COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

160 Marshall Hail Jr.
This leads to 18~~+8x-56 = 0 whose roots are x = -2 and x = 14/g.
Since x is a rational integer it is x = es(z) = @a(t) = -2 and the characters
for b2 = z are completely determined.
The remaining 3640 elements are equally divided between the classes
for b and b-l since c(b) = c(b-l) whence there are 1820 in each and
c(b) = c(b-l) = 16. As all characters except ~2 and ~3 are real we have
&b> = ei(b-‘) in all other cases and row orthogonality determines all of
these. Complex conjugacy and the fact that c(b) = 16 determines e2 and
e3 for b and b-l.
From the table of characters we see that S(2) is non-Abelian since
c(b) = 16, and so the center Z(2) of S(2) is of order 2,4, or 8. An element of
order 5 or 13 in N(2) would normalize and so centralize Z(2) contrary to
the fact that C(5) = S(5) and C(13) = S(13). Hence the number of S(2)‘s
is a multiple of 65 and so is either 65 or 455. If the number were 455, since
455 * 1 (mod 4) there would be two S(2)‘s PO n P2 = K where Kis of order
32 and No(K)conta an odd number of S(2)‘s. As 32- 1 = 31 is not a multiple
of 5, 7, or 13 an element of odd order normalizing Kcentralizes some
2-element, contrary to the fact C(5) = S(5), C(7) = S(7)and C(13) = S(13).
Hence the number of S(2)‘s is not 455 but is 65 and N(2) is of order 64.7 and
is a Frobenius group as C(7) = S(7). Thus S(2) has a center Z(2) of order
8 and S(2)/Z(2) is also of order 8 and both are acted upon by S(7) without
fixed points. Each of 65 S(2)‘s has 63 elements besides the identity giving
4095 2-elements which must be distinct since G has 455 involutions and
3640 elements of order 4. Hence the S(2)‘s have trivial intersection and in
the representation of G on the 65 cosets of N(2), every 2-element fixes exactly
one letter. G1 = N(2) is a Frobenius group on the 64 letters it moves, being
the regular representation of S(2) normalized by an element c of order 7.
An automorphism m of order 7 on an elementary Abelian group of
order 8 satisfies either PBCmtl = I or x?+““+~= 1 for every element. The
group S(2) of order 64 is determined up to isomorphism by the fact that it is
non-Abelian and has a center Z(2) and factor group S(2)/Z(2) both elementary
of order 8 and having an automorphism of order 7 which is fixed point
free. Here c, of order 7, induces the automorphism, and on S(2)/Z(2) we have
the relation xC3++l = 1 and on q2)&c2+1 & . " 1. If the same relation
held in both places S(2) would be Abelian. Interchanging the relations
amounts to replacing c by c-l.
Take bl = b of order 4 and b2, = el is in Z(2). The automorphism in-
duced by c (X -+ c-lxc) is given by
bl -t bz, el + e2,
bz - bs, e2 - e3,
b3 -f b&z, e3 --L 8163.
(5.28)
These give rise to relations
Simple groups of order less than one million
b; = el = (bz, bd,
b$ = e2 = (bl, b&b2, bd,
bg = e3 = (bl, Wbl, Wbz, bd,
using the commutator notation (x, y) = x-ly-lxy.
161
(5.29)
We may now give the permutations on letters 0, . . . , 63 generating N(2).
c = (O)(l, 2, 3,4, 5, 6, 7)(8, 9, 10, 11, 12, 13, 14)(15, 16, 17, 18, 19, 20, 21)
(22, 23, 24, 25, 26, 27,28)(29,30,31,32,33,34, 35)(36,37,38,39,40,41,42)
(43,44,45,46, 47, 48, 49)(50 51,52,53,54,55,56)(57,58,59,60,61,62,63).
b = bl = (0, 8, 1, 15)(2, 22, 6, 50)(3, 29,4, 36)(5, 43, 7, 57)(9, 39, 58, 25)
(10,28, 52, 14)(11, 37,46, 51)(12, 62, 40, 27)(13, 33, 34, 19)(16, 60, 44, 18)
(17,49, 24,42)(20,61, 41, 54)(21,45, 56, 31)(23, 53, 30, 32)(26,48, 47, 55)
(35, 59, 63, 38).
(5.30)
The Suzuki group G will be obtained by adjoining a further letter ~0 and
finding an involution z (necessarily conjugate to b2) interchanging OJ and 0
and so normalizing the group (c). z’ must also have the property that for
any u E G, = N(2) zur = my for appropriate x, y E N(2). With a certain
amount of trial this determines z (up to a conjugate by c) as
z = (03, O)(l, 57)(2, 63)(3, 62)(4, 61)(5, 60)(6, 59)(7, 58)(8, 26)(9,25)
(10,24)(11, 23)(12, 22)(13, 28)(14, 27)(15, 37)(16, 36)(17, 42)(18, 41)
(19,40)(20, 39)(21, 38)(29, 52)(30, 51)(31, 50)(32, 56)(33, 55)(34, 54)
(35, 53)(43)(44, 49)(45, 48)(46,47).
(5.31)
6. Construction of a new simple group of order 604,800. In his announcement
“Still one more new simple group of finite order” Zvonimir Janko
gives a character table for a simple group of order 604,800 if such a group
exists. In this group there are two classes of involutions, one with a centralizer
of order 1920, the other with a centralizer of order 240. He gives the
character table for such a group.
The group has 21 classes and three of the characters are as follows: