 .ab/ 6 odd1 C 3 .abc/ 8 even2 C 2 .ab/.cd/ 3 even4 .abcd/ 6 oddNote that A 4 contains exactly 3 elements of order 2, namely those of type 2 C 2, and thattogether with 1 they form a subgroup V . This group is a union of conjugacy classes, and istherefore a normal subgroup of S 4 .THEOREM 4.33 (GALOIS) The group A n is simple if n 5REMARK 4.34 For n D 2, A n is trivial, and for n D 3, A n is cyclic of order 3, and hencesimple; for n D 4 it is nonabelian and nonsimple — it contains the normal, even characteristic,subgroup V (see 4.32).LEMMA 4.35 Let N be a normal subgroup of A n (n 5/; if N contains a cycle of lengththree, then it contains all cycles of length three, and so equals A n (by 4.28).PROOF. Let be the cycle of length three in N , and let be a second cycle of length threein A n . We know from (4.30) that D gg 1 for some g 2 S n . If g 2 A n , then this showsthat is also in N . If not, because n 5, there exists a transposition t 2 S n disjoint from. Then tg 2 A n and D tt1 D tgg 1 t1 ;and so again 2 N .The next lemma completes the proof of the Theorem.LEMMA 4.36 Every normal subgroup N of A n , n 5, N ¤ 1, contains a cycle of length3.✷

66 4. GROUPS ACTING ON SETSPROOF. Let 2 N , ¤ 1. If is not a 3-cycle, we shall construct another element 0 2 N , 0 ¤ 1, which fixes more elements of f1;2;:::;ng than does . If 0 is not a 3-cycle, thenwe can apply the same construction. After a finite number of steps, we arrive at a 3-cycle.Suppose is not a 3-cycle. When we express it as a product of disjoint cycles, either itcontains a cycle of length 3 or else it is a product of transpositions, say(i) D .i 1 i 2 i 3 :::/ or(ii) D .i 1 i 2 /.i 3 i 4 /.In the first case, moves two numbers, say i 4 , i 5 , other than i 1 , i 2 , i 3 , because ¤def.i 1 i 2 i 3 /, .i 1 :::i 4 /. Let D .i 3 i 4 i 5 /. Then 1 D 1 D .i 1 i 2 i 4 :::/ 2 N , and is distinctfrom (because it acts differently on i 2 ). Thus 0 D def 1 1 ¤ 1, but 0 D 1 1 fixesi 2 and all elements other than i 1 ;:::;i 5 fixed by — it therefore fixes more elements than.In the second case, form , 1 , 0 as in the first case with i 4 as in (ii) and i 5 anyelement distinct from i 1 ;i 2 ;i 3 ;i 4 . Then 1 D .i 1 i 2 /.i 4 i 5 / is distinct from becauseit acts differently on i 4 . Thus 0 D 1 1 ¤ 1, but 0 fixes i 1 and i 2 , and all elements¤ i 1 ;:::;i 5 not fixed by — it therefore fixes at least one more element than . ✷COROLLARY 4.37 For n 5, the only normal subgroups of S n are 1; A n , and S n .PROOF. If N is normal in S n , then N \A n is normal in A n . Therefore either N \A n D A nor N \ A n D f1g. In the first case, N A n , which has index 2 in S n , and so N D A n orS n . In the second case, the map x 7! xA n WN ! S n =A n is injective, and so N has order 1or 2, but it can’t have order 2 because no conjugacy class in S n (other than f1g) consists ofa single element.✷ASIDE 4.38 There exists a description of the conjugacy classes in A n , from which it is possible todeduce its simplicity for n 5 (see Exercise 4-11).ASIDE 4.39 A group G is said to be solvable if there exist subgroupsG D G 0 G 1 G i1 G i G r D f1gsuch that each G i is normal in G i 1 and each quotient G i 1 =G i is commutative. Thus A n (also S n /is not solvable if n 5. Let f .X/ 2 QŒX be of degree n.In Galois theory, one attaches to f a subgroup G f of the group of permutations of the rootsof f , and shows that the roots of f can be obtained from the coefficients of f by the algebraicoperations of addition, subtraction, multiplication, division, and the extraction of mth roots if andonly if G f is solvable (Galois’s theorem). For every n, there exist lots of polynomials f of degreen with G f S n , and hence (when n 5) lots of polynomials not solvable in radicals.The Todd-Coxeter algorithm.Let G be a group described by a finite presentation, and let H be a subgroup described by agenerating set. Then the Todd-Coxeter algorithm 4 is a strategy for writing down the set of4 To solve a problem, an algorithm must always terminate in a finite time with the correct answer to theproblem. The Todd-Coxeter algorithm does not solve the problem of determining whether a finite presentationdefines a finite group (in fact, there is no such algorithm). It does, however, solve the problem of determiningthe order of a finite group from a finite presentation of the group (use the algorithm with H the trivial subgroup1.)

The Todd-Coxeter algorithm. 67left cosets of H in G together with the action of G on the set. I illustrate it with an example(from Artin 1991, 6.9, which provides more details, but note that he composes permutationsin the reverse direction from us).Let G D ha;b;c j a 3 ;b 2 ;c 2 ;cbai and let H be the subgroup generated by c (strictlyspeaking, H is the subgroup generated by the element of G represented by the reducedword c). The operation of G on the set of cosets is described by the action of the generators,which must satisfy the following rules:(i) Each generator (a;b;c in our example) acts as a permutation.(ii) The relations (a 3 ;b 2 ;c 2 ;cba in our example) act trivially.(iii) The generators of H (c in our example) fix the coset 1H .(iv) The operation on the cosets is transitive.The strategy is to introduce cosets, denoted 1;2;::: with 1 D 1H , as necessary.Rule (iii) tells us simply that c1 D c. We now apply the first two rules. Since wedon’t know what a1 is, let’s denote it 2: a1 D 2. Similarly, let a2 D 3. Now a3 D a 3 1,which according to (ii) must be 1. Thus, we have introduced three (potential) cosets 1, 2, 3,permuted by a as follows:1 7! a2 7! a3 7! a1:What is b1? We don’t know, and so it is prudent to introduce another coset 4 D b1. Nowb4 D 1 because b 2 D 1, and so we have1 b7! 4 b7! 1:We still have the relation cba. We know a1 D 2, but we don’t know what b2 is, and so weset b2 D 5:1 7! a2 7! b5:By (iii) c1 D 1, and by (ii) applied to cba we have c5 D 1. Therefore, according to (i) wemust have 5 D 1; we drop 5, and so now b2 D 1. Since b4 D 1 we must have 4 D 2, and sowe can drop 4 also. What we know can be summarized by the table:a a a b b c c a b c1 2 3 1 2 1 1 1 2 1 12 3 1 2 1 2 2 3 23 1 2 3 3 3 1 2 3The bottom right corner, which is forced by (ii), tells us that c2 D 3. Hence also c3 D 2,and this then determines the rest of the table:a a a b b c c a b c1 2 3 1 2 1 1 1 2 1 12 3 1 2 1 2 3 2 3 3 23 1 2 3 3 3 2 3 1 2 3We find that we have three cosets on which a;b;c act asa D .123/ b D .12/ c D .23/:More precisely, we have written down a map G ! S 3 that is consistent with the aboverules. A theorem (Artin 1991, 9.10) now says that this does in fact describe the action of

68 4. GROUPS ACTING ON SETSG on G=H . Since the three elements .123/, .12/, and .23/ generate S 3 , this shows that theaction of G on G=H induces an isomorphism G ! S 3 , and that H is a subgroup of order2.In Artin 1991, 6.9, it is explained how to make this procedure into an algorithm which,when it succeeds in producing a consistent table, will in fact produce the correct table.This algorithm is implemented in Gap — see the worksheet.Primitive actions.Let G be a group acting on a set X, and let be a partition of X. We say that is stabilizedby G ifA 2 H) gA 2 :It suffices to check the condition for a set of generators for G.EXAMPLE 4.40 (a) The subgroup G D h.1234/i of S 4 stabilizes the partition ff1;3g;f2;4ggof f1;2;3;4g.(b) Identify X D f1;2;3;4g with the set of vertices of the square on which D 4 actsin the usual way, namely, with r D .1234/, s D .2;4/. Then D 4 stabilizes the partitionff1;3g;f2;4gg (opposite vertices stay opposite).(c) Let X be the set of partitions of f1;2;3;4g into two sets, each with two elements.Then S 4 acts on X, and Ker.S 4 ! Sym.X// is the subgroup V defined in (4.32).The group G always stabilizes the trivial partitions of X, namely, the set of all oneelementsubsets of X, and fXg. When it stabilizes only those partitions, we say that theaction is primitive; otherwise it is imprimitive. A subgroup of Sym.X/ (e.g., of S n ) is saidto be primitive if it acts primitively on X. Obviously, S n itself is primitive, but Example4.40b shows that D 4 , regarded as a subgroup of S 4 in the obvious way, is not primitive.EXAMPLE 4.41 A doubly transitive action is primitive: if it stabilizedffx;x 0 ;:::g;fy;:::g:::g,then there would be no element sending .x;x 0 / to .x;y/.REMARK 4.42 The G-orbits form a partition of X that is stabilized by G. If the action isprimitive, then the partition into orbits must be one of the trivial ones. Henceaction primitive H) action transitive or trivial.For the remainder of this section, G is a finite group acting transitively on a set X with atleast two elements.PROPOSITION 4.43 The group G acts imprimitively if and only if there is a proper subsetA of X with at least 2 elements such that,for each g 2 G, either gA D A or gA \ A D ;: (25)PROOF. H): The partition stabilized by G contains such an A.(H: From such an A, we can form a partition fA;g 1 A;g 2 A;:::g of X, which is stabilizedby G.✷

Exercises 69A subset A of X satisfying (25) is called block.PROPOSITION 4.44 Let A be a block in X with jAj 2 and A ¤ X. For any x 2 A,Stab.x/ Stab.A/ G:PROOF. We have Stab.A/ Stab.x/ becausegx D x H) gA \ A ¤ ; H) gA D A:Let y 2 A, y ¤ x. Because G acts transitively on X, there is a g 2 G such that gx D y.Then g 2 Stab.A/, but g … Stab.x/:Let y … A. There is a g 2 G such that gx D y, and then g … Stab.A/:✷THEOREM 4.45 The group G acts primitively on X if and only if, for one (hence all) x inX, Stab.x/ is a maximal subgroup of G.PROOF. If G does not act primitively on X, then (see 4.43) there is a block A X with atleast two elements, and so (4.44) shows that Stab.x/ will not be maximal for any x 2 A.Conversely, suppose that there exists an x in X and a subgroup H such thatStab.x/ H G.Then I claim that A D H x is a block ¤ X with at least two elements.Because H ¤ Stab.x/, H x ¤ fxg, and so fxg A X.If g 2 H , then gA D A. If g … H , then gA is disjoint from A: for suppose ghx D h 0 xsome h 0 2 H ; then h 0 1 gh 2 Stab.x/ H , say h 0 1 gh D h 00 , and g D h 0 h 00 h 1 2 H . ✷Exercises4-1 Let H 1 and H 2 be subgroups of a group G. Show that the maps of G-sets G=H 1 !G=H 2 are in natural one-to-one correspondence with the elements gH 2 of G=H 2 such thatH 1 gH 2 g 1 .4-2 (a) Show that a finite group can’t be equal to the union of the conjugates of a propersubgroup.(b) Give an example of a proper subset S of a finite group G such that G D S g2G gSg 1 .4-3 Prove that any noncommutative group of order p 3 , p an odd prime, is isomorphic toone of the two groups constructed in (3.14, 3.15).4-4 Let p be the smallest prime dividing .G W 1/ (assumed finite). Show that any subgroupof G of index p is normal.4-5 Show that a group of order 2m, m odd, contains a subgroup of index 2. (Hint: UseCayley’s theorem 1.22)

70 4. GROUPS ACTING ON SETS4-6 For n 5, show that the k-cycles in S n generate S n or A n according as k is even orodd.4-7 Let G D GL 3 .F 2 /.(a) Show that .G W 1/ D 168.(b) Let X be the set of lines through the origin in F 3 2; show that X has 7 elements, andthat there is a natural injective homomorphism G ,! Sym.X/ D S 7 .(c) Use Jordan canonical forms to show that G has six conjugacy classes, with 1, 21, 42,56, 24, and 24 elements respectively. [Note that if M is a free F 2 Œ˛-module of rankone, then End F2 Œ˛.M / D F 2 Œ˛.](d) Deduce that G is simple.4-8 Let G be a group. If Aut.G/ is cyclic, prove that G is commutative; if further, G isfinite, prove that G is cyclic.4-9 Show that S n is generated by .12/;.13/;:::;.1n/; also by .12/;.23/;:::;.n 1n/.4-10 Let K be a conjugacy class of a finite group G contained in a normal subgroup Hof G. Prove that K is a union of k conjugacy classes of equal size in H , where k D .G WH C G .x// for any x 2 K.4-11 (a) Let 2 A n . From Exercise 4-10 we know that the conjugacy class of in S neither remains a single conjugacy class in A n or breaks up as a union of two classes of equalsize. Show that the second case occurs ” does not commute with an odd permutation” the partition of n defined by consists of distinct odd integers.(b) For each conjugacy class K in A 7 , give a member of K, and determine jKj.4-12 Let G be the group with generators a;b and relations a 4 D 1 D b 2 , aba D bab.(a) Use the Todd-Coxeter algorithm (with H D 1) to find the image of G under thehomomorphism G ! S n , n D .G W 1/, given by Cayley’s Theorem 1.11. [No need toinclude every step; just an outline will do.](b) Use Sage/Gap to check your answer.4-13 Show that if the action of G on X is primitive and effective, then the action of anynormal subgroup H ¤ 1 of G is transitive.4-14 (a) Check that A 4 has 8 elements of order 3, and 3 elements of order 2. Hence it hasno element of order 6.(b) Prove that A 4 has no subgroup of order 6 (cf. 1.30). (Use 4.23.)(c) Prove that A 4 is the only subgroup of S 4 of order 12.4-15 Let G be a group with a subgroup of index r. Prove:(a) If G is simple, then .G W 1/ divides rŠ.(b) If r D 2;3; or 4, then G can’t be simple.(c) There exists a nonabelian simple group with a subgroup of index 5.

Exercises 714-16 Prove that S n is isomorphic to a subgroup of A nC2 .4-17 Let H and K be subgroups of a group G. A double coset of H and K in G is a setof the formHaK D fhak j h 2 H , k 2 Kgfor some a 2 G.(a) Show that the double cosets of H and K in G partition G.(b) Let H \ aKa 1 act on H K by b.h;k/ D .hb;a 1 b 1 ak/. Show that the orbitsfor this action are exactly the fibres of the map .h;k/ 7! hakWH K ! HaK.(c) (Double coset counting formula). Use (b) to show thatjHaKj DjH jjKjjH \ aKa 1 j :

CHAPTER 5The Sylow Theorems; ApplicationsIn this chapter, all groups are finite.Let G be a group and let p be a prime dividing .GW1/. A subgroup of G is called aSylow p-subgroup of G if its order is the highest power of p dividing .G W 1/. In otherwords, H is a Sylow p-subgroup of G if it is a p-group and its index in G is prime to p.The Sylow theorems state that there exist Sylow p-subgroups for all primes p dividing.GW1/, that the Sylow p-subgroups for a fixed p are conjugate, and that every p-subgroupof G is contained in such a subgroup; moreover, the theorems restrict the possible numberof Sylow p-subgroups in G.The Sylow theoremsIn the proofs, we frequently use that if O is an orbit for a group H acting on a set X, andx 0 2 O, then the map H ! X, h 7! hx 0 induces a bijectionsee (4.7). ThereforeH=Stab.x 0 / ! OI.H W Stab.x 0 // D jOj:In particular, when H is a p-group, jOj is a power of p: either O consists of a singleelement, or jOj is divisible by p. Since X is a disjoint union of the orbits, we can conclude:LEMMA 5.1 Let H be a p-group acting on a finite set X, and let X H be the set of pointsfixed by H ; thenjXj jX H j .mod p/:When the lemma is applied to a p-group H acting on itself by conjugation, we find that.Z.H / W 1/ .H W 1/mod pand so pj.Z.H /W1/ (cf. the proof of 4.16).THEOREM 5.2 (SYLOW I) Let G be a finite group, and let p be prime. If p r j.G W 1/, thenG has a subgroup of order p r :73

74 5. THE SYLOW THEOREMS; APPLICATIONSPROOF. According to (4.17), it suffices to prove this with p r the highest power of p dividing.G W 1/, and so from now on we assume that .G W 1/ D p r m with m not divisible by p.LetX D fsubsets of G with p r elementsg;with the action of G defined byG X ! X;.g;A/ 7! gA defD fga j a 2 Ag:Let A 2 X, and letH D Stab.A/ defD fg 2 G j gA D Ag:For any a 0 2 A, h 7! ha 0 WH ! A is injective (cancellation law), and so .H W 1/ jAj D p r .In the equation.G W 1/ D .G W H /.H W 1/we know that .G W 1/ D p r m, .H W 1/ p r , and that .G W H / is the number of elements inthe orbit of A. If we can find an A such that p doesn’t divide the number of elements in itsorbit, then we can conclude that (for such an A), H D StabA has order p r .The number of elements in X isjXj D p r mp rD .pr m/.p r m 1/.p r m i/.p r m p r C 1/p r .p r 1/.p r i/.p r p r :C 1/Note that, because i < p r , the power of p dividing p r m i is the power of p dividingi. The same is true for p r i. Therefore the corresponding terms on top and bottom aredivisible by the same powers of p, and so p does not divide jXj. Because the orbits form apartition of X,jXj D X jO i j; O i the distinct orbits;and so at least one of the jO i j is not divisible by p.✷EXAMPLE 5.3 Let F p D Z=pZ, the field with p elements, and let G D GL n .F p /. Then n matrices in G are precisely those whose columns form a basis for F n p . Thus, the firstcolumn can be any nonzero vector in F n p , of which there are pn 1; the second columncan be any vector not in the span of the first column, of which there are p n p; and so on.Therefore, the order of G is.p n 1/.p n p/.p n p 2 /.p n p n 1 /;and so the power of p dividing .G W 1/ is p 1C2CC.n 1/ . Consider the upper triangularmatrices with 1’s down the diagonal:011 0 1 0 0 1 :BC@:: : : A0 0 0 1They form a subgroup U of order p n 1 p n 2 p, which is therefore a Sylow p-subgroupG.

