COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

156 Marshall Hall Jr. Simple groups of order less than one million 157
Hence, by orthogonality between the a column and the b column in (5.12),
@l(b) = - 1. By the Brauer-Fowler formula in (3.17) we have ciik = 7 if
Xi = xi = b and xk = a. Here this gives
7 = 25200
1
1+= . (5.14)
@I2 ( 1
This gives c(b)2 = 3840 which is a conflict since 3840 is not a square. Hence
we may exclude the degrees in (5.12).
For the degrees in (5.13) if we restrict el to N(7) we see, from the values
@i(l) = 8, cl(a) = 1, that we must have
@l/N(7)= ~i+il/+/-Jl+~2+~89
where li and S are any combination of 1.0 and 11. We conclude that
(5.15)
PI(b) = 2,0, -2. (5.16)
The matrix M(b) which has cl(b) as its trace has 8 eigenvalues which are
+ 1 or - 1, say r (+ 1)‘s and t (- 1)‘s where r + t = 8. The determinant of
M(b) (which may be taken in diagonal form) is (- 1)‘. As the determinant
of M is a one-dimensional representation of G, which is simple, it must be
1 for every element. Hence t is even and cl(b) = r- t = 8 -2t z 0 (mod 4).
Thus cl(b) = 0 (mod 4) and from (5.16) this makes cl(b) = 0. The 7-conjugate
characters &,I!&, 0s are equal for b and so, by orthogonality with the
a-column, 6,(b) = f&(b) = 83(b) = 1. In this case with Xi = xj = b, xk=a
the Brauer-Fowler formula becomes
-= c(N2
7 = 25200
(5.17)
giving c(b)3 = 3200 which is not a square, and so the degrees (5.13) are
also to be excluded.
Thus every possibility has been excluded and we conclude that there is no
simple group of order 25200.
There are other cases, not illustrated here, in which the restrictions to
V(p) such as ~151 V(7) = oo+ 7, u 2 are very useful. For example if there is an
involution z in V(7) we must have ,u2(r) = -2 in order for the determinant
to be + 1. In this case z is the only involution in V(7) and so in the center
of N(7). Also XE.(Z) = - 13 and so c(z)=-169. If say IN(7)l = 84, then
H = Co(r) properly contains N(7). This may force H to be G or to contain
a number of S(7)‘s such as 15, which is impossible by the writer’s results
in (14).
EXAMPLE 6. The Suzuki group.
g = 29120 = 64.5.7.13.
/
Here the Suzuki group Su(8) of order 29120 = 64.5.7 * 13 will be constructed
directly from its order using the Brauer theory of modular characters.
The only divisors of g of the form 1 + 13k are 14, 40, and 560. By the
Brauer-Reynolds results (1 I), 14 S(13)‘s is possible only for PSLa(13).
There is no further factorization 12~40+(13u+1)(13v-1) so that 40
S(13)‘s is impossible. For 560 S(13)‘s we have the factorizations
6720 = 12.560 = 14.480 = 40.168 = 105*64.HereIN(13)1=52=4.13. For
4 = 2 there are no degrees 1 + S&+ S& = 0 with f;:12* 560, SOfO 3 - 2(13),
S& E l(13). For q = 4 we have 1+S0f0+S$s+&&s+84f4 = 0 with
514.560 and SOfO E -4(13), Sif; z 1(13), i = 2, 3, 4. Here the possible
values for S&o are -4, -160, -56, 35 of which the 4 is too small since
4~813-1) = 6. For SiJ;, the possible values are 560, 14, 40, - 64. Since
one of thef’s must be odd we must have Sofa = 35, and we find the only
combination to be
1+35-64+2(14) = 0. (5.18)
As 7 { IN(13)l it now follows that 13 { IiV(7)l and so the number of S(7)‘s
is a multiple of 13. For S(7) we must have 4 = 2 since 3 does not divide g.
Here 1 +Sofo+S2fi = 0 with Sofa E -2 (mod 7), Sf; E 1 (mod 7). As
degrees not in (5.18) are multiples of 13 this forces &fi = 64, and so we
have SOHO = - 65 and the S(7) degrees are
, l-65+64 = 0. (5.19)
Since these degrees must divide q( 1 + rp) = 2( 1 + 7r) the only possibility is
1 + 7r = 65 * 32 = 2080 S(7)‘s, N(7) = 14, q = 2 w = 1. Since degrees not
in (5.18) or (5.19) are multiples of both 7 and 13, and as g = 29120 is the
sum of the squares of all degrees, there is exactly one further irreducible
character and this is of degree 91.
At this stage we have a partial character table for G, where an S( 13) = (a)
with a13 = 1 and N(13) is defined by
al3 = 1, b4 = 1, b-lab = as. (5.20)
There are three classes of elements of order 13 with representatives a, a3, a9
and as c(a) = 13 each of these contains 2240 elements. For an S(7) = (c)
we have as defining relations for N(7)
c’ = 1 x2 = 1, xcx = c-i!
9 11 (5.21)
and each of the three classes with representatives c, c2, c” contains 4160
elements. The partial table follows: