COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
166<br />
Marshall Hall Jr.<br />
Since t moves all letters it must interchange the orbits of odd length. In<br />
particular if t = (03, Ai) then Ai is one of the numbers 51,. . . ,57,65,. . , ,71,<br />
72,. . ., 78. As tat = a-l,<br />
where(-,Ai,-,Ai...)<br />
(02,03,04,05,06,07,08) (6.15)<br />
(-, Ai, -, Ai, . . . . . . . a)<br />
is in some order (51, 57, 56, 55, 54, 53, 52),<br />
(65,71,70,69,68,67,66) or (72,78,77,76,75,74,73) and also b2dl = zi .<br />
0 i<br />
We have<br />
b2a4 = (OO)(Ol, 13, 35,26,04, 24, 15, 30)(02, 20,21, 14)<br />
(03, 10, 18, 19, 33, 05, 11, 32)(06, 36, 16, 17)<br />
(07, 23, 34, 22,09,27,08, 25)( 12, 31, 29,28)<br />
(37, 41, 60, 88, 56, 53, 90, 61)(38, 71, 66, 40, 44, 62, 59, 50)<br />
(39, 79)(42, 57, 73, 78, 52,43,45,49)(46, 51, 48,93, 84, 89, 81, 99)<br />
(47, 72)(54, 65, 55, 85, 67, 96, 70, 80)(58, 69, 77, 86, 83, 98, 95, 75)<br />
(63,76,97, 94, 82, 92, 74, 68)(64)(87,91)<br />
(6.16)<br />
Here Ai = 73, Aj = 78 is the only pair from the 21 orbit of (a, d) in<br />
Ai<br />
the 63 orbit of H satisfying b2d =<br />
0<br />
A. and a-l = (. . . Ai,-,Aj. . .).<br />
Hence from (6.13) we must have t = (03,‘73). The equation (6.13) and the<br />
relations tat = a-l, tc = ct now completely determine t:<br />
t = (00,01)(02, 74)(03, 73)(04, 72)(05, 78)(06, 77)(07, 76)(08, 75)(09, 34)<br />
(10, 33)(11, 32)(12, 31)(13, 30)(14, 36)(15, 35)(16, 71)(17,70)(18, 69)<br />
(19, 68)(20, 67)(21, 66)(22, 65)(23, 53)(24, 52)(25, 51)(26, 57)(27, 56)<br />
(28, 55)(29, 54)(37,91)(38, 90)(39, 89)(40, 88)(41, 87)(42, 86)(43, 92)<br />
(44,99)(45,98)(46,97)(47,96)(48,95)(49,94)(50,93)(58, 85)(59, 84)<br />
(6.17)<br />
(60, 83)(61, 82)(62, 81)(63, 80)(64, 79).<br />
We now have a permutation group G on 100 letters 00, 01,. . . ,99,<br />
G = (a, b, an t) d a group<br />
H = (a, b) fixing 00 where it is known that<br />
H is the simple group of order 6048. Let M = Gee be the subgroup of<br />
G fixing 00. It is well known that M is generated by the 300 elements<br />
xia Xp-“, xtb z-l, xit Xif-ly<br />
(6.18)<br />
, i=OO,..., 99, are coset representatives of M in<br />
G and 7 = xj is the coset representative of My = MXj. Here, with the help<br />
Simple groups of order less than one million 167<br />
of Mr. Peter Swinnerton-Dyer and the Titan computer at the Cambridge<br />
University Mathematical Laboratory, it was shown that each of the 300<br />
permutations in (6.18) lies in H = (a, b). Thus M C H, and so M = H,<br />
whence Goo = H and G is of order 604,800.<br />
It remains to be shown that G is simple. In G the normafizer of the group<br />
(a) contains the element t interchanging 00 and 01. As these are the only<br />
letters fixed by (a) it follows that [N&(a))] = 2. As <strong>IN</strong>H((a))I = 21 it<br />
follows that <strong>IN</strong>&(a))/ = 42 and (a) = S(7) is its own centralizer in G.<br />
In a chief series for G one of the factors has order a multiple of 6048.<br />
If this is not a minimal normal subgroup, then a minimal normal subgroup<br />
K has order 2, 4, 5, or 25. But then an S(7) normalizing K must also centralize<br />
K, which is false since S(7) is its own centralizer. Hence a minimal<br />
normal subgroup K has order a multiple of 6048 and also 14,400 since<br />
it must contain all 14,400 S(7)‘s. Thus either [GX] = 2 or G = K and<br />
G is simple. If [G:K] = 2 then N,(7) is of order 21 and so tcj K but HE K.<br />
But mapping G/K onto the group + 1, - 1 we map H- + 1, t - - 1 and<br />
this conflicts with the relation (6.12). Hence G is simple.<br />
As this is written some questions remain unanswered. The original<br />
character table given by Janko has been shown by Walter Feit to be in<br />
error. Janko has made corrections to his table. The character table of the<br />
group constructed here has not yet been calculated. And it has not been<br />
established that there is a unique simple group or order 604,800.<br />
[Added in proof by the Editor: The uniqueness of the simple group of<br />
order 604,800 has since been established ; see Marshall Hall Jr. and David<br />
Wales: The simple group of order 604 800, J. Algebra 9 (1968), 417-450.]<br />
REFERENCES<br />
1. E. ARTTN: Geometric Algebra (Interscience, New York, 1957).<br />
2. R. BRAUER: Investigations on group characters. Ann. of Math. 42 (1941), 936-958.<br />
3. R. BRAUER: On groups whose order contains a prime number to the first power,<br />
part I, Am. J. Math. 64 (1942), 401-420; part II, Am. J. Math. 64 (1942), 421-440.<br />
4. R. BRAUER: On the connections between the ordinary and the modular characters of<br />
groups of finite order. Ann. of Math. 42 (1941). 926-935.<br />
5. R. BRAIJER: On permutation groups of prime’degree and related classes of groups.<br />
Ann. of Math. 44 (1943), 57-79.<br />
6. R. BRAUER: Zur Darstellungstheorie der Gruppen endlicher Ordnung, part I,<br />
Math. Zeit. 63 (1956), 406444; part II, Math. Zeit. 72 (1959), 25-46.<br />
7. R. BRAUER: Some applications of the theory of blocks of characters of finite groups,<br />
J. of Algebra, part I, 1(1964), 152-167; part II, 1(1964), 307-334; part III, 3 (1966),<br />
225-255.<br />
8. R. BRAUER and K. A. FOWLER: On groups of even order. Ann. of Math. 62 (1955),<br />
5655.583.<br />
9. R. BRAUER and C. NESBITT: On the modular characters of groups. Ann. of Math. 42<br />
(1941), 556-590.<br />
10. R. BRAUER and W. F. REYNOLDS: On a problem of E. Artin. Ann. of Math. 68 (1958).<br />
713-720.<br />
CPA 12