The Sylow theorems 75REMARK 5.4 The theorem gives another proof of Cauchy’s theorem (4.13). If a prime pdivides .GW1/, then G will have a subgroup H of order p, and any g 2 H , g ¤ 1, is anelement of G of order p.REMARK 5.5 The proof of Theorem 5.2 can be modified to show directly that for eachpower p r of p dividing .G W 1/ there is a subgroup H of G of order p r . One again writes.G W 1/ D p r m and considers the set X of all subsets of order p r . In this case, the highestpower p r 0of p dividing jXj is the highest power of p dividing m, and it follows that thereis an orbit in X whose order is not divisible by p r 0C1 . For an A in such an orbit, the samecounting argument shows that Stab.A/ has p r elements. We recommend that the readerwrite out the details.THEOREM 5.6 (SYLOW II) Let G be a finite group, and let jGj D p r m with m not divisibleby p.(a) Any two Sylow p-subgroups are conjugate.(b) Let s p be the number of Sylow p-subgroups in G; then s p 1 mod p and s p jm.(c) Every p-subgroup of G is contained in a Sylow p-subgroup.Let H be a subgroup of G. Recall (4.6, 4.8) that the normalizer of H in G isN G .H / D fg 2 G j gHg 1 D H g;and that the number of conjugates of H in G is .G W N G .H //.LEMMA 5.7 Let P be a Sylow p-subgroup of G, and let H be a p-subgroup. If H normalizesP , i.e., if H N G .P /, then H P . In particular, no Sylow p-subgroup of Gother than P normalizes P .PROOF. Because H and P are subgroups of N G .P / with P normal in N G .P /, HP is asubgroup, and H=H \P ' HP=P (apply 1.46). Therefore .HP W P / is a power of p (hereis where we use that H is a p-group), but.HP W 1/ D .HP W P /.P W 1/;and .P W 1/ is the largest power of p dividing .G W 1/, hence also the largest power of pdividing .HP W 1/. Thus .HP W P / D p 0 D 1, and H P .✷PROOF (OF SYLOW II) (a) Let X be the set of Sylow p-subgroups in G, and let G act onX by conjugation,.g;P / 7! gP g 1 WG X ! X:Let O be one of the G-orbits: we have to show O is all of X.Let P 2 O, and let P act on O through the action of G. This single G-orbit may breakup into several P -orbits, one of which will be fP g. In fact this is the only one-point orbitbecausefQg is a P -orbit ” P normalizes Q;which we know (5.7) happens only for Q D P . Hence the number of elements in everyP -orbit other than fP g is divisible by p, and we have that jOj 1 mod p.

76 5. THE SYLOW THEOREMS; APPLICATIONSSuppose there exists a P … O. We again let P act on O, but this time the argumentshows that there are no one-point orbits, and so the number of elements in every P -orbit isdivisible by p. This implies that #O is divisible by p, which contradicts what we proved inthe last paragraph. There can be no such P , and so O is all of X.(b) Since s p is now the number of elements in O, we have also shown that s p 1 (modp/.Let P be a Sylow p-subgroup of G. According to (a), s p is the number of conjugatesof P , which equals.G W N G .P // D.G W 1/.N G .P / W 1/ D .G W 1/.N G .P / W P / .P W 1/ D m.N G .P / W P / :This is a factor of m.(c) Let H be a p-subgroup of G, and let H act on the set X of Sylow p-subgroups byconjugation. Because jXj D s p is not divisible by p, X H must be nonempty (Lemma 5.1),i.e., at least one H -orbit consists of a single Sylow p-subgroup. But then H normalizes Pand Lemma 5.7 implies that H P .✷COROLLARY 5.8 A Sylow p-subgroup is normal if and only if it is the only Sylow p-subgroup.PROOF. Let P be a Sylow p-subgroup of G. If P is normal, then (a) of Sylow II impliesthat it is the only Sylow p-subgroup. The converse statement follows from (3.7c) (whichshows, in fact, that P is even characteristic).✷COROLLARY 5.9 Suppose that a group G has only one Sylow p-subgroup for each primep dividing its order. Then G is a direct product of its Sylow p-subgroups.PROOF. Let P 1 ;:::;P k be Sylow subgroups of G, and let jP i j D p r ii; the p i are distinctprimes. Because each P i is normal in G, the product P 1 P k is a normal subgroup of G.We shall prove by induction on k that it has order p r 11 pr k. If k D 1, there is nothing tokprove, and so we may suppose that k 2 and that P 1 P k 1 has order p r 11 pr k 1k 1 . ThenP 1 P k 1 \ P k D 1; therefore (1.51) shows that .P 1 P k 1 /P k is the direct product ofP 1 P k 1 and P k , and so has order p r 11 pr k. Now (1.52) applied to the full set of Sylowksubgroups of G shows that G is their direct product.EXAMPLE 5.10 Let G D GL.V / where V is a vector space of dimension n over F p . Thereis a geometric description of the Sylow subgroups of G. A full flag F in V is a sequenceof subspacesV D V n V n 1 V i V 1 f0gwith dimV i D i. Given such a flag F , let U.F / be the set of linear maps ˛WV ! V suchthat(a) ˛.V i / V i for all i, and(b) the endomorphism of V i =V i1 induced by ˛ is the identity map.✷

Alternative approach to the Sylow theorems 77I claim that U.F / is a Sylow p-subgroup of G. Indeed, we can construct a basis fe 1 ;:::;e n gfor V such fe 1 g is basis for V 1 , fe 1 ;e 2 g is a basis for V 2 , and so on. Relative to this basis,the matrices of the elements of U.F / are exactly the elements of the group U of (5.3).Let g 2 GL n .F/. Then gF D deffgV n ;gV n 1 ;:::g is again a full flag, and U.gF / Dg U.F / g 1 . From (a) of Sylow II, we see that the Sylow p-subgroups of G are preciselythe groups of the form U.F / for some full flag F .ASIDE 5.11 Some books use different numberings for Sylow’s theorems. I have essentially followedthe original (Sylow 1872).Alternative approach to the Sylow theoremsWe briefly forget that we have proved the Sylow theorems.THEOREM 5.12 Let G be a group, and let P be a Sylow p-subgroup of G. For any subgroupH of G, there exists an a 2 G such that H \ aP a 1 is a Sylow p-subgroup of H .PROOF. Recall (Exercise 4-17) that G is a disjoint union of the double cosets for H andP , and sojGj D X jHaP j D X jH jjP ja a jH \ aP a 1 jwhere the sum is over a set of representatives for the double cosets. On dividing by jP j wefind thatjGjjP j D X jH ja jH \ aP a 1 j ;and so there exists an a such that .H WH \ aP a 1 / is not divisible by p. For such an a,H \ aP a 1 is a Sylow p-subgroup of H .✷PROOF (OF SYLOW I) According to Cayley’s theorem (1.22), G embeds into S n , and S nembeds into GL n .F p / (see 7.1b below). As GL n .F p / has a Sylow p-subgroup (see 5.3), soalso does G.✷PROOF (OF SYLOW II(a,c)) Let P be a Sylow p-subgroup of G, and let P 0 be a p-subgroupof G. Then P 0 is the unique Sylow p-subgroup of P 0 , and so the theorem with H D P 0shows that aP a 1 P 0 for some a. This implies (a) and (c) of Sylow II.✷ExamplesWe apply what we have learnt to obtain information about groups of various orders.5.13 (GROUPS OF ORDER 99) Let G have order 99. The Sylow theorems imply that Ghas at least one subgroup H of order 11, and in fact s 11ˇˇ 9911 and s 11 1 mod 11. Itfollows that s 11 D 1, and H is normal. Similarly, s 9 j11 and s 9 1 mod 3, and so theSylow 3-subgroup is also normal. Hence G is isomorphic to the direct product of its Sylowsubgroups (5.9), which are both commutative (4.18), and so G commutative.Here is an alternative proof. Verify as before that the Sylow 11-subgroup N of G isnormal. The Sylow 3-subgroup Q maps bijectively onto G=N , and so G D N ⋊ Q. It

78 5. THE SYLOW THEOREMS; APPLICATIONSremains to determine the action by conjugation of Q on N . But Aut.N / is cyclic of order10 (see 3.5), and so there is only the trivial homomorphism Q ! Aut.N /. It follows thatG is the direct product of N and Q.5.14 (GROUPS OF ORDER pq, p;q PRIMES, p < q) Let G be such a group, and let P andQ be Sylow p and q subgroups. Then .G W Q/ D p, which is the smallest prime dividing.G W 1/, and so (see Exercise 4-4) Q is normal. Because P maps bijectively onto G=Q, wehave thatG D Q ⋊ P;and it remains to determine the action of P on Q by conjugation.The group Aut.Q/ is cyclic of order q 1 (see 3.5), and so, unless pjq 1, G D Q P .If pjq 1, then Aut.Q/ (being cyclic) has a unique subgroup P 0 of order p. In fact P 0consists of the mapsx 7! x i ; fi 2 Z=qZ j i p D 1g:Let a and b be generators for P and Q respectively, and suppose that the action of a on Qby conjugation is x 7! x i 0; i 0 ¤ 1 (in Z=qZ). Then G has generators a;b and relationsa p ; b q ; aba 1 D b i 0:Choosing a different i 0 amounts to choosing a different generator a for P , and so gives anisomorphic group G.In summary: if p ∤ q 1, then the only group of order pq is the cyclic group C pq ; ifpjq 1, then there is also a nonabelian group given by the above generators and relations.5.15 (GROUPS OF ORDER 30) Let G be a group of order 30. Thens 3 D 1;4;7;10;::: and divides 10Is 5 D 1;6;11;::: and divides 6:Hence s 3 D 1 or 10, and s 5 D 1 or 6. In fact, at least one is 1, for otherwise there would be20 elements of order 3 and 24 elements of order 5, which is impossible. Therefore, a Sylow3-subgroup P or a Sylow 5-subgroup Q is normal, and so H D PQ is a subgroup of G.Because 3 doesn’t divide 5 1 D 4, (5.14) shows that H is commutative, H C 3 C 5 .HenceG D .C 3 C 5 / ⋊ C 2 ;and it remains to determine the possible homomorphisms WC 2 ! Aut.C 3 C 5 /. But sucha homomorphism is determined by the image of the nonidentity element of C 2 , whichmust be an element of order 2. Let a, b, c generate C 3 , C 5 , C 2 . ThenAut.C 3 C 5 / D Aut.C 3 / Aut.C 5 /;and the only elements of AutC 3 and AutC 5 of order 2 are a 7! a 1 and b 7! b 1 . Thusthere are exactly 4 homomorphisms , and .c/ is one of the following elements: a 7! ab 7! b a 7! a a 7! a1b 7! b 1 b 7! b a 7! a1b 7! b 1 :The groups corresponding to these homomorphisms have centres of order 30, 3 (generatedby a), 5 (generated by b), and 1 respectively, and hence are nonisomorphic. We have shown

Examples 79that (up to isomorphism) there are exactly 4 groups of order 30. For example, the third onour list has generators a;b;c and relationsa 3 ; b 5 ; c 2 ; ab D ba; cac 1 D a 1 ; cbc 1 D b:5.16 (GROUPS OF ORDER 12) Let G be a group of order 12, and let P be its Sylow 3-subgroup. If P is not normal, then P doesn’t contain a nontrivial normal subgroup of G,and so the map (4.2, action on the left cosets)' W G ! Sym.G=P / S 4is injective, and its image is a subgroup of S 4 of order 12. From Sylow II we see that Ghas exactly 4 Sylow 3-subgroups, and hence it has exactly 8 elements of order 3. But allelements of S 4 of order 3 are in A 4 (see the table in 4.32), and so '.G/ intersects A 4 in asubgroup with at least 8 elements. By Lagrange’s theorem '.G/ D A 4 , and so G A 4 .Now assume that P is normal. Then G D P ⋊ Q where Q is the Sylow 4-subgroup. IfQ is cyclic of order 4, then there is a unique nontrivial map Q.D C 4 / ! Aut.P /.D C 2 /,and hence we obtain a single noncommutative group C 3 ⋊ C 4 . If Q D C 2 C 2 , there areexactly 3 nontrivial homomorphism WQ ! Aut.P /, but the three groups resulting are allisomorphic to S 3 C 2 with C 2 D Ker. (The homomorphisms differ by an automorphismof Q, and so we can also apply Lemma 3.18.)In total, there are 3 noncommutative groups of order 12 and 2 commutative groups.5.17 (GROUPS OF ORDER p 3 ) Let G be a group of order p 3 , with p an odd prime, andassume G is not commutative. We know from (4.17) that G has a normal subgroup N oforder p 2 .If every element of G has order p (except 1), then N C p C p and there is a subgroupQ of G of order p such that Q \ N D f1g. HenceG D N ⋊ Qfor some homomorphism WQ ! N . The order of Aut.N / GL 2 .F p / is .p 2 1/.p 2 p/(see 5.3), and so its Sylow p-subgroups have order p. By the Sylow theorems, they areconjugate, and so Lemma 3.19 shows that there is exactly one nonabelian group in thiscase.Suppose G has elements of order p 2 , and let N be the subgroup generated by such anelement a. Because .G W N / D p is the smallest (in fact only) prime dividing .G W 1/, N isnormal in G (Exercise 4-4). We next show that G contains an element of order p not in N .We know Z.G/ ¤ 1, and, because G isn’t commutative, that G=Z.G/ is not cyclic(4.19). Therefore .Z.G/ W 1/ D p and G=Z.G/ C p C p . In particular, we see that forall x 2 G, x p 2 Z.G/. Because G=Z.G/ is commutative, the commutator of any pair ofelements of G lies in Z.G/, and an easy induction argument shows that.xy/ n D x n y n n.n 1/Œy;x 2 ; n 1:Therefore .xy/ p D x p y p , and so x 7! x p WG ! G is a homomorphism. Its image is containedin Z.G/, and so its kernel has order at least p 2 . Since N contains only p 1 elementsof order p, we see that there exists an element b of order p outside N . Hence

80 5. THE SYLOW THEOREMS; APPLICATIONSG D hai ⋊ hbi C p 2 ⋊ C p , and it remains to observe (3.19) that the nontrivial homomorphismsC p ! Aut.C p 2/ C p C p 1 give isomorphic groups.Thus, up to isomorphism, the only noncommutative groups of order p 3 are those constructedin (3.14, 3.15).5.18 (GROUPS OF ORDER 2p n , 4p n ; AND 8p n , p ODD) Let G be a group of order 2 m p n ,1 m 3, p an odd prime, 1 n. We shall show that G is not simple. Let P be a Sylowp-subgroup and let N D N G .P /, so that s p D .G W N /.From Sylow II, we know that s p j2 m , s p D 1;p C 1;2p C 1;:::. If s p D 1, P is normal.If not, there are two cases to consider:(i) s p D 4 and p D 3, or(ii) s p D 8 and p D 7:In the first case, the action by conjugation of G on the set of Sylow 3-subgroups 1 definesa homomorphism G ! S 4 , which, if G is simple, must be injective. Therefore .G W 1/j4Š,and so n D 1; we have .G W 1/ D 2 m 3. Now the Sylow 2-subgroup has index 3, and so wehave a homomorphism G ! S 3 . Its kernel is a nontrivial normal subgroup of G.In the second case, the same argument shows that .G W 1/j8Š, and so n D 1 again. Thus.G W 1/ D 56 and s 7 D 8. Therefore G has 48 elements of order 7, and so there can be onlyone Sylow 2-subgroup, which must therefore be normal.Note that groups of order pq r , p;q primes, p < q are not simple, because Exercise 4-4shows that the Sylow q-subgroup is normal. An examination of cases now reveals that A 5is the smallest noncyclic simple group.5.19 (GROUPS OF ORDER 60) Let G be a simple group of order 60. We shall show that Gis isomorphic to A 5 . Let P be a Sylow 2-subgroup and N D N G .P /, so that s 2 D .G W N /.According to the Sylow theorems, s 2 D 1;3;5; or 15:(a) The case s 2 D 1 is impossible, because P would be normal (see 5.8).(b) The case s 2 D 3 is impossible, because the kernel of G ! Sym.G=N / would be anontrivial normal subgroup of G.(c) In the case s 2 D 5, we get an inclusion G ,! Sym.G=N / D S 5 , which realizes G asa subgroup of index 2 in S 5 , but we saw in (4.37) that, for n 5, A n is the only subgroupof index 2 in S n .(d) In the case s 2 D 15, a counting argument (using that s 5 D 6) shows that there existtwo Sylow 2-subgroups P and Q intersecting in a group of order 2. The normalizer Nof P \ Q contains P and Q, and so it has index 1, 3, or 5 in G. The first two cases areimpossible for the same reasons as in (a) and (b). If .GWN / D 5, the argument in (c) givesan isomorphism G A 5 ; but this is impossible because s 2 .A 5 / D 5.Exercises5-1 Show that a finite group (not necessarily commutative) is cyclic if, for each n > 0, itcontains at most n elements of order dividing n.1 Equivalently, the usual map G ! Sym.G=N /.

CHAPTER 6Subnormal Series; Solvable andNilpotent GroupsSubnormal Series.Let G be a group. A chain of subgroupsG D G 0 G 1 G i G iC1 G n D f1g:is called a subnormal series if G i is normal in G i 1 for every i, and it is called a normalseries if G i is normal in G for every i. 1 The series is said to be without repetitions if all theinclusions G i 1 G i are proper (i.e., G i 1 ¤ G i ). Then n is called the length of the series.The quotient groups G i 1 =G i are called the quotient (or factor) groups of the series.A subnormal series is said to be a composition series if it has no proper refinement thatis also a subnormal series. In other words, it is a composition series if G i is maximal amongthe proper normal subgroups G i 1 for each i. Thus a subnormal series is a compositionseries if and only if each quotient group is simple and nontrivial. Obviously, every finitegroup has a composition series (usually many): choose G 1 to be a maximal proper normalsubgroup of G; then choose G 2 to be a maximal proper normal subgroup of G 1 , etc.. Aninfinite group may or may not have a finite composition series.Note that from a subnormal serieswe obtain a sequence of exact sequencesG D G 0 F G 1 F F G i F G iC1 F F G n D f1g1 ! G n 1 ! G n 2 ! G n 2 =G n 1 ! 11 ! G iC1 ! G i ! G i =G iC1 ! 11 ! G 1 ! G 0 ! G 0 =G 1 ! 1:Thus G is built up out of the quotients G 0 =G 1 ;G 1 =G 2 ;:::;G n 1 by forming successive extensions.In particular, since every finite group has a composition series, it can be regarded1 Some authors write “normal series” where we write “subnormal series” and “invariant series” where wewrite “normal series”.81

82 6. SUBNORMAL SERIES; SOLVABLE AND NILPOTENT GROUPSas being built up out of simple groups. The Jordan-Hölder theorem, which is the main topicof this section, says that these simple groups are independent of the composition series (upto order and isomorphism).Note that if G has a subnormal series G D G 0 F G 1 F F G n D f1g, then.G W 1/ D Y 1in .G i 1 W G i / D Y 1in .G i 1=G i W 1/:EXAMPLE 6.1 (a) The symmetric group S 3 has a composition seriesS 3 F A 3 F 1with quotients C 2 , C 3 :(b) The symmetric group S 4 has a composition seriesS 4 F A 4 F V F h.13/.24/i F 1;where V C 2 C 2 consists of all elements of order 2 in A 4 (see 4.32). The quotients areC 2 , C 3 , C 2 , C 2 .(c) Any full flag in F n p , p a prime, is a composition series. Its length is n, and itsquotients are C p ;C p ;:::;C p :(d) Consider the cyclic group C m D hai. For any factorization m D p 1 p r of m intoa product of primes (not necessarily distinct), there is a composition seriesC m F C m F C mp 1 p 1 p 2F k k khai ha p 1i ha p 1p 2iThe length is r, and the quotients are C p1 ;C p2 ;:::;C pr .(e) Suppose G is a direct product of simple groups, G D H 1 H r . Then G has acomposition seriesG F H 2 H r F H 3 H r F of length r and with quotients H 1 ;H 2 ;:::;H r . Note that for any permutation of f1;2;:::rg,there is another composition series with quotients H .1/ ;H .2/ ;:::;H .r/ .(f) We saw in (4.37) that for n 5, the only normal subgroups of S n are S n , A n , f1g,and in (4.33) that A n is simple. Hence S n F A n F f1g is the only composition series for S n .THEOREM 6.2 (JORDAN-HÖLDER) 2 Let G be a finite group. IfG D G 0 F G 1 F F G s D f1gG D H 0 F H 1 F F H t D f1gare two composition series for G, then s D t and there is a permutation of f1;2;:::;sgsuch that G i =G iC1 H .i/ =H .i/C1 .PROOF. We use induction on the order of G.Case I: H 1 D G 1 . In this case, we have two composition series for G 1 , to which we canapply the induction hypothesis.2 Jordan showed that corresponding quotients had the same order, and Hölder that they were isomorphic.

Solvable groups 83Case II: H 1 ¤ G 1 . Because G 1 and H 1 are both normal in G, the product G 1 H 1 is anormal subgroup of G. It properly contains both G 1 and H 1 , which are maximal normalsubgroups of G, and so G 1 H 1 D G. ThereforeG=G 1 D G 1 H 1 =G 1 ' H 1 =G 1 \ H 1 (see 1.46).Similarly G=H 1 ' G 1 =G 1 \ H 1 . Let K 2 D G 1 \ H 1 ; then K 2 is a maximal normal subgroupin both G 1 and H 1 , andG=G 1 ' H 1 =K 2 ; G=H 1 ' G 1 =K 2 : (26)Choose a composition seriesWe have the picture:K 2 F K 3 F F K u :G 1 F G 2 F F G s G K 2 F F K u : H 1 F H 2 F F H tOn applying the induction hypothesis to G 1 and H 1 and their composition series in thediagram, we find thatQuotients.G F G 1 F G 2 F / D fG=G 1 ;G 1 =G 2 ;G 2 =G 3 ;:::g (definition) fG=G 1 ;G 1 =K 2 ;K 2 =K 3 ;:::g (induction) fH 1 =K 2 ;G=H 1 ;K 2 =K 3 ;:::g (apply (26)) fG=H 1 ;H 1 =K 2 ;K 2 =K 3 ;:::g (reorder) fG=H 1 ;H 1 =H 2 ;H 2 =H 3 ;:::g (induction)D Quotients.G F H 1 F H 2 F / (definition).✷Note that the theorem applied to a cyclic group C m implies that the factorization of aninteger into a product of primes is unique.REMARK 6.3 There are infinite groups having finite composition series (there are eveninfinite simple groups). For such a group, let d.G/ be the minimum length of a compositionseries. Then the Jordan-Hölder theorem extends to show that all composition series havelength d.G/ and have isomorphic quotient groups. The same proof works except that youhave to use induction on d.G/ instead of jGj and verify that a normal subgroup of a groupwith a finite composition series also has a finite composition series (Exercise 6-1).The quotients of a composition series are sometimes called composition factors.Solvable groupsA subnormal series whose quotient groups are all commutative is called a solvable series.A group is solvable (or soluble) if it has a solvable series. Alternatively, we can say thata group is solvable if it can be obtained by forming successive extensions of commutativegroups. Since a commutative group is simple if and only if it is cyclic of prime order, we

84 6. SUBNORMAL SERIES; SOLVABLE AND NILPOTENT GROUPSsee that G is solvable if and only if for one (hence every) composition series the quotientsare all cyclic groups of prime order.Every commutative group is solvable, as is every dihedral group. The results in Chapter5 show that every group of order < 60 is solvable. By contrast, a noncommutative simplegroup, e.g., A n for n 5, will not be solvable.THEOREM 6.4 (FEIT-THOMPSON) Every finite group of odd order is solvable. 3PROOF. The proof occupies an entire issue of the Pacific Journal of Mathematics (Feit andThompson 1963).✷In other words, every noncommutative finite simple group has even order, and thereforecontains an element of order 2. For the role this theorem played in the classification of thefinite simple groups, see p. 51. For a more recent look at the Feit-Thompson theorem, seeGlauberman 1999. 1 EXAMPLE 6.5 Consider the subgroups B Dand U Dof GL0 0 12 .F /,some field F . Then U is a normal subgroup of B, and B=U ' F F , U ' .F;C/.Hence B is solvable.PROPOSITION 6.6 (a) Every subgroup and every quotient group of a solvable group issolvable.(b) An extension of solvable groups is solvable.PROOF. (a) Let G F G 1 F F G n be a solvable series for G, and let H be a subgroup of G.The homomorphismx 7! xG iC1 W H \ G i ! G i =G iC1has kernel .H \ G i / \ G iC1 D H \ G iC1 . Therefore, H \ G iC1 is a normal subgroup ofH \ G i and the quotient H \ G i =H \ G iC1 injects into G i =G iC1 , which is commutative.We have shown thatH F H \ G 1 F F H \ G nis a solvable series for H .Let NG be a quotient group of G, and let NG i be the image of G i in NG. Thenis a solvable series for NG.3 Burnside (1897, p. 379) wrote:NG F NG 1 F F NG n D f1gNo simple group of odd order is at present known to exist. An investigation as to the existenceor non-existence of such groups would undoubtedly lead, whatever the conclusion might be, toresults of importance; it may be recommended to the reader as well worth his attention. Also,there is no known simple group whose order contains fewer than three different primes. . . .Significant progress in the first problem was not made until Suzuki, M., The nonexistence of a certain type ofsimple group of finite order, 1957. However, the second problem was solved by Burnside himself, who provedusing characters that any group whose order contains fewer than three different primes is solvable (see Alperinand Bell 1995, p. 182).

Solvable groups 85(b) Let N be a normal subgroup of G, and let NG D G=N . We have to show that if Nand NG are solvable, then so also is G. LetNG F NG 1 F F NG n D f1gN F N 1 F F N m D f1gbe solvable series for NG and N , and let G i be the inverse image of NG i in G. ThenG i =G iC1 ' NG i = NG iC1 (see 1.48), and soG F G 1 F F G n .D N / F N 1 F F N mis a solvable series for G.✷COROLLARY 6.7 A finite p-group is solvable.PROOF. We use induction on the order the group G. According to (4.16), the centre Z.G/of G is nontrivial, and so the induction hypothesis implies that G=Z.G/ is solvable. BecauseZ.G/ is commutative, (b) of the proposition shows that G is solvable.✷ThusLet G be a group. Recall that the commutator of x;y 2 G isŒx;y D xyx 1 y 1 D xy.yx/ 1Œx;y D 1 ” xy D yx;and G is commutative if and only if every commutator equals 1.EXAMPLE 6.8 For any finite-dimensional vector space V over a field k and any full flagF D fV n ;V n 1 ;:::g in V , the groupB.F / D f˛ 2 Aut.V / j ˛.V j / V j all j gis solvable. Indeed, let U.F / be the group defined in Example 5.10. Then B.F /=U.F / iscommutative, and, when k D F p , U.F / is a p-group. This proves that B.F / is solvablewhen k D F p , and in the general case one defines subgroups B 0 B 1 of B.F / withB i D f˛ 2 B.F / j ˛.V j / V j i all j gand notes that the commutator of two elements of B i lies in B iC1 .For any homomorphism 'WG ! H'.Œx;y/ D '.xyx 1 y 1 / D Œ'.x/;'.y/;i.e., ' maps the commutator of x;y to the commutator of '.x/;'.y/. In particular, we seethat if H is commutative, then ' maps all commutators in G to 1.The group G 0 D G .1/ generated by the commutators in G is called the commutator orfirst derived subgroup of G.

86 6. SUBNORMAL SERIES; SOLVABLE AND NILPOTENT GROUPSPROPOSITION 6.9 The commutator subgroup G 0 is a characteristic subgroup of G; it is thesmallest normal subgroup of G such that G=G 0 is commutative.PROOF. An automorphism ˛ of G maps the generating set for G 0 into G 0 , and hence mapsG 0 into G 0 . Since this is true for all automorphisms of G, G 0 is characteristic.Write g 7! Ng for the homomorphism g 7! gG 0 WG ! G=G 0 . Then Œ Ng; N h D Œg;h, whichis 1 because Œg;h 2 G 0 . Hence Œ Ng; N h D 1 for all Ng, N h 2 G=G 0 , which shows that G=G 0 iscommutative.Let N be a second normal subgroup of G such that G=N is commutative. Then Œg;h 7!1 in G=N , and so Œg;h 2 N . Since these elements generate G 0 , N G 0 . ✷For n 5; A n is the smallest normal subgroup of S n giving a commutative quotient.Hence .S n / 0 D A n .The second derived subgroup of G is .G 0 / 0 ; the third is G .3/ D .G 00 / 0 ; and so on. Sincea characteristic subgroup of a characteristic subgroup is characteristic (3.7a), each derivedgroup G .n/ is a characteristic subgroup of G. Hence we obtain a normal seriesG G .1/ G .2/ ;which is called the derived series of G. For example, when n 5, the derived series of S nisS n A n A n A n :PROPOSITION 6.10 A group G is solvable if and only if its kth derived subgroup G .k/ D 1for some k.PROOF. If G .k/ D 1, then the derived series is a solvable series for G. Conversely, letG D G 0 F G 1 F G 2 F F G s D 1be a solvable series for G. Because G=G 1 is commutative, G 1 G 0 . Now G 0 G 2 is asubgroup of G 1 , and fromwe see thatG 0 =G 0 \ G 2'! G 0 G 2 =G 2 G 1 =G 2G 1 =G 2 commutative H) G 0 =G 0 \ G 2 commutative H) G 00 G 0 \ G 2 G 2 :Continuing in the fashion, we find that G .i/ G i for all i, and hence G .s/ D 1.✷Thus, a solvable group G has a canonical solvable series, namely the derived series, inwhich all the groups are normal in G. The proof of the proposition shows that the derivedseries is the shortest solvable series for G. Its length is called the solvable length of G.ASIDE 6.11 Not every element of the commutator subgroup of a group is itself a commutator, butthe smallest groups where this occurs have order 94. This was shown by a computer search throughthe libraries of small groups.

Nilpotent groups 87Nilpotent groupsLet G be a group. Recall that we write Z.G/ for the centre of G. Let Z 2 .G/ G be thesubgroup of G corresponding to Z.G=Z.G// G=Z.G/. Thusg 2 Z 2 .G/ ” Œg;x 2 Z.G/ for all x 2 G:Continuing in this fashion, we get a sequence of subgroups (ascending central series)wheref1g Z.G/ Z 2 .G/ g 2 Z i .G/ ” Œg;x 2 Z i 1 .G/ for all x 2 G:If Z m .G/ D G for some m, then G is said to be nilpotent, and the smallest such m is calledthe (nilpotency) class of G. For example, all finite p-groups are nilpotent (apply 4.16).Only the group f1g has class 0, and the groups of class 1 are exactly the commutativegroups. A group G is of class 2 if and only if G=Z.G/ is commutative — such a group issaid to be metabelian.EXAMPLE 6.12 (a) A nilpotent group is obviously solvable, but the converse is false. Forexample, for a field F , let a bB Dˇˇˇˇa;b;c 2 F; ac ¤ 0:0 cThen Z.B/ D faI j a ¤ 0g, and the centre of B=Z.B/ is trivial. Therefore B=Z.B/ is notnilpotent, but we saw in 8(6.5) 0that it is1solvable. 98019< 1 =< 1 0 =(b) The group G D @ 0 1 A:; is metabelian: its centre is @ 0 1 0A: ; , and0 0 10 0 1G=Z.G/ is commutative.(c) Any nonabelian group G of order p 3 is metabelian. In fact, G 0 D Z.G/ has orderp (see 5.17), and G=G 0 is commutative (4.18). In particular, the quaternion and dihedralgroups of order 8, Q and D 4 , are metabelian. The dihedral group D 2 n is nilpotent of classn — this can be proved by induction, using that Z.D 2 n/ has order 2, and D 2 n=Z.D 2 n/ D 2 n 1. If n is not a power of 2, then D n is not nilpotent (use Theorem 6.18 below).PROPOSITION 6.13 (a) A subgroup of a nilpotent group is nilpotent.(b) A quotient of a nilpotent group is nilpotent.PROOF. (a) Let H be a subgroup of a nilpotent group G. Clearly, Z.H / Z.G/ \ H .Assume (inductively) that Z i .H / Z i .G/\H ; then Z iC1 .H / Z iC1 .G/\H , because(for h 2 H )h 2 Z iC1 .G/ H) Œh;x 2 Z i .G/ all x 2 G H) Œh;x 2 Z i .H / all x 2 H:(b) Straightforward.✷

88 6. SUBNORMAL SERIES; SOLVABLE AND NILPOTENT GROUPSREMARK 6.14 It should be noted that if H is a subgroup of G, then Z.H / may be biggerthan Z.G/. For example, the centre of a 0H Dˇˇˇˇab ¤ 0 GL0 b2 .F /:is H itself, but the centre of GL 2 .F / consists only of the scalar matrices.PROPOSITION 6.15 A group G is nilpotent of class m if and only iffor all g 1 ;:::;g mC1 2 G:Œ:::ŒŒg 1 ;g 2 ;g 3 ;:::;;g mC1 D 1PROOF. Recall, g 2 Z i .G/ ” Œg;x 2 Z i 1 .G/ for all x 2 G:Assume G is nilpotent of class m; thenG D Z m .G/ H) Œg 1 ;g 2 2 Z m1 .G/ all g 1 ;g 2 2 GH) ŒŒg 1 ;g 2 ;g 3 2 Z m 2 .G/ all g 1 ;g 2 ;g 3 2 GFor the converse, let g 1 2 G. ThenH) ŒŒŒg 1 ;g 2 ;g 3 ;:::;g m 2 Z.G/ all g 1 ;:::;g m 2 GH) ŒŒŒg 1 ;g 2 ;g 3 ;:::;g mC1 D 1 all g 1 ;:::;g m 2 G:ŒŒ:::ŒŒg 1 ;g 2 ;g 3 ;:::;g m ;g mC1 D 1 for all g 1 ;g 2 ;:::;g mC1 2 GH) Œ:::ŒŒg 1 ;g 2 ;g 3 ;:::;g m 2 Z.G/; for all g 1 ;:::;g m 2 GH) Œ:::ŒŒg 1 ;g 2 ;g 3 ;:::;g m 1 2 Z 2 .G/; for all g 1 ;:::;g m 1 2 GH) g 1 2 Z m .G/ all g 1 2 G: ✷An extension of nilpotent groups need not be nilpotent, i.e.,N and G=N nilpotent G nilpotent. (27)For example, the subgroup U of the group B in Examples 6.5 and 6.12 is commutative andB=U is commutative, but B is not nilpotent.However, the implication (27) holds when N is contained in the centre of G. In fact,we have the following more precise result.COROLLARY 6.16 For any subgroup N of the centre of G,G=N nilpotent of class m H) G nilpotent of class m C 1:PROOF. Write for the map G ! G=N . Then.Œ:::ŒŒg 1 ;g 2 ;g 3 ;:::;g m ;g mC1 / D Œ:::ŒŒg 1 ;g 2 ;g 3 ;:::;g m ;g mC1 D 1all g 1 ;:::;g mC1 2 G. Hence Œ:::ŒŒg 1 ;g 2 ;g 3 ;:::;g m ;g mC1 2 N Z.G/, and soŒ:::ŒŒg 1 ;g 2 ;g 3 ;:::;g mC1 ;g mC2 D 1 all g 1 ;:::;g mC2 2 G:✷

Nilpotent groups 89COROLLARY 6.17 A finite p-group is nilpotent.PROOF. We use induction on the order of G. Because Z.G/ ¤ 1, G=Z.G/ nilpotent, whichimplies that G is nilpotent.✷Recall that an extensionis central if .N / Z.G/. Then:1 ! N ! G ! Q ! 1the nilpotent groups are those that can be obtained from commutative groupsby successive central extensions.Contrast:the solvable groups are those that can be obtained from commutative groupsby successive extensions (not necessarily central).THEOREM 6.18 A finite group is nilpotent if and only if it is equal to a direct product ofits Sylow subgroups.PROOF. A direct product of nilpotent groups is obviously nilpotent, and so the “if” directionfollows from the preceding corollary. For the converse, let G be a finite nilpotent group.According to (5.9) it suffices to prove that all Sylow subgroups are normal. Let P be such asubgroup of G, and let N D N G .P /. The first lemma below shows that N G .N / D N , andthe second then implies that N D G, i.e., that P is normal in G.✷LEMMA 6.19 Let P be a Sylow p-subgroup of a finite group G. For any subgroup H ofG containing N G .P /, we have N G .H / D H .PROOF. Let g 2 N G .H /, so that gHg 1 D H . Then H gP g 1 D P 0 , which is a Sylowp-subgroup of H . By Sylow II, hP 0 h 1 D P for some h 2 H , and so hgP g 1 h 1 P .Hence hg 2 N G .P / H , and so g 2 H:✷LEMMA 6.20 Let H be proper subgroup of a finite nilpotent group G; then H ¤ N G .H /.PROOF. The statement is obviously true for commutative groups, and so we can assumeG to be noncommutative. We use induction on the order of G. Because G is nilpotent,Z.G/ ¤ 1. Certainly the elements of Z.G/ normalize H , and so if Z.G/ H , we haveH Z.G/ H N G .H /. Thus we may suppose Z.G/ H . Then the normalizer of Hin G corresponds under (1.47) to the normalizer of H=Z.G/ in G=Z.G/, and we can applythe induction hypothesis.✷REMARK 6.21 For a finite abelian group G we recover the fact that G is a direct productof its p-primary subgroups.PROPOSITION 6.22 (FRATTINI’S ARGUMENT) Let H be a normal subgroup of a finitegroup G, and let P be a Sylow p-subgroup of H . Then G D H N G .P /.

90 6. SUBNORMAL SERIES; SOLVABLE AND NILPOTENT GROUPSPROOF. Let g 2 G. Then gP g 1 gHg 1 D H , and both gP g 1 and P are Sylow p-subgroups of H . According to Sylow II, there is an h 2 H such that gP g 1 D hP h 1 , andit follows that h 1 g 2 N G .P / and so g 2 H N G .P /.✷THEOREM 6.23 A finite group is nilpotent if and only if every maximal proper subgroupis normal.PROOF. We saw in Lemma 6.20 that for any proper subgroup H of a nilpotent group G,H N G .H /. Hence,H maximal H) N G .H / D G;i.e., H is normal in G.Conversely, suppose every maximal proper subgroup of G is normal. We shall checkthe condition of Theorem 6.18. Thus, let P be a Sylow p-subgroup of G. If P is notnormal in G, then there exists a maximal proper subgroup H of G containing N G .P /.Being maximal, H is normal, and so Frattini’s argument shows that G D H N G .P / D H— contradiction. ✷Groups with operatorsRecall that the set Aut.G/ of automorphisms of a group G is again a group. Let A bea group. A pair .G;'/ consisting of a group G together with a homomorphism 'WA !Aut.G/ is called an A-group, or G is said to have A as a group of operators.Let G be an A-group, and write ˛x for '.˛/x. Then(a) .˛ˇ/ x D ˛.ˇ x/(b) ˛.xy/ D ˛x ˛y(c) 1 x D x(' is a homomorphism);('.˛/ is a homomorphism);(' is a homomorphism).Conversely, a map .˛;x/ 7! ˛x W AG ! G satisfying (a), (b), (c) arises from a homomorphismA ! Aut.G/. Conditions (a) and (c) show that x 7! ˛x is inverse to x 7! .˛ 1/ x,and so x 7! ˛x is a bijection G ! G. Condition (b) then shows that it is an automorphismof G. Finally, (a) shows that the map '.˛/ D .x 7! ˛x/ is a homomorphism A ! Aut.G/.Let G be a group with operators A. A subgroup H of G is admissible or A-invariant ifx 2 H H) ˛x 2 H , all ˛ 2 A:An intersection of admissible groups is admissible. If H is admissible, so also are itsnormalizer N G .H / and centralizer C G .H /:An A-homomorphism (or admissible homomorphism) of A-groups is a homomorphismWG ! G 0 such that .˛g/ D ˛.g/ for all ˛ 2 A, g 2 G:EXAMPLE 6.24 (a) A group G can be regarded as a group with f1g as group of operators.In this case all subgroups and homomorphisms are admissible, and so the theory of groupswith operators includes the theory of groups without operators.(b) Consider G acting on itself by conjugation, i.e., consider G together with the homomorphismg 7! i g WG ! Aut.G/:In this case, the admissible subgroups are the normal subgroups.(c) Consider G with A D Aut.G/ as group of operators. In this case, the admissiblesubgroups are the characteristic subgroups.

Groups with operators 91Almost everything we have proved for groups also holds for groups with operators. Inparticular, the Theorems 1.45, 1.46, and 1.47 hold for groups with operators. In each case,the proof is the same as before except that admissibility must be checked.THEOREM 6.25 For any admissible homomorphism WG ! G 0 of A-groups, N defD Ker./is an admissible normal subgroup of G, .G/ is an admissible subgroup of G 0 , and factorsin a natural way into the composite of an admissible surjection, an admissible isomorphism,and an admissible injection:G ↠ G=N ' ! .G/ ,! G 0 :THEOREM 6.26 Let G be a group with operators A, and let H and N be admissiblesubgroups with N normal. Then H \ N is a normal admissible subgroup of H , HNis an admissible subgroup of G, and h.H \ N / 7! hH is an admissible isomorphismH=H \ N ! HN=N:THEOREM 6.27 Let 'WG ! NG be a surjective admissible homomorphism of A-groups.Under the one-to-one correspondence H $ HNbetween the set of subgroups of G containingKer.'/ and the set of subgroups of NG (see 1.47), admissible subgroups correspond toadmissible subgroups.Let 'WA ! Aut.G/ be a group with A operating. An admissible subnormal series is achain of admissible subgroups of GG G 1 G 2 G rwith each G i normal in G i 1 . Define similarly an admissible composition series. The quotientsof an admissible subnormal series are A-groups, and the quotients of an admissiblecomposition series are simple A-groups, i.e., they have no normal admissible subgroupsapart from the obvious two.The Jordan-Hölder theorem continues to hold for A-groups. In this case the isomorphismsbetween the corresponding quotients of two composition series are admissible. Theproof is the same as that of the original theorem, because it uses only the isomorphismtheorems, which we have noted also hold for A-groups.EXAMPLE 6.28 (a) Consider G with G acting by conjugation. In this case an admissiblesubnormal series is a sequence of subgroupsG D G 0 G 1 G 2 G s D f1g;with each G i normal in G, i.e., a normal series. The action of G on G i by conjugationpasses to the quotient, to give an action of G on G i =G iC1 . The quotients of two admissiblecomposition series are isomorphic as G-groups.(b) Consider G with A D Aut.G/ as operator group. In this case, an admissible subnormalseries is a sequenceG D G 0 G 1 G 2 G s D f1gwith each G i a characteristic subgroup of G, and the quotients of two admissible compositionseries are isomorphic as Aut.G/-groups.

92 6. SUBNORMAL SERIES; SOLVABLE AND NILPOTENT GROUPSKrull-Schmidt theoremA group G is indecomposable if G ¤ 1 and G is not isomorphic to a direct product of twonontrivial groups, i.e., ifG H H 0 H) H D 1 or H 0 D 1:EXAMPLE 6.29 (a) A simple group is indecomposable, but an indecomposable group neednot be simple: it may have a normal subgroup. For example, S 3 is indecomposable but hasC 3 as a normal subgroup.(b) A finite commutative group is indecomposable if and only if it is cyclic of primepowerorder.Of course, this is obvious from the classification, but it is not difficult to prove it directly.Let G be cyclic of order p n , and suppose that G H H 0 . Then H and H 0 must be p-groups, and they can’t both be killed by p m , m < n. It follows that one must be cyclicof order p n , and that the other is trivial. Conversely, suppose that G is commutative andindecomposable. Since every finite commutative group is (obviously) a direct product ofp-groups with p running over the primes, G is a p-group. If g is an element of G of highestorder, one shows that hgi is a direct factor of G, G hgi H , which is a contradiction.(c) Every finite group can be written as a direct product of indecomposable groups(obviously).THEOREM 6.30 (KRULL-SCHMIDT) Suppose that G is a direct product of indecomposablesubgroups G 1 ;:::;G s and of indecomposable subgroups H 1 ;:::;H t :G ' G 1 G s ; G ' H 1 H t :Then s D t, and there is a re-indexing such that G i H i . Moreover, given r, we canarrange the numbering so thatG D G 1 G r H rC1 H t :PROOF. See Rotman 1995, 6.36.✷EXAMPLE 6.31 Let G D F p F p , and think of it as a two-dimensional vector space overF p . LetG 1 D h.1;0/i; G 2 D h.0;1/iI H 1 D h.1;1/i; H 2 D h.1; 1/i:Then G D G 1 G 2 , G D H 1 H 2 , G D G 1 H 2 .REMARK 6.32 (a) The Krull-Schmidt theorem holds also for an infinite group provided itsatisfies both chain conditions on subgroups, i.e., ascending and descending sequences ofsubgroups of G become stationary.(b) The Krull-Schmidt theorem also holds for groups with operators. For example,let Aut.G/ operate on G; then the subgroups in the statement of the theorem will all becharacteristic.(c) When applied to a finite abelian group, the theorem shows that the groups C mi in adecomposition G D C m1 ::: C mr with each m i a prime power are uniquely determinedup to isomorphism (and ordering).

Exercises 93Exercises6-1 Let G be a group (not necessarily finite) with a finite composition seriesG D G 0 G 1 G n D 1;and let N be a normal subgroup of G. Show thatN D N \ G 0 N \ G 1 N \ G n D 1becomes a composition series for N once the repetitions have been omitted.6-2 If G 1 and G 2 are groups such that G 0 1 G0 2 and G 1=G 0 1 G 2=G 0 2 , are G 1 and G 2necessarily isomorphic? (Here 0 denotes the commutator subgroup.)

CHAPTER 7Representations of Finite GroupsThroughout this chapter, G is a finite group and F is a field. All vector spaces are finitedimensional.An F -algebra is a ring A containing F in its centre and finite dimensional as an F -vector space. We do not assume A to be commutative; for example, A could be the matrixalgebra M n .F /. Let e 1 ;:::;e n be a basis for A as an F -vector space; then e i e j DPk ak ij e kfor some a k ij 2 F , called the structure constants of A relative to the basis .e i/ i ;once a basis has been chosen, the algebra A is uniquely determined by its structure constants.All A-modules are finite dimensional when regarded as F -vector spaces. For an A-module V , mV denotes the direct sum of m copies of V .The opposite A opp of an F -algebra A is the same F -algebra as A but with the multiplicationreversed, i.e., A opp D .A;C; 0 / with a 0 b D ba. In other words, there is a one-to-onecorrespondence a $ a 0 WA $ A opp which is an isomorphism of F -vector spaces and has theproperty that a 0 b 0 D .ba/ 0 .An A-module M is simple if it is nonzero and contains no submodules except 0 and M ,and it is semisimple if it is isomorphic to a direct sum of simple modules.Matrix representationsA matrix representation of degree n of G over F is a homomorphism G ! GL n .F /.The representation is said to be faithful if the homomorphism is injective. Thus a faithfulrepresentation identifies G with group of n n matrices.EXAMPLE 7.1 (a) There is a representation Q ! GL 2 .C/ of the quaternion group Q Dha;bi sending a to p p 0 11 0and b to0 11 0. In fact, that is how we originally definedQ in (1.18).(b) Let G D S n . For each 2 S n , let I./ be the matrix obtained from the identitymatrix by using to permute the rows. Then, for any n n matrix A, I./A is obtainedfrom A by using to permute the rows. In particular, I./I. 0 / D I. 0 /, and so 7! I./is a representation of S n . Clearly, it is faithful. As every finite group embeds into S n forsome n (Cayley’s theorem, see 1.22), this shows that every finite group has a faithful matrixrepresentation.(c) Let G D C n D hi. If F contains a nth root of 1, say , then there is representation i 7! i WC n ! GL 1 .F / D F . The representation is faithful if and only if has order95

96 7. REPRESENTATIONS OF FINITE GROUPSexactly n. If n D p is prime and F has characteristic p, then X p 1 D .X 1/ p , and so 1is the only pth root of 1 in F . In this case, the representation is trivial, but there is a faithfulrepresentation i 17!i0 1WC p ! GL 2 .F /:ASIDE 7.2 Recall that the Burnside problem asks whether every finitely generated group with finiteexponent is finite (see p. 37). Burnside proved that the problem has a positive answer for subgroupsof GL n .C/. Therefore, no infinite finitely generated group with finite exponent has a faithful representationover C.Roots of 1 in fieldsAs the last example indicates, the representations of a group over a field F depend on theroots of 1 in the field. The nth roots of 1 in a field F form a subgroup n .F / of F , whichis cyclic (see 1.56).If the characteristic of F divides n, then j n .F /j < n. Otherwise, X n 1 has distinctroots (a multiple root would have to be a root of its derivative nX n 1 ), and we can alwaysarrange that j n .F /j D n by extending F , for example, by replacing a subfield F of C withF Œ where D e 2i=n , or by replacing F with F ŒX=.g.X// where g.X/ is an irreduciblefactor of X n 1 not dividing X m 1 for any proper divisor m of n:An element of order n in F is called a primitive nth root of 1. To say that F containsa primitive nth root of 1, , means that n .F / is a cyclic group of order n and that generates it (and it implies that either F has characteristic 0 or it has characteristic a primenot dividing n).Linear representationsRecall (4.1) that we have defined the notion of a group G acting a set. When the set is anF -vector space V , we say that the action is linear if the mapg V WV ! V , x 7! gx;is linear for each g 2 G. Then g V has inverse the linear map .g 1 / V , and g 7! g V WG !GL.V / is a homomorphism. Thus, from a linear action of G on V , we obtain a homomorphismof groups G ! GL.V /; conversely, every such homomorphism defines a linearaction of G on V . We call a homomorphism G ! GL.V / a linear representation of G onV . Note that a linear representation of G on F n is just a matrix representation of degree n.EXAMPLE 7.3 (a) Let G D C n D hi, and assume that F contains a primitive nth root of1, say . Let G ! GL.V / be a linear representation of G. Then . L / n D . n / L D 1, and sothe minimum polynomial of L divides X n 1. As X n 1 has n distinct roots 0 ;:::; n 1in F , the vector space V decomposes into a direct sum of eigenspacesV D M 0in 1 V i; V idefD fv 2 V j v D i vg.Conversely, every such direct sum decomposition of G arises from a representation of G.

Maschke’s theorem 97(b) Let G be a commutative group of exponent n, and assume that F contains a primitiventh root of 1. LetG _ D Hom.G;F / D Hom.G; n .F //To give a representation of G on a vector space V is the same as to give a direct sumdecompositionV D M V def; V D fv 2 V j v D ./vg.2G _When G is cyclic, this is a restatement of (a), and the general case follows easily (decomposeV with respect to the action of one cyclic factor of G; then decompose each summandwith respect to the action of a second cyclic factor of G; and so on).Maschke’s theoremLet G ! GL.V / be a linear representation of G on an F -vector space V . A subspace Wof V is said to be G-invariant if gW W for all g 2 G. An F -linear map ˛WV ! V 0 ofvector spaces on which G acts linearly is said to be G-invariant if˛.gv/ D g.˛v/ for all g 2 G;v 2 V:Finally, a bilinear form WV V ! F is said to be G-invariant if.gv;gv 0 / D .v;v 0 / for all g 2 G, v;v 0 2 V:THEOREM 7.4 (MASCHKE) Let G ! GL.V / be a linear representation of G. If the characteristicof F does not divide jGj, then every G-invariant subspace W of V has a G-invariant complement, i.e., there exists a G-invariant subspace W 0 such that V D W ˚ W 0 .Note that the theorem always applies when F has characteristic zero.The condition on the characteristic is certainly necessary: let G D hi be the cyclicgroup of order p, where p is the characteristic of F , and let acts on V D F 2 as the matrix1 10 1(see 7.3b); the subspace .0 / is G-invariant, and this complementary subspaces arethose of the form F a b, b ¤ 0; none of them is G-invariant.Because of the importance of the ideas involved, we present two proofs of Maschke’stheorem.PROOF OF MASCHKE’S THEOREM (CASE F D R OR C)LEMMA 7.5 Let be a symmetric bilinear form on V , and let W be a subspace of V . If and W are G-invariant, then so also is W ? D deffv 2 V j .w;v/ D 0 for all w 2 W g.PROOF. Let v 2 W ? and let g 2 G. For any w 2 W , .w;gv/ D .g 1 w;v/ because isG-invariant, and .g 1 w;v/ D 0 because W is G-invariant. This shows that gv 2 W ? . ✷Recall from linear algebra that if is nondegenerate, then V D W ˚ W ? . Therefore,in order to prove Maschke’s theorem, it suffices to show that there exists a G-invariantnondegenerate symmetric bilinear from WV V ! F .

98 7. REPRESENTATIONS OF FINITE GROUPSLEMMA 7.6 For any symmetric bilinear form on V ,N.v;w/ defD X g2G .gv;gw/is a G-invariant symmetric bilinear form on V .PROOF. The form is obviously bilinear and symmetric, and for g 0 2 G,N.g 0 v;g 0 w/ defD X g2G .gg 0v;gg 0 w/;which equals P g2G .gv;gw/ because, as g runs over G, so also does gg 0.✷Unfortunately, we can’t conclude that N is nondegenerate when is (otherwise wecould prove that all F ŒG-modules are semisimple, with no restriction on F or G).LEMMA 7.7 Let F D R. If is a positive definite symmetric bilinear form on V , then soalso is N.PROOF. If N is positive definite, then for any nonzero v in V ,N.v;v/ D X .gv;gv/ > 0:g2G ✷This completes the proof of Maschke’s theorem when F D R, because there certainlyexist positive definite symmetric bilinear forms on V . A similar argument using hermitianforms applies when F D C (or, indeed, when F is any subfield of C).ASIDE 7.8 A representation of a group G on a real vector space V is unitary if there exists aG-invariant positive definite symmetric bilinear form on V . Lemma 7.6 shows that every unitaryrepresentation is semisimple, and Lemma 7.7 shows that every real representation of a finite groupis unitary.PROOF OF MASCHKE’S THEOREM (GENERAL CASE)An endomorphism of an F -vector space V is called a projector if 2 D . The minimumpolynomial of a projector divides X 2 X D X.X 1/, and so V decomposes into a directsum of eigenspaces,V0 ./ D fv 2 V j v D 0g D Ker./V D V 0 ./ ˚ V 1 ./, whereV 1 ./ D fv 2 V j v D vg D Im./:Conversely, a decomposition V D V 0 ˚ V 1 arises from a projector .v 0 ;v 1 / 7! .0;v 1 /.Now suppose that G acts linearly on V . If a projector is G-invariant, then V 1 ./ andV 0 ./ are obviously G-invariant. Thus, to prove the theorem it suffices to show that W isthe image of a G–invariant projector .We begin by choosing an F -linear projector with image W , which certainly exists,and we modify it to obtain a G-invariant projector N with the same image. For v 2 V , letN.v/ D 1 XjGjg .g 1 v/ :g2G

The group algebra; semisimplicity 99This makes sense because jGj 1 2 F , and it defines an F -linear map NWV ! V . Letw 2 W ; then g 1 w 2 W , and soN.w/ D 1jGjXg2G g.g 1 w/ D 1 Xw D w: (28)jGj g2GThe image of N is contained in W , because Im./ W and W is G-invariant, and soN 2 .v/ defD N. N .v// .28/D N.v/for any v 2 V . Thus, N is a projector, and (28) shows that Im. N/ W , and hence Im. N/ DW . It remains to show that N is G-invariant. For g 0 2 VN.g 0 v/ D 1jGjXg2G g .g 1 g 0 v/ 1 XD g 0jGj.g 1g2G0 g/ .g 1 g 0 v/ ;which equals g 0 N .v/ because, as g runs over G, so also does g 10 g.The group algebra; semisimplicityThe group algebra F ŒG of G is defined to be the F -vector space with basis the elementsof G endowed with the multiplication extending that on G. Thus,˘ an element of F ŒG is a sum P g2G c gg, c g 2 F ,˘ two elements P g2G c gg and P g2G c0 g g of F ŒG are equal if and only if c g D cg0for all g, and P P ˘ g2G gg c g2G c0 g g D P g2G c00 g g; c00 g D P g 1 g 2 Dg c g 1cg 0 2:A linear actiong;v 7! gvWG V ! Vof G on an F -vector space extends uniquely to an action of F ŒG on V ,Xg2G c gg;v 7! X g2G c ggvWF ŒG V ! V;which makes V into an F ŒG-module. The submodules for this action are exactly the G-invariant subspaces.Let G ! GL.V / be a linear representation of G. When V is simple (resp. semisimple)as an F ŒG-module, the representation is usually said to be irreducible (resp. completelyreducible). However, I will call them simple (resp. semisimple) representations.PROPOSITION 7.9 If the characteristic of F does not divide jGj, then every F ŒG-moduleis a direct sum of simple submodules.PROOF. Let V be a F ŒG-module. If V is simple, then there is nothing to prove. Otherwise,it contains a nonzero proper submodule W . According to Maschke’s theorem, V D W ˚W 0 with W 0 an F ŒG-submodule. If W and W 0 are simple, then the proof is complete;otherwise, we can continue the argument, which terminates in a finite number of stepsbecause V has finite dimension as an F -vector space.✷As we have observed, the linear representations of G can be regarded as F ŒG-modules.Thus, to understand the linear representations of G, we need to understand the F ŒGmodules,and for this we need to understand the structure of the F -algebra F ŒG. In the nextthree sections we study F -algebras and their modules; in particular, we prove the famousWedderburn theorems concerning F -algebras whose modules are all semisimple.

100 7. REPRESENTATIONS OF FINITE GROUPSSemisimple modulesIn this section, A is an F -algebra.PROPOSITION 7.10 Every A-module V admits a filtrationV D V 0 V 1 V s D f0gsuch that the quotients V i =V iC1 are simple A-modules. IfV D W 0 W 1 W t D f0gis a second such filtration, then s D t and there is a permutation of f1;:::;sg such thatV i =V iC1 W .i/ =W .i/C1 for all i.PROOF. This is a variant of the Jordan-Hölder theorem (6.2), which can be proved by thesame argument.✷COROLLARY 7.11 SupposeV V 1 ˚ ˚ V s W 1 ˚ ˚ W twith all the A-modules V i and W j simple. Then s D t and there is a permutation off1;:::;sg such that V i W .i/ .PROOF. Each decomposition defines a filtration, to which the proposition can be applied. ✷PROPOSITION 7.12 Let V be an A-module. If V is a sum of simple submodules, sayV D P i2I S i (the sum need not be direct), then for any submodule W of V , there is asubset J of I such thatV D W ˚ Mi2J S i:PROOF. Let J be maximal among the subsets of I such the sum S JdefD P j 2J S j is directand W \ S J D 0. I claim that W C S J D V (hence V is the direct sum of W and the S jwith j 2 J ). For this, it suffices to show that each S i is contained in W C S J . Because S iis simple, S i \ .W C S J / equals S i or 0. In the first case, S i W C S J , and in the secondS J \ S i D 0 and W \ .S J C S i / D 0, contradicting the definition of I .✷COROLLARY 7.13 The following conditions on an A-module V are equivalent:(a) V is semisimple;(b) V is a sum of simple submodules;(c) every submodule of V has a complement.PROOF. The proposition shows that (b) implies (c), and the argument in the proof of (7.9)shows that (c) implies (a). It is obvious that (a) implies (b).✷COROLLARY 7.14 Sums, submodules, and quotient modules of semisimple modules aresemisimple.PROOF. Each is a sum of simple modules.✷

Simple F -algebras and their modules 101Simple F -algebras and their modulesAn F -algebra A is said to be simple if it contains no two-sided ideals except 0 and A. Weshall make frequent use of the following observation:The kernel of a homomorphism f WA ! B of F -algebras is an ideal in A notcontaining 1; therefore, if A is simple, then f is injective.EXAMPLE 7.15 We consider the matrix algebra M n .F /. Let e ij be the matrix with 1 inthe .i;j /th position and zeros elsewhere.(a) Let I be a two-sided ideal in M n .F /, and suppose that I contains a nonzero matrixM D .m ij / with, say, m i0 j 0¤ 0. Ase ii0 M e j0 j D m i0 j 0e ijand e ii0 M e j0 j 2 I , we see that I contains all the matrices e ij and so equalsM n .F /. We have shown that M n .F / is simple.(b) For M;N 2 M n .F /, the j th column of M N is M N j where N j is the j th columnof N . Therefore, for a given matrix N ,Nj D 0 ) .M N / j D 0(29)N j ¤ 0 ) .M N / j can be arbitrary:For 1 i n, let L.i/ be the set of matrices whose j th columns are zero for j ¤ iand whose ith column is arbitrary. For example, when n D 4,80190 0 0ˆ=C@ 0 0 0A M 4 .F /:ˆ:>;0 0 0It follows from (29) that L.i/ is a minimal left ideal in M n .F /. Note that M n .F / isa direct sumM n .F / D L.1/ ˚ ˚ L.n/of minimal left ideals.EXAMPLE 7.16 An F -algebra is said to be a division algebra if every nonzero elementa has an inverse, i.e., there exists a b such that ab D 1 D ba. Thus a division algebrasatisfies all the axioms to be a field except commutativity (and for this reason is sometimescalled a skew field). Clearly, a division algebra has no nonzero proper ideals, left, right, ortwo-sided, and so is simple.If D is a division algebra, then the argument in (7.15a) shows that the algebra M n .D/is simple.EXAMPLE 7.17 For a;b 2 F , let H.a;b/ be the F -algebra with basis 1;i;j;k (as anF -vector space) and with the multiplication determined byi 2 D a; j 2 D b; ij D k D j i(so ik D iij D aj etc.). Then H.a;b/ is an F -algebra, called a quaternion algebra overF . For example, if F D R, then H. 1; 1/ is the usual quaternion algebra. One can showthat H.a;b/ is either a division algebra or it is isomorphic to M 2 .F /. In particular, it issimple.

102 7. REPRESENTATIONS OF FINITE GROUPS7.18 Much of linear algebra does not require that the field be commutative. For example,the usual arguments show that a finitely generated module V over a division algebra D hasa basis, and that all bases have the same number n of elements — n is called the dimensionof V . In particular, all finitely generated D-modules are free.7.19 Let A be an F -algebra, and let A A denote A regarded as a left A-module. Rightmultiplication x 7! xa on A A by an element a of A is an A-linear endomorphism of A A.Moreover, every A-linear map 'W A A ! A A is of this form with a D '.1/. Thus,End A . A A/ ' A(as F -vector spaces).Let ' a be the map x 7! xa. Thenand soMore generally,.' a ı ' a 0/.1/ defD ' a .' a 0.1// D ' a .a 0 / D a 0 a D ' a 0 a.1/;End A . A A/ ' A oppEnd A .V / ' A oppfor any A-module V that is free of rank 1, and(as F -algebras).End A .V / ' M n .A opp /for any free A-module V of rank n (cf. 7.32 below).CENTRALIZERSLet A be an F -subalgebra of an F -algebra B. The centralizer of A in B isIt is again an F -subalgebra of B.C B .A/ D fb 2 B j ba D ab for all a 2 Ag:EXAMPLE 7.20 In the following examples, the centralizers are taken in M n .F /.(a) Let A be the set of scalar matrices in M n .F /, i.e., A D F I n . Clearly, C.A/ DM n .F /.(b) Let A D M n .F /. Then C.A/ is the centre of M n .F /, which we now compute. Lete ij be the matrix with 1 in the .i;j /th position and zeros elsewhere, so thateim if j D le ij e lm D0 if j ¤ l:Let ˛ D .a ij / 2 M n .F /. Then ˛ D P i;j a ij e ij , and so ˛e lm D P i a ile im ande lm˛ D P j a mj e lj . If ˛ is in the centre of M n .F /, then ˛e lm D e lm˛, and soa il D 0 for i ¤ l, a mj D 0 for j ¤ m, and a l l D a mm . It follows that the centre ofM n .F / is set of scalar matrices F I n . Thus C.A/ D F I n .(c) Let A be the set of diagonal matrices in M n .F /. In this case, C.A/ D A.Notice that in all three cases, C.C.A// D A.

Simple F -algebras and their modules 103THEOREM 7.21 (DOUBLE CENTRALIZER THEOREM) Let A be an F -algebra, and let Vbe a faithful semisimple A-module. Then C.C.A// D A (centralizers taken in End F .V /).PROOF. Let D D C.A/ and let B D C.D/. Clearly A B, and the reverse inclusionfollows from the next lemma when we take v 1 ;:::;v n to generate V as a F -vector space. ✷LEMMA 7.22 For any v 1 ;:::; v n 2 V and b 2 B, there exists an a 2 A such thatav 1 D bv 1 ; av 2 D bv 2 ; :::; av n D bv n :PROOF. We first prove this for n D 1. Note that Av 1 is an A-submodule of V , and so (see7.13) there exists an A-submodule W of V such that V D Av 1 ˚ W . Let WV ! V be themap .av 1 ;w/ 7! .av 1 ;0/ (projection onto Av 1 ). It is A-linear, hence lies in D, and has theproperty that .v/ D v if and only if v 2 Av 1 . Now.bv 1 / D b.v 1 / D bv 1 ;and so bv 1 2 Av 1 , as required.We now prove the general case. Let W be the direct sum of n copies of V with A actingdiagonally, i.e.,a.v 1 ;:::;v n / D .av 1 ;:::;av n /; a 2 A; v i 2 V:Then W is again a semisimple A-module (7.14). The centralizer of A in End F .W / consistsof the matrices . ij / 1i;j n , ij 2 End F .V /, such that . ij a/ D .a ij / for all a 2 A, i.e.,such that ij 2 D (cf. 7.32). In other words, the centralizer of A in End F .A/ is M n .D/. Anargument as in Example 7.20(b), using the matrices e ij .ı/ with ı in the ij th position andzeros elsewhere, shows that the centralizer of M n .D/ in End F .W / consists of the diagonalmatrices 01ˇ 0 00 ˇ 0B@:: : ::C: A0 0 ˇwith ˇ 2 B. We now apply the case n D 1 of the lemma to A, W , b, and the vector.v 1 ;:::;v n / to complete the proof. ✷THEOREM 7.23 Every simple F -algebra is isomorphic to M n .D/ for some n and somedivision F -algebra D.PROOF. Choose a simple A-module S, for example, any minimal left ideal of A. Then Aacts faithfully on S, because the kernel of A ! End F .S/ will be a two-sided ideal of A notcontaining 1, and hence is 0.Let D be the centralizer of A in the F -algebra End F .S/ of F -linear maps S ! S.According to the double centralizer theorem (7.21), the centralizer of D in End F .S/ isA, i.e., A D End D .S/. Schur’s lemma (7.24 below) implies that D is a division algebra.Therefore S is a free D-module (7.18), say, S D n , and so End D .S/ M n .D opp / (see7.19). ✷

104 7. REPRESENTATIONS OF FINITE GROUPSLEMMA 7.24 (SCHUR’S LEMMA) For every F -algebra A and simple A-module S, End A .S/is a division algebra.PROOF. Let be an A-linear map S ! S. Then Ker./ is an A-submodule of S, and soit is either S or 0. In the first case, is zero, and in the second it is an isomorphism, i.e., ithas an inverse that is also A-linear.✷MODULES OVER SIMPLE F -ALGEBRASFor any F -algebra A, the submodules of A A are the left ideals in A, and the simple submodulesof A A are the minimal left ideals.PROPOSITION 7.25 Any two minimal left ideals of a simple F -algebra are isomorphic asleft A-modules, and A A is a direct sum of its minimal left ideals.PROOF. After Theorem 7.23, we may assume that A D M n .D/ for some division algebraD. We saw in (7.16) that the minimal left ideals in M n .D/ are those of the form L.fj g/.Clearly A D L 1j n L.fj g/ and each L.fj g/ is isomorphic to Dn with its natural actionof M n .D/.✷THEOREM 7.26 Let A be a simple F -algebra, and let S be a simple A-module. Then everyA-module is isomorphic to a direct sum of copies of S.PROOF. Let S 0 be a minimal left ideal of A. The proposition shows that A A S0 nn. Let e 1 ;:::;e r be a set of generators for V as an A-module. The mapfor some.a 1 ;:::;a r / 7! X a i e irealizes V as a quotient of a direct sum of r copies of A A, and hence as a quotient of nrS 0 .Thus, V is a sum of simple submodules each isomorphic to S 0 , and so Proposition 7.12shows that V mS 0 for some m.✷COROLLARY 7.27 Let A be a simple F -algebra. Then any two simple A-modules areisomorphic, and any two A-modules having the same dimension over F are isomorphic.PROOF. Obvious from the Theorem.✷COROLLARY 7.28 The integer n in Theorem 7.23 is uniquely determined by A, and D isuniquely determined up to isomorphism.PROOF. If A M n .D/, then D End A .S/ for any simple A-module S. Thus, the statementfollows from Theorem 7.26.✷

Semisimple F -algebras and their modules 105CLASSIFICATION OF THE DIVISION ALGEBRAS OVER FAfter Theorem 7.23, to classify the simple algebras over F , it remains to classify the divisionalgebras over F .PROPOSITION 7.29 When F is algebraically closed, the only division algebra over F is Fitself.PROOF. Let D be division algebra over F . For any ˛ 2 D, the F -subalgebra F Œ˛ of Dgenerated by ˛ is a field (because it is an integral domain of finite degree over F ). Therefore˛ 2 F .✷ASIDE 7.30 The classification of the isomorphism classes of division algebras over a field F isone the most difficult and interesting problems in algebra and number theory. For F D R, the onlydivision algebra is the usual quaternion algebra. For F finite, the only division algebra with centreF is F itself (theorem of Wedderburn).A division algebra over F whose centre is F is said to be central (formerly normal). Brauershowed that the set of isomorphism classes of central division algebras over a field form a group,called (by Hasse and Noether) the Brauer group 1 of the field. The statements in the last paragraphshow that the Brauer groups of algebraically closed fields and finite fields are zero, and the Brauergroup of R has order 2. The Brauer groups of Q and its finite extensions were computed by Albert,Brauer, Hasse, and Noether in the 1930s as a consequence of class field theory.Semisimple F -algebras and their modulesAn F -algebra A is said to be semisimple if every A-module is semisimple. Theorem 7.26shows that simple F -algebras are semisimple, and Maschke’s theorem shows that the groupalgebra F ŒG is semisimple when the order of G is not divisible by the characteristic of F(see 7.9).EXAMPLE 7.31 Let A be a finite product of simple F -algebras. Then every minimal leftideal of a simple factor of A is a simple A-submodule of A A. Therefore, A A is a direct sumof simple A-modules, and so is semisimple. Since every A-module is a quotient of a directsum of copies of A A, this shows that A is semisimple.Before stating the main result of this section, we recall some elementary module theory.7.32 Let A be an F -algebra, and consider modulesM D M 1 ˚ ˚ M nN D N 1 ˚ ˚ N m :Let ˛ be an A-linear map M ! N . For x j 2 M j , let˛.0;:::;0;x j ;0;:::;0/ D .y 1 ;:::;y m /:1 The tensor product D ˝F D 0 of two central simple algebras over F is again a central simple algebra, andhence is isomorphic to M r .D 00 / for some central simple algebra D 00 . DefineŒDŒD 0 D ŒD 00 :This product is associative because of the associativity of tensor products, the isomorphism class of F is anidentity element, and ŒD opp is an inverse for ŒD.

106 7. REPRESENTATIONS OF FINITE GROUPSThen x j 7! y i is an A-linear map M j ! N i , which we denote ˛ij . Thus, ˛ defines anm n matrix whose ij th coefficient is an A-linear map M j ! N i . Conversely, every suchmatrix .˛ij / defines an A-linear map M ! N , namely,0 1 0101 01x 1 ˛11 ˛1j ˛1n x 1 ˛11 .x 1 / C C ˛1n .x n /:: : ::def:x j7!˛i1 ˛ij ˛j nx jD˛i1 .x 1 / C C ˛in .x n /:B C BCBC BC@ : A @ : : : A@: A @ : Ax n ˛m1 ˛mj ˛mn x n ˛m1 .x 1 / C C ˛mn .x n /Thus, we seeHom A .M;N / ' Hom A .M j ;N i / 1j n, 1im(30)(isomorphism of F -vector spaces). When M D N , this becomes an isomorphism of F -algebras. For example, if M is a direct sum of m copies of M 0 , then(m m matrices with coefficients in the ring End A .M 0 /).End A .M / ' M m .End A .M 0 // (31)THEOREM 7.33 Let V be a finite dimensional F -vector space, and let A be an F -subalgebraof End F .V /. If V is semisimple as an A-module, then the centralizer of A in End F .V / isa product of simple F -algebras (hence it is a semisimple F -algebra).PROOF. By assumption, we can write V L i r iS i where the S i are simple A-modules,no two of which are isomorphic. The centralizer of A in End F .V / is End A .V /, andEnd A .V / End A . L i r iS i /. Because Hom A .S j ;S i / D 0 for i ¤ j ,End A . M r i S i / ' Y i End A.r i S i / by (30)' Y i M r i.D i / by (31)where D i D End A .S i /. According to Schur’s lemma (7.24), D i is a division algebra, andtherefore M ri .D i / is a simple F -algebra (see 7.16).✷THEOREM 7.34 Every semisimple F -algebra is isomorphic to a product of simple F -algebras.PROOF. Choose an A-module V on which A acts faithfully, for example, V D A A. Then Ais equal to its double centralizer C.C.A// in End F .V / (see 7.21). According to Theorem7.33, C.A/ is semisimple, and so C.C.A// is a product of simple algebras. ✷Modules over a semisimple F -algebraLet A D B C be a product of F -algebras. A B-module M becomes an A-module withthe action.b;c/m D bm:THEOREM 7.35 Let A be a semisimple F -algebra, say, A D A 1 A t with the A isimple. For each A i , let S i be a simple A i -module (cf. 7.27).

The representations of G 107(a) Each S i is a simple A-module, and every simple A-module is isomorphic to exactlyof of the S i .(b) Every A-module is isomorphic to L r i S i for some r i 2 N, and two modules L r i S iand L ri 0S i are isomorphic if and only if r i D ri 0 for all i.PROOF. (a) It is obvious that each S i is simple when regarded as an A-module, and that notwo of them are isomorphic. It follows from (7.25) that A A L r i S i for some r i 2 N. Let Sbe a simple A-module, and let x be a nonzero element of S. Then the map a 7! axW A A ! Sis surjective, and so its restriction to some S i in A A is nonzero, and hence an isomorphism.(b) The first part follows from (a) and the definition of a semisimple ring, and the secondpart follows from (7.11).✷The representations of GPROPOSITION 7.36 The dimension of the centre of F ŒG as an F -vector space is the numberof conjugacy classes in G.PROOF. Let C 1 ;:::;C t be the conjugacy classes in G, and, for each i, let c i be the elementPa2C ia in F ŒG. We shall prove the stronger statement:centre of F ŒG D F c 1 ˚ ˚ F c t (32)As c 1 ;:::;c t are obviously linearly independent, it suffices to show that they span the centre.For any g 2 G and P a2G m aa 2 F ŒG,X g m aa g 1 D X m agag 1 :a2G a2GThe coefficient of a in the right hand sum is m g 1 ag, and soX g m aa g 1 D X ma2G a2Gg 1 aga:This shows that P a2G m aa lies in the centre of F ŒG if and only if the function a 7! m a isconstant on conjugacy classes, i.e., if and only if P a2G m aa 2 P i F c i.✷REMARK 7.37 An element P a2G m aa of F ŒG can be regarded as a map a 7! m a WG !F . In this way, F ŒG ' Map.G;F /. The action of G on F ŒG corresponds to the action.gf /.a/ D f .g 1 a/ of g 2 G on f WG ! F . In the above proof, we showed that theelements of the centre of F ŒG correspond exactly to the functions f WG ! F that areconstant on each conjugacy class. Such functions are called class functions.In the remainder of this chapter, we assume that F is an algebraically closed field ofcharacteristic zero (e.g., C)PROPOSITION 7.38 The group algebra F ŒG is isomorphic to a product of matrix algebrasover F .PROOF. Recall that, when F has characteristic zero, Maschke’s theorem (7.9) implies thatF ŒG is semisimple, and so is a product of simple algebras (7.35). Each of these is a matrixalgebra over a division algebra (7.23), but the only division algebra over an algebraicallyclosed field is the field itself (7.29).✷

108 7. REPRESENTATIONS OF FINITE GROUPSThe representation G ! GL. F ŒG F ŒG/ is called the regular representation.THEOREM 7.39 (a) The number of isomorphism classes of simple F ŒG-modules is equalto the number of conjugacy classes in G.(b) The multiplicity of any simple representation S in the regular representation is equalto its degree dim F S.(c) Let S 1 ;:::;S t be a set of representatives for the isomorphism classes of simple F G-modules, and let f i D dim F S i . ThenXf 21itiD jGj:PROOF. (a) Under our hypothesis, F ŒG M f1 .F /M ft .F / for some integers f 1 ;:::;f t .According to Theorem 7.35, the number of isomorphism classes of simple F ŒG-modulesis the number of factors t. The centre of a product of F -algebras is the product of theircentres, and so the centre of F ŒG is isomorphic to tF . Therefore t is the dimension of thecentre of F , which we know equals the number of conjugacy classes of G.(b) With the notations of (7.15), M f .F / ' L.1/ ˚ ˚ L.f /.(c) The equality is simply the statementX1it dim F M fi .F / D dim F F ŒG:✷The characters of GRecall that the trace Tr V .˛/ of an endomorphism ˛WV ! V of a vector space V is P a iiwhere .a ij / is the matrix of ˛ with respect to some basis for V . It is independent of thechoice of the basis (the traces of conjugate matrices are equal).From each representation of g 7! g V WG ! GL.V /, we obtain a function V on G; V .g/ D Tr V .g V /;called the character of . Note that V depends only on the isomorphism class of theF ŒG-module V , and that V is a class function. The character is said to be simple (orirreducible) if it is defined by a simple F G-module. The principal character 1 is thatdefined by the trivial representation of G (so 1 .g/ D 1 for all g 2 G), and the regularcharacter reg is that defined by the regular representation. On computing reg .g/ by usingthe elements of G as a basis for F ŒG, one see that reg .g/ is the number of elements a ofG such that ga D a, and so jGj if g D e reg .g/ D0 otherwise.When V has dimension 1, the character V of G is said to be linear. In this case, GL.V / 'F , and so V .g/ D .g/. Therefore, V is a homomorphism G ! F , and so thisdefinition of “linear character” essentially agrees with the earlier one.LEMMA 7.40 For any G-modules V and V 0 , V ˚V 0 D V C V 0:

The characters of G 109PROOF. Compute the matrix of g L with respect to a basis of V ˚ V 0 that is made by combininga basis for V with a basis for V 0 .✷Let S 1 ;:::;S t be a set of representatives for the isomorphism classes of simple F G-modules with S 1 chosen to be the trivial representation, and let 1 ;:::; t be the correspondingcharacters.PROPOSITION 7.41 The functions 1 ;:::; t are linearly independent over F , i.e., if c 1 ;:::;c t 2F are such that P i c i i .g/ D 0 for all g 2 G, then the c i are all zero.PROOF. Write F ŒG M f1 .F / M ft .F /, and let e j D .0;:::;0;1;0;:::;0/. Thene j acts as 1 on S j and as 0 on S j for i ¤ j , and sofj D dim i .e j / DF S j if i D j0 otherwise.(33)Therefore,from which the claim follows.Xi c i i .e j / D c j f j ,✷PROPOSITION 7.42 Two F ŒG-modules are isomorphic if and only if their characters areequal.PROOF. We have already observed that the character of a representation depends onlyon its isomorphism class. Conversely, if V D L 1it c iS i , c i 2 N, then its character is V D P 1it c i i , and (33) shows that c i D V .e i /=f i . Therefore V determines themultiplicity with which each S i occurs in V , and hence it determines the isomorphismclass of V .✷ASIDE 7.43 The proposition is false if F is allowed to have characteristic p ¤ 0. For example, the1 irepresentation i 7! WC0 1 p ! GL 2 .F / of (7.1c) is not trivial, but it has the same characteras the trivial representation. The proposition is false even when the characteristic of F doesn’tdivide the order of the group, because, for any representation G ! GL.V /, the character of therepresentation of G on pV is identically zero.Any function G ! F that can be expressed as a Z-linear combination of characters iscalled a virtual character. 2PROPOSITION 7.44 The simple characters of G form a Z-basis for the virtual charactersof G.PROOF. Let 1 ;:::; t be the simple characters of G. Then the characters of G are exactlythe class functions that can be expressed in the form P m i i , m i 2 N, and so the virtualcharacters are exactly the class functions that can be expressed P m i i , m i 2 Z. Thereforethe simple characters certainly generate the Z-module of virtual characters, and Proposition7.41 shows that they are linearly independent over Z (even over F ). ✷2 Some authors call it a generalized character, but this is to be avoided: there is more than one way togeneralize the notion of a character.

110 7. REPRESENTATIONS OF FINITE GROUPSPROPOSITION 7.45 The simple characters of G form an F -basis for the class functions onG.PROOF. The class functions are the functions from the set of conjugacy classes in G toF . As this set has t elements, they form an F -vector space of dimension t. As the simplecharacters are a set of t linearly independent elements of this vector space, they must forma basis.✷We now assume that F is a subfield of C stable under complex conjugation c 7! Nc.For class functions f 1 and f 2 on G, define.f 1 jf 2 / D 1 XjGjf 1.a/f 2 .a/:a2GLEMMA 7.46 The pairing . j / is an inner product on the F -space of class functions on G.PROOF. We have to check:˘ .f 1 C f 2 jf / D .f 1 jf / C .f 2 jf / for all class functions f 1 ;f 2 ;f ;˘ .cf 1 jf 2 / D c.f 1 ;f 2 / for c 2 F and class functions f 1 ;f 2 ;˘ .f 2 jf 1 / D .f 1 jf 2 / for all class functions f 1 ;f 2 ;˘ .f jf / > 0 for all nonzero class functions f .All of these are obvious from the definition.✷For an F ŒG-module V , V G denotes the submodule of elements fixed by G:LEMMA 7.47 Let be the elementa projector with image V G .PROOF. For any g 2 G,V G D fv 2 V j gv D v for all g 2 Gg1PjGj a2G a of F ŒG. For any F ŒG-module V , V isg D 1jGjXa2G ga D 1 Xa D ; (34)jGj a2Gfrom which it follows that D (in the F -algebra F ŒG). Therefore, for any F ŒGmoduleV , 2 V D V and so V is a projector. If v is in its image, say v D v 0 , thengv D gv 0.34/D v 0 D vand so v lies in V G . Conversely, if v 2 V G , the obviously v D 1v is in the image of .PROPOSITION 7.48 For any F ŒG-module V ,dim F V G D 1 XjGj V .a/:a2GjGjPa2G av D v, and so✷

The character table of a group 111PROOF. Let be as in Lemma 7.47. Because V is a projector, V is the direct sum ofits 0-eigenspace and its 1-eigenspace, and we showed that the latter is V G . Therefore,Tr V . V / D dim F V G . On the other hand, because the trace is a linear function,Tr V . V / D 1jGjXa2G Tr V .a V / D 1 XjGj V .a/:a2G✷THEOREM 7.49 For any F ŒG-modules V and W;dim F Hom F ŒG .V;W / D . V j W /.PROOF. The group G acts on the space Hom F .V;W / of F -linear maps V ! W by therule,.g'/.v/ D g.'.gv//; g 2 G; ' 2 Hom F .V;W /; v 2 V;and Hom F .V;W / G D Hom F G .V;W /.COROLLARY 7.50 If and 0 are simple characters, then.j 0 1 if D 0/ D0 otherwise.Therefore the simple characters form an orthonormal basis for the space of class functionson G.The character table of a groupTo be written.ExamplesTo be written.Exercises7-1 Let C be an n r matrix with coefficients in a field F . Show thatfM 2 M n .F / j MC D 0gis a left ideal in M n .F /, and that every left ideal is of this form for some C .7-2 This exercise shows how to recover a finite group G from its category of representationsover a field k. Let S be a finite set, and let A be the set of maps S ! k.(a) Show that A becomes a commutative ring with the product.f 1 f 2 /.g/ D f 1 .g/f 2 .g/; f 1 , f 2 2 A; g 2 G:Moreover, when we identify c 2 k with the constant function, A becomes a k-algebra.✷

112 7. REPRESENTATIONS OF FINITE GROUPS(b) Show thatA ' Y s2S k s(product of copies of k indexed by the elements of S);and that the k s are exactly the minimal k-subalgebras of A. Deduce that End k-alg .A/ 'Sym.S/.(c) Let .f 1 ;f 2 / 2 A A act on S S by .f 1 ;f 2 /.s 1 ;s 2 / D f 1 .s 1 /f 2 .s 2 /; show that thisdefines a bijection A ˝ A ' Map.S S;k/. Now take S D G.(d) Show that the map r A WG ! End k-linear .A/,.r A .g/f /.g 0 / D f .gg 0 /; f 2 A; g;g 0 2 Gis a representation of G (this is the regular representation).(e) Define WA ! A ˝ A by .f /.g 1 ;g 2 / D f .g 1 g 2 /. Show that, for any homomorphism˛WA ! A of k-algebras such .1˝˛/ı D ı˛, there exists a unique elementg 2 G such that ˛.f / D gf for all f 2 A. [Hint: Deduce from (b) that there exists abijection WG ! G such that .˛f /.g/ D f .g/ for all g 2 G. From the hypothesison ˛, deduce that .g 1 g 2 / D g 1 .g 2 / for all g 1 ;g 2 2 G.R/. Hence .g/ D g .e/for all g 2 G. Deduce that ˛.f / D .e/f for all f 2 A.](f) Show that the following maps are G-equivarianteWk ! A (trivial representation on k; r A on A/mWA ˝ A ! A (r A ˝ r A on A ˝ A; r A on A/WA ! A ˝ A(r A on A; 1 ˝ r A on A ˝ A/:(g) Suppose that we are given, for each finite dimensional representation .V;r V /, a k-linear map V . If the family . V / satisfies the conditionsi) for all representations V , W , V ˝W D V ˝ W Iii) for k with its trivial representation, k D id k ;iii) for all G-equivariant maps ˛WV ! W , W ı ˛ D ˛ ı V Ithen there exists a unique g 2 G.R/ such that V D r V .g/ for all V . [Hint: show that A satisfies the conditions of (d).]NOTES For a historical account of the representation theory of finite groups, emphasizing the workof “the four principal contributors to the theory in its formative stages: Ferdinand Georg Frobenius,William Burnside, Issai Schur, and Richard Brauer”, see Curtis 1999.

APPENDIX AAdditional Exercises34. Prove that a finite group G having just one maximal subgroup must be a cyclic p-group,p prime.35. Let a and b be two elements of S 76 . If a and b both have order 146 and ab D ba, whatare the possible orders of the product ab?37. Suppose that the group G is generated by a set X.(a) Show that if gxg 1 2 X for all x 2 X; g 2 G, then the commutator subgroup of G isgenerated by the set of all elements xyx 1 y 1 for x;y 2 X.(b) Show that if x 2 D 1 for all x 2 X, then the subgroup H of G generated by the set ofall elements xy for x;y 2 X has index 1 or 2.38. Suppose p 3 and 2p 1 are both prime numbers (e.g., p D 3;7;19;31;:::/. Prove, ordisprove by example, that every group of order p.2p 1/ is commutative.39. Let H be a subgroup of a group G. Prove or disprove the following:(a) If G is finite and P is a Sylow p-subgroup, then H \ P is a Sylow p-subgroup ofH .(b) If G is finite, P is a Sylow p-subgroup, and H N G .P /, then N G .H / D H .(c) If g is an element of G such that gHg 1 H , then g 2 N G .H /.40. Prove that there is no simple group of order 616.41. Let n and k be integers 1 k n. Let H be the subgroup of S n generated by thecycle .a 1 :::a k /. Find the order of the centralizer of H in S n . Then find the order of thenormalizer of H in S n . [The centralizer of H is the set of g 2 G such ghg 1 D h for allh 2 H . It is again a subgroup of G.]42. Prove or disprove the following statement: if H is a subgroup of an infinite group G,then for all x 2 G, xH x 1 H H) x 1 H x H .43. Let H be a finite normal subgroup of a group G, and let g be an element of G. Supposethat g has order n and that the only element of H that commutes with g is 1. Show that:(a) the mapping h 7! g 1 h 1 gh is a bijection from H to H ;(b) the coset gH consists of elements of G of order n.113

114 A. ADDITIONAL EXERCISES44. Show that if a permutation in a subgroup G of S n maps x to y, then the normalizers ofthe stabilizers Stab.x/ and Stab.y/ of x and y have the same order.45. Prove that if all Sylow subgroups of a finite group G are normal and abelian, then thegroup is abelian.46. A group is generated by two elements a and b satisfying the relations: a 3 D b 2 , a m D 1,b n D 1 where m and n are positive integers. For what values of m and n can G be infinite.47. Show that the group G generated by elements x and y with defining relations x 2 Dy 3 D .xy/ 4 D 1 is a finite solvable group, and find the order of G and its successive derivedsubgroups G 0 , G 00 , G 000 .48. A group G is generated by a normal set X of elements of order 2. Show that the commutatorsubgroup G 0 of G is generated by all squares of products xy of pairs of elementsof X.49. Determine the normalizer N in GL n .F / of the subgroup H of diagonal matrices, andprove that N=H is isomorphic to the symmetric group S n .50. Let G be a group with generators x and y and defining relations x 2 , y 5 , .xy/ 4 . Whatis the index in G of the commutator group G 0 of G.51. Let G be a finite group, and H the subgroup generated by the elements of odd order.Show that H is normal, and that the order of G=H is a power of 2.52. Let G be a finite group, and P a Sylow p-subgroup. Show that if H is a subgroup ofG such that N G .P / H G, then(a) the normalizer of H in G is H ;(b) .G W H / 1 (mod p).53. Let G be a group of order 33 25. Show that G is solvable. (Hint: A first step is to finda normal subgroup of order 11 using the Sylow theorems.)54. Suppose that ˛ is an endomorphism of the group G that maps G onto G and commuteswith all inner automorphisms of G. Show that if G is its own commutator subgroup, then˛x D x for all x in G.55. Let G be a finite group with generators s and t each of order 2. Let n D .G W 1/=2.(a) Show that G has a cyclic subgroup of order n. Now assume n odd.(b) Describe all conjugacy classes of G.(c) Describe all subgroups of G of the form C.x/ D fy 2 Gjxy D yxg, x 2 G.(d) Describe all cyclic subgroups of G.(e) Describe all subgroups of G in terms of (b) and (d).(f) Verify that any two p-subgroups of G are conjugate .p prime).56. Let G act transitively on a set X. Let N be a normal subgroup of G, and let Y be theset of orbits of N in X. Prove that:(a) There is a natural action of G on Y which is transitive and shows that every orbit ofN on X has the same cardinality.(b) Show by example that if N is not normal then its orbits need not have the samecardinality.

11557. Prove that every maximal subgroup of a finite p-group is normal of prime index .p isprime).58. A group G is metacyclic if it has a cyclic normal subgroup N with cyclic quotient G=N .Prove that subgroups and quotient groups of metacyclic groups are metacyclic. Prove ordisprove that direct products of metacyclic groups are metacylic.59. Let G be a group acting doubly transitively on X, and let x 2 X. Prove that:(a) The stabilizer G x of x is a maximal subgroup of G.(b) If N is a normal subgroup of G, then either N is contained in G x or it acts transitivelyon X.60. Let x;y be elements of a group G such that xyx 1 D y 5 , x has order 3, and y ¤ 1 hasodd order. Find (with proof) the order of y.61. Let H be a maximal subgroup of G, and let A be a normal subgroup of H and suchthat the conjugates of A in G generate it.(a) Prove that if N is a normal subgroup of G, then either N H or G D NA.(b) Let M be the intersection of the conjugates of H in G. Prove that if G is equal to itscommutator subgroup and A is abelian, then G=M is a simple group.62. (a) Prove that the centre of a nonabelian group of order p 3 , p prime, has order p.(b) Exhibit a nonabelian group of order 16 whose centre is not cyclic.63. Show that the group with generators ˛ and ˇ and defining relations˛2 D ˇ2 D .˛ˇ/ 3 D 1is isomorphic with the symmetric group S 3 of degree 3 by giving, with proof, an explicitisomorphism.64. Prove or give a counter-example:(a) Every group of order 30 has a normal subgroup of order 15.(b) Every group of order 30 is nilpotent.65. Let t 2 Z, and let G be the group with generators x;y and relations xyx 1 D y t ,x 3 D 1.(a) Find necessary and sufficient conditions on t for G to be finite.(b) In case G is finite, determine its order.66. Let G be a group of order pq, p ¤ q primes.(a) Prove G is solvable.(b) Prove that G is nilpotent ” G is abelian ” G is cyclic.(c) Is G always nilpotent? (Prove or find a counterexample.)67. Let X be a set with p n elements, p prime, and let G be a finite group acting transitivelyon X. Prove that every Sylow p-subgroup of G acts transitively on X.68. Let G D ha;b;c j bc D cb, a 4 D b 2 D c 2 D 1; aca 1 D c, aba 1 D bci. Determinethe order of G and find the derived series of G.69. Let N be a nontrivial normal subgroup of a nilpotent group G. Prove that N \ Z.G/ ¤1.70. Do not assume Sylow’s theorems in this problem.

116 A. ADDITIONAL EXERCISES(a) Let H be a subgroup of a finite group G, and P a Sylow p-subgroup of G. Provethat there exists an x 2 G such that xP x 01 \ H is a Sylow 1 p-subgroup of H .1 :::(b) Prove that the group of n n matrices B0 1 C@ ::: A is a Sylow p-subgroup of0 1GL n .F p /.(c) Indicate how (a) and (b) can be used to prove that any finite group has a Sylow p-subgroup.71. Suppose H is a normal subgroup of a finite group G such that G=H is cyclic of ordern, where n is relatively prime to .G W 1/. Prove that G is equal to the semidirect productH ⋊ S with S a cyclic subgroup of G of order n.72. Let H be a minimal normal subgroup of a finite solvable group G. Prove that H isisomorphic to a direct sum of cyclic groups of order p for some prime p.73. (a) Prove that subgroups A and B of a group G are of finite index in G if and only ifA \ B is of finite index in G.(b) An element x of a group G is said to be an FC-element if its centralizer C G .x/ has finiteindex in G. Prove that the set of all F C elements in G is a normal.74. Let G be a group of order p 2 q 2 for primes p > q. Prove that G has a normal subgroupof order p n for some n 1.75. (a) Let K be a finite nilpotent group, and let L be a subgroup of K such that LıK D K,where ıK is the derived subgroup. Prove that L D K. [You may assume that a finite groupis nilpotent if and only if every maximal subgroup is normal.](b) Let G be a finite group. If G has a subgroup H such that both G=ıH and H arenilpotent, prove that G is nilpotent.76. Let G be a finite noncyclic p-group. Prove that the following are equivalent:(a) .G W Z.G// p 2 .(b) Every maximal subgroup of G is abelian.(c) There exist at least two maximal subgroups that are abelian.77. Prove that every group G of order 56 can be written (nontrivially) as a semidirectproduct. Find (with proofs) two non-isomorphic non-abelian groups of order 56.78. Let G be a finite group and ' W G ! G a homomorphism.(a) Prove that there is an integer n 0 such that ' n .G/ D ' m .G/ for all integers m n.Let ˛ D ' n .(b) Prove that G is the semi-direct product of the subgroups Ker˛ and Im˛.(c) Prove that Im˛ is normal in G or give a counterexample.79. Let S be a set of representatives for the conjugacy classes in a finite group G and let Hbe a subgroup of G. Show that S H H) H D G.80. Let G be a finite group.(a) Prove that there is a unique normal subgroup K of G such that (i) G=K is solvableand (ii) if N is a normal subgroup and G=N is solvable, then N K.(b) Show that K is characteristic.(c) Prove that K D ŒK;K and that K D 1 or K is nonsolvable.

APPENDIX BSolutions to the ExercisesThese solutions fall somewhere between hints and complete solutions. Students were expectedto write out complete solutions.1-1 By inspection, the only element of order 2 is c D a 2 D b 2 . Since gcg 1 also has order2, it must equal c, i.e., gcg 1 D c for all g 2 Q. Thus c commutes with all elements of Q,and f1;cg is a normal subgroup of Q. The remaining subgroups have orders 1, 4, or 8, andare automatically normal (see 1.36a).1-2 The product ab D 1 1, and0 1 n 1 1 1D0 1n0 11-3 Consider the subsets fg;g 1 g of G. Each set has exactly 2 elements unless g has order1 or 2, in which case it has 1 element. Since G is a disjoint union of these sets, there mustbe a (nonzero) even number of sets with 1 element, and hence at least one element of order2.1-4 The symmetric group S n contains a subgroup that is a direct product of subgroups S n1 ,. . . , S nr .1-5 Because the group G=N has order n, .gN / n D 1 for every g 2 G (see 1.27). But.gN / n D g n N , and so g n 2 N . For the second statement, consider the subgroup f1;sg ofD 3 . It has index 3 in D 3 , but the element t has order 2, and so t 3 D t … f1;sg.1-6 (a) Let a;b 2 G. We are given that a 2 D b 2 D .ab/ 2 D e. In particular, abab D e. Onmultiplying this on right by ba, we find that ab D ba. (b) Show by induction that0 1n01n.n 1/1 a b 1 na nb C2ac@ 0 1 cAD @ 0 1 nc A:0 0 1 0 0 11-7 Let H be a subgroup of a finite group G, and assume that H contains at least oneelement from each conjugacy class of G. Then G is the union of the conjugates of H , butthere are at most .GWH / of these, and they overlap (at least) in f1g, and so this can happenonly if H D G. (According to Serre 2003, this result goes back to Jordan in the 1870s.)117.

118 B. SOLUTIONS TO THE EXERCISES1-8 Commensurability is obviously reflexive and symmetric, and so it suffices to provetransitivity. We shall use that if a subgroup H of a group G has finite index in G, thenH \ G 0 has finite index in G 0 for any subgroup G 0 of G (because the natural map G 0 =H \G 0 ! G=H is injective). Using this, it follows that if H 1 and H 3 are both commensurablewith H 2 , then H 1 \ H 2 \ H 3 is of finite index in H 1 \ H 2 and in H 2 \ H 3 (and thereforealso in H 1 and H 3 ). As H 1 \ H 3 H 1 \ H 2 \ H 3 , it also has finite index in each of H 1and H 3 .1-9 By assumption, the set G is nonempty, so let a 2 G. Because G satisfies the cancellationlaw, the map x 7! axWG ! G is a permutuation of G, and some power of thispermutation is the identity permutation. Therefore, for some n 1, a n x D x for all x 2 G,and so a n is a left neutral element. By counting, one sees that every element has a leftinverse, and so we can apply (1.10a).2-1 The key point is that hai D ha 2 i ha n i. Apply (1.50) to see that D 2n breaks up as aproduct.2-2 Note first that any group generated by a commuting set of elements must be commutative,and so the group G in the problem is commutative. According to (2.8), any mapfa 1 ;:::;a n g ! A with A commutative extends uniquely to homomorphism G ! A, and soG has the universal property that characterizes the free abelian group on the generators a i .2-3 (a) If a ¤ b, then the word a ab 1 b 1 is reduced and ¤ 1. Therefore, if a n b n D1, then a D b. (b) is similar. (c) The reduced form of x n , x ¤ 1, has length at least n.2-4 (a) Universality. (b) C 1 C 1 is commutative, and the only commutative free groupsare 1 and C 1 . (c) Suppose a is a nonempty reduced word in x 1 ;:::;x n , say a D x i (orxi1 ). For j ¤ i, the reduced form of Œx j ;a D defx j axj1 a 1 can’t be empty, and so a andx j don’t commute.2-5 The unique element of order 2 is b 2 . Since gb 2 g 1 also has order 2 for any g 2 Q n ,we see that gb 2 g 1 D b 2 , and so b 2 lies in the centre. [Check that it is the full centre.]The quotient group Q n =hb 2 i has generators a and b, and relations a 2n 2 D 1, b 2 D 1,bab 1 D a 1 , which is a presentation for D 2 n 2 (see 2.9).2-6 (a) A comparison of the presentation D n D hr;s j r n ;s 2 ;srsr D 1i with that for Gsuggests putting r D ab and s D a. Check (using 2.8) that there are homomorphisms:D n ! G; r 7! ab; s 7! a; G ! D n ; a 7! s; b 7! s 1 r.The composites D n ! G ! D n and G ! D n ! G are the both the identity map on generatingelements, and therefore (2.8 again) are identity maps. (b) Omit.2-7 The hint gives ab 3 a 1 D bc 3 b 1 . But b 3 D 1. So c 3 D 1. Since c 4 D 1, this forcesc D 1. From acac 1 D 1 this gives a 2 D 1. But a 3 D 1. So a D 1. The final relation thengives b D 1.

1192-8 The elements x 2 , xy, y 2 lie in the kernel, and it is easy to see that hx;yjx 2 ;xy;y 2 ihas order (at most) 2, and so they must generate the kernel (at least as a normal group —the problem is unclear). One can prove directly that these elements are free, or else applythe Nielsen-Schreier theorem (2.6). Note that the formula on p. 34 (correctly) predicts thatthe kernel is free of rank 2 2 2 C 1 D 32-9 We have to show that if s and t are elements of a finite group satisfying t 1 s 3 t D s 5 ,then the given element g is equal to 1. Because the group is finite, s n D 1 for some n. If3jn, the proof is easy, and so we suppose that gcd.3;n/ D 1. But thenand so s 3r D s. HenceNow,tg D s 1 .t3r C nr 0 D 1, some r;r 0 2 Z;1 st D t1 s 1 t/s.t1 s 3r t D .t1 s 3 t/ r D s 5r :1 st/ D s 1 s 5r ss 5r D 1;as required. [In such a question, look for a pattern. Note that g has two conjugates in it, asdoes the relation for G, and so it is natural to try to relate them.]3-1 Let N be the unique subgroup of order 2 in G. Then G=N has order 4, but there is nosubgroup Q G of order 4 with Q \ N D 1 (because every group of order 4 contains agroup of order 2), and so G ¤ N ⋊ Q for any Q. A similar argument applies to subgroupsN of order 4.3-2 For any g 2 G, gMg 1 is a subgroup of order m, and therefore equals M . Thus M(similarly N ) is normal in G, and MN is a subgroup of G. The order of any element ofM \ N divides gcd.m;n/ D 1, and so equals 1. Now (1.51) shows that M N MN ,which therefore has order mn, and so equals G.3-3 Show that GL 2 .F 2 / permutes the 3 nonzero vectors in F 2 F 2 (2-dimensional vectorspace over F 2 ).3-4 The following solutions were suggested by readers. We write the quaternion group asQ D f˙1;˙i;˙j;˙kg:(A) Take a cube. Write the six elements of Q of order 4 on the six faces with i opposite i,etc.. Each rotation of the cube induces an automorphism of Q, and Aut.Q/ is the symmetrygroup of the cube, S 4 . (B) The group Q has one element of order 2, namely 1, and sixelements of order 4, namely, ˙i, ˙j , ˙k. Any automorphism ˛ of Q must map 1 toitself and permute the elements of order 4. Note that ij D k, jk D i, ki D j , so ˛ mustsend the circularly ordered set i;j;k to a similar set, i.e., to one of the eight sets in thefollowing table:i j k i j ki j k i j ki k j i k ji k j i k jBecause ˛. 1/ D 1, ˛ must permute the rows of the table, and it is not difficult to see thatall permutations are possible.

120 B. SOLUTIONS TO THE EXERCISES3-5 The pair80198019< 1 0 b = < a 0 0 =N D @ 0 1 cA: ; and Q D @ 0 a 0A:;0 0 10 0 dsatisfies the conditions (i), (ii), (iii) of (3.8). For example, for (i) (Maple says that)0a 010b 1 010b a 01b@ 0 a c A@0 1 cA@0 a c A0 0 d 0 0 1 0 0 d10b1 0d C 1 1.b C ab/dD @c0 1d C 1 .c C ac/ Ad0 0 1It is not a direct product of the two groups because it is not commutative.3-6 Let g generate C 1 . Then the only other generator is g 1 , and the only nontrivialautomorphism is g 7! g 1 . Hence Aut.C 1 / D f˙1g. The homomorphism S 3 ! Aut.S 3 /is injective because Z.S 3 / D 1, but S 3 has exactly 3 elements a 1 ;a 2 ;a 3 of order 2 and 2elements b;b 2 of order 3. The elements a 1 ;b generate S 3 , and there are only 6 possibilitiesfor ˛.a 1 /, ˛.b/, and so S 3 ! Aut.S 3 / is also onto.3-7 (a) The element g o.q/ 2 N , and so has order dividing jN j. (c) The element g D.1;4;3/.2;5/, and so this is obvious. (d) By the first part, ..1;0;:::;0/;q/ p D ..1;:::;1/;1/,and .1;:::;1/ has order p in .C p / p . (e) We have .n;q/.n;q/ D .nn 1 ;qq/ D .1;1/:3-8 Let n q 2 Z.G/. Then.n q/.1 q 0 / D n qq 0.1 q 0 /.n q/ D q 0 nq 0 1 q 0 qall q 0 2 QH) n 2 C N .Q/q 2 Z.Q/and.n q/.n 0 1/ D nqn 0 q 1 q.n 0 1/.n q/ D n 0 n qn 0 2 NThe converse and the remaining statements are easy.H) n 1 n 0 n D qn 0 q 1 :4-1 Let 'WG=H 1 ! G=H 2 be a G-map, and let '.H 1 / D gH 2 . For a 2 G, '.aH 1 / Da'.H 1 / D agH 2 . When a 2 H 1 , '.aH 1 / D gH 2 , and so agH 2 D gH 2 ; hence g 1 ag 2H 2 , and so a 2 gH 2 g 1 . We have shown H 1 gH 2 g 1 . Conversely, if g satisfies thiscondition, the aH 1 7! agH 2 is a well-defined map of G-sets.4-2 Let H be a proper subgroup of G, and let N D N G .H /. The number of conjugates ofH is .G W N / .G W H / (see 4.8). Since each conjugate of H has .H W 1/ elements and theconjugates overlap (at least) in f1g, we see thatˇˇ[gHg 1ˇˇˇ < .G W H /.H W 1/ D .G W 1/:For the second part, choose S to be a set of representatives for the conjugacy classes.4-3 According to 4.17, 4.18, there is a normal subgroup N of order p 2 , which is commutative.Now show that G has an element c of order p not in N , and deduce that G D N ⋊hci,etc..

1214-4 Let H be a subgroup of index p, and let N be the kernel of G ! Sym.G=H / — it isthe largest normal subgroup of G contained in H (see 4.22). If N ¤ H , then .H W N / isdivisible by a prime q p, and .G W N / is divisible by pq. But pq doesn’t divide pŠ —contradiction.4-5 Embed G into S 2m , and let N D A 2m \ G. Then G=N ,! S 2m =A 2m D C 2 , and so.G W N / 2. Let a be an element of order 2 in G, and let b 1 ;:::;b m be a set of rightcoset representatives for hai in G, so that G D fb 1 ;ab 1 ;:::;b m ;ab m g. The image of a inS 2m is the product of the m transpositions .b 1 ;ab 1 /;:::;.b m ;ab m /, and since m is odd, thisimplies that a … N .4-6 The set X of k-cycles in S n is normal, and so the group it generates is normal (1.38).But, when n 5, the only nontrivial normal subgroups of S n are A n and S n itself. If k isodd, then X is contained in A n , and if k is even, then it isn’t.4-7 (a) The number of possible first rows is 2 3 1; of second rows 2 3 2; of third rows2 3 2 2 ; whence .G W 1/ D 7 6 4 D 168. (b) Let V D F 3 2 . Then jV j D 23 D 8. Each linethrough the origin contains exactly one point ¤ origin, and so jXj D 7. (c) We make a listof possible characteristic and minimal polynomials:Characteristic poly. Min’l poly. Size Order of element in class1 X 3 C X 2 C X C 1 X C 1 1 12 X 3 C X 2 C X C 1 .X C 1/ 2 21 23 X 3 C X 2 C X C 1 .X C 1/ 3 42 44 X 3 C 1 D .X C 1/.X 2 C X C 1/ Same 56 35 X 3 C X C 1 (irreducible) Same 24 76 X 3 C X 2 C 1 (irreducible) Same 24 7Here size denotes the number of elements in the conjugacy class. Case 5: Let ˛ be anendomorphism with characteristic polynomial X 3 C X C 1. Check from its minimal polynomialthat ˛7 D 1, and so ˛ has order 7. Note that V is a free F 2 Œ˛-module of rank one,and so the centralizer of ˛ in G is F 2 Œ˛ \ G D h˛i. Thus jC G .˛/j D 7, and the number ofelements in the conjugacy class of ˛ is 168=7 D 24. Case 6: Exactly the same as Case 5.Case 4: Here V D V 1 ˚ V 2 as an F 2 Œ˛-module, andEnd F2 Œ˛.V / D End F2 Œ˛.V 1 / ˚ End F2 Œ˛.V 2 /:Deduce that jC G .˛/j D 3, and so the number of conjugates of ˛ is 1683D 56. Case 3: HereC G .˛/ D F 2 Œ˛ \ G D h˛i, which has order 4. Case 1: Here ˛ is the identity element.Case 2: Here V D V 1 ˚ V 2 as an F 2 Œ˛-module, where ˛ acts as 1 on V 1 and has minimalpolynomial X 2 C 1 on V 2 . Either analyse, or simply note that this conjugacy class containsall the remaining elements. (d) Since 168 D 2 3 3 7, a proper nontrivial subgroup H ofG will have order2;4;8;3;6;12;24;7;14;28;56;21;24, or 84:If H is normal, it will be a disjoint union of f1g and some other conjugacy classes, and so.N W 1/ D 1 C P c i with c i equal to 21, 24, 42, or 56, but this doesn’t happen.4-8 Since G=Z.G/ ,! Aut.G/, we see that G=Z.G/ is cyclic, and so by (4.19) that G iscommutative. If G is finite and not cyclic, it has a factor C p r C p s etc..

122 B. SOLUTIONS TO THE EXERCISES4-9 Clearly .ij / D .1j /.1i/.1j /. Hence any subgroup containing .12/;.13/;::: contains alltranspositions, and we know S n is generated by transpositions.4-10 Note that C G .x/\H D C H .x/, and so H=C H .x/ H C G .x/=C G .x//. Prove eachclass has the same number c of elements. ThenjKj D .G W C G .x// D .G W H C G .x//.H C G .x/ W C G .x// D kc:4-11 (a) The first equivalence follows from the preceding problem. For the second, note that commutes with all cycles in its decomposition, and so they must be even (i.e., have oddlength); if two cycles have the same odd length k, one can find a product of k transpositionswhich interchanges them, and commutes with ; conversely, show that if the partition of ndefined by consists of distinct integers, then commutes only with the group generatedby the cycles in its cycle decomposition. (b) List of conjugacy classes in S 7 , their size,parity, and (when the parity is even) whether it splits in A 7 .Cycle Size Parity Splits in A 7 ‹ C 7 ./ contains1 .1/ 1 E N2 .12/ 21 O3 .123/ 70 E N .67/4 .1234/ 210 O5 .12345/ 504 E N .67/6 .123456/ 840 O7 .1234567/ 720 E Y 720 doesn’t divide 25208 .12/.34/ 105 E N .67/9 .12/.345/ 420 O10 .12/.3456/ 630 E N .12/11 .12/.3456/ 504 O12 .123/.456/ 280 E N .14/.25/.36/13 .123/.4567/ 420 O14 .12/.34/.56/ 105 O15 .12/.34/.567/ 210 E N .12/4-12 According to Gap, n D 6, a 7! .13/.26/.45/, b 7! .12/.34/.56/.4-13 Since Stab.gx 0 / D g Stab.x 0 /g 1 , if H Stab.x 0 / then H Stab.x/ for all x, andso H D 1, contrary to hypothesis. Now Stab.x 0 / is maximal, and so H Stab.x 0 / D G,which shows that H acts transitively.5-1 Let p be a prime dividing jGj and let P be a Sylow p-subgroup, of order p m say. Theelements of P all have order dividing p m , and it has at most1 C p C C p m 1 D pm 1p 1 < pmelements of order dividing p m 1 ; therefore P must have an element of order p m , and so itis cyclic. Each Sylow p-subgroup has exactly p m elements of order dividing p m , and sothere can be only one. Now (5.9) shows that G is a product of its Sylow subgroups.

6-2 No, D 4 and the quaternion group have isomorphic commutator subgroups and quotientgroups but are not isomorphic. Similarly, S n and A n C 2 are not isomorphic when n 5.123

APPENDIX CTwo-Hour Examination1. Which of the following statements are true (give brief justifications for each of (a), (b),(c), (d); give a correct set of implications for (e)).(a) If a and b are elements of a group, then a 2 D 1; b 3 D 1 H) .ab/ 6 D 1.(b) The following two elements are conjugate in S 7 : 1 2 3 4 5 6 7;3 4 5 6 7 2 1 1 2 3 4 5 6 7:2 3 1 5 6 7 4(c) If G and H are finite groups and G A 594 H A 594 ; then G H .(d) The only subgroup of A 5 containing .123/ is A 5 itself.(e) Nilpotent H) cyclic H) commutative H) solvable (for a finite group).2. How many Sylow 11-subgroups can a group of order 110 D 2 5 11 have? Classifythe groups of order 110 containing a subgroup of order 10. Must every group of order 110contain a subgroup of order 10?3. Let G be a finite nilpotent group. Show that if every commutative quotient of G is cyclic,then G itself is cyclic. Is the statement true for nonnilpotent groups?4. (a) Let G be a subgroup of Sym.X/, where X is a set with n elements. If G is commutativeand acts transitively on X, show that each element g ¤ 1 of G moves every elementof X. Deduce that .G W 1/ n.(b) For each m 1, find a commutative subgroup of S 3m of order 3 m .(c) Show that a commutative subgroup of S n has order 3 n 3 .5. Let H be a normal subgroup of a group G, and let P be a subgroup of H . Assume thatevery automorphism of H is inner. Prove that G D H N G .P /.6. (a) Describe the group with generators x and y and defining relation yxy 1 D x 1 .(b) Describe the group with generators x and y and defining relations yxy 1 D x 1 ,xyx 1 D y 1 .You may use results proved in class or in the notes, but you should indicate clearly whatyou are using.125

126 C. TWO-HOUR EXAMINATIONSOLUTIONS1. (a) False: in ha;bja 2 ;b 3 i, ab has infinite order.(b) True, the cycle decompositions are (1357)(246), (123)(4567).(c) True, use the Krull-Schmidt theorem.(d) False, the group it generates is proper.(e) Cyclic H) commutative H) nilpotent H) solvable.2. The number of Sylow 11-subgroups s 11 D 1;12;::: and divides 10. Hence there is onlyone Sylow 11-subgroup P . HaveG D P ⋊ H; P D C 11 ; H D C 10 or D 5 :Now have to look at the maps W H ! Aut.C 11 / D C 10 . Yes, by the Schur-Zassenhauslemma.3. Suppose G has class > 1. Then G has quotient H of class 2. Consider1 ! Z.H / ! H ! H=Z.H / ! 1:Then H is commutative by (4.17), which is a contradiction. Therefore G is commutative,and hence cyclic.Alternatively, by induction, which shows that G=Z.G/ is cyclic.No! In fact, it’s not even true for solvable groups (e.g., S 3 ).4. (a) If gx D x, then ghx D hgx D hx. Hence g fixes every element of X, and so g D 1.Fix an x 2 X; then g 7! gx W G ! X is injective. [Note that Cayley’s theorem gives anembedding G ,! S n , n D .G W 1/.](b) Partition the set into subsets of order 3, and let G D G 1 G m .(c) Let O 1 ;:::;O r be the orbits of G, and let G i be the image of G in Sym.O i /. ThenG ,! G 1 G r , and so (by induction),.G W 1/ .G 1 W 1/.G r W 1/ 3 n 13 3nr3 D 3n3 :5. Let g 2 G, and let h 2 H be such that conjugation by h on H agrees with conjugationby g. Then gP g 1 D hP h 1 , and so h 1 g 2 N G .P /.6. (a) It’s the group .G D hxi ⋊ hyi D C 1 ⋊ C 1with WC 1 ! Aut.C 1 / D ˙1. Alternatively, the elements can be written uniquely in theform x i y j , i;j 2 Z, and yx D x 1 y.(b) It’s the quaternion group. From the two relations getand so x 2 D y 2 . The second relation impliesyx D x 1 y; yx D xy 1xy 2 x 1 D y 2 ;D y 2 ;and so y 4 D 1.Alternatively, the Todd-Coxeter algorithm shows that it is the subgroup of S 8 generatedby .1287/.3465/ and .1584/.2673/.

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128 BIBLIOGRAPHYRONAN, M. 2006. Symmetry and the monster. One of the greatest quests of mathematics.Oxford University Press, Oxford.ROTMAN, J. J. 1995. An introduction to the theory of groups, volume 148 of GraduateTexts in Mathematics. Springer-Verlag, New York, fourth edition.SCHUPP, P. E. 1987. A characterization of inner automorphisms. Proc. Amer. Math. Soc.101:226–228.SERRE, J.-P. 1980. Trees. Springer-Verlag, Berlin. Translated from the French by JohnStillwell.SERRE, J.-P. 2003. On a theorem of Jordan. Bull. Amer. Math. Soc. (N.S.) 40:429–440.SOLOMON, R. 2001. A brief history of the classification of the finite simple groups. Bull.Amer. Math. Soc. (N.S.) 38:315–352.SYLOW, M. L. 1872. Théorèmes sur les groupes de substitutions. Math. Ann. 5:584–594.WILD, M. 2005. The groups of order sixteen made easy. Amer. Math. Monthly 112:20–31.

Indexactiondoubly transitive, 55effective, 53faithful, 53free, 55imprimitive, 68k-fold transitive, 55left, 53linear, 96primitive, 68right, 54transitive, 55trivial, 53algebradivision, 101group, 99quaternion, 101semisimple, 105simple, 101algorithmTodd-Coxeter, 38, 66automorphisminner, 41of a bilinear form, 10of a group, 41outer, 41basisfor a commutative group, 25block, 69centralizerof element, 56of subgroup, 113centralizer, of a subalgebra, 102centreof a group, 12classnilpotency, 87class equationclass, 58commutator, 24, 35composition factors, 83conjugacy class, 54cosetleft, 16right, 16Coxeter system, 38cycle, 62dimension, 102disjoint cyclesdisjoint, 62elementneutral, 8elementary divisors, 26equivariant map, 54exact sequence, 48exponentof a group, 30, 37extensioncentral, 48isomorphic, 48split, 49extension, of groups, 48faithful representation, 95flagfull, 76Frattini’s argument, 89G-map, 54G-set, 53generates, 13generatorsof a group, 35group, 74-, 14A-, 90abelian, 9additive, 7alternating, 62Burnside, 37commutative, 9complete, 42Coxeter, 38129

130 INDEXcyclic, 13dihedral, 13finite reflection, 39finitely presented, 37free, 33free abelian, 35general linear, 10indecomposable, 92isotropy, 55metabelian, 87metacyclic, 115multiplicative, 7nilpotent, 87of rigid motions, 54of symmetries, 9orthogonal, 10p, 8permutation, 9primitive, 68quaternion, 14generalized, 35quotient, 21reflection, 39simple, 19soluble, 83solvable, 66, 83special linear, 20symplectic, 10with operators, 90group.factor, 21groupsof order 12, 79of order 2 m p n , m 3., 80of order 2p, 58of order 30, 78of order 60, 80of order 99, 77of order p, 17of order p 2 , 59of order p 3 , 79of order pq, 78of small order, 14homogeneous, 55homomorphismadmissible, 90of groups, 15indexof a subgroup, 17invariant factors, 26inverse, 8inversion, of a permutation, 61isomorphismof G-sets, 54of groups, 8, 15kernel, 20Klein Viergruppe, 14lengthof a subnormal series, 81solvable, 86length of a cycle, 62map, of G-sets, 54modulesemisimple, 95simple, 95monoidfree, 31negative, 8normalizerof a subgroup, 56opposite algebra, 95orbit, 54orderof a group, 8of an element, 9partitionof a natural number, 64stabilized, 68permutationeven, 61odd, 61presentationof a group, 35problemBurnside, 37restricted Burnside, 37word, 37productdirect, 9, 23knit, 48semidirect, 44Zappa-Szép, 48projector, 98quotients, of a normal seriesof a normal series, 81rankof a commutative group, 26of a Coxeter system, 38of a free group, 34

Index 131reduced formof a word, 32reflection, 39relations, 35defining, 35representationlinear, 96matrix, 95seriesadmissible subnormal, 91ascending central, 87composition, 81derived, 86normal, 81solvable, 83subnormal, 81without repetitions, 81signature, 61skew field, 101stabilizerof a subset, 56of an element, 55stable subsetstable, 54subgroup, 12A-invariant, 90admissible, 90characteristic, 43commutator, 85first derived, 85generated by a set, 13normal, 18normal generated by a set, 20second derived, 86Sylow p-, 73torsion, 10subsetnormal, 20supportof a cycle, 62isomorphism, 22Jordan-Hölder, 82Krull-Schmidt, 92Lagrange, 17Maschke, 97Nielsen-Schreier, 34nilpotency condition, 89primitivity condition, 69Schur-Zassenhaus, 49structure of commutative groups, 26structure of Coxeter groups, 39Sylow I, 73Sylow II, 75Sylow subgroups of subgroups, 77transposition, 14word, 31reduced, 32wordsequivalent, 33tablemultiplication, 11theoremCauchy, 58Cayley, 16centre of a p-group, 59commutative groups have bases, 25correspondence, 22double centralizer, 103factorization of homomorphisms, 22Feit-Thompson, 84Galois, 65