COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

154 Marshall Hall Jr.
36 S(7)%, /N(7) 1 = 700 = 4.25~7
50 S(7)%, ]N(7) 1 = 504 = 8.9.7
120 S(7)‘s, IN(7) \ = 210 = 2.3.5.7
225 S(7)%, lN(7) ) = 112 = 16.7
400 S(7)%, 1 N(7) 1 = 63 = 9 -7
1800 S(7)‘s, /N(7) 1 = 14 = 2.7
We shall handle these separately. Except for 1800 S(7)‘s we have wri.
In &(7) for 36 or 50 S(7)‘s we have a character of degree 6 or 8 in every
case. By the Stanton condition in (8) we may exclude 36 or 50 S(7)‘s.
For 120 S(7)‘s, /N(7) 1 = 2 * 3 * 5 * 7 and there are characters of degree 6 or
8 to be excluded by the Stanton condition except for
q = 2, w = 15, degrees i-16+15 = 0 (l+b&+s& = 0);
q = 6, w = 5, degrees
l-48+2(36)+2(-20)+15 = 0,
if 120+2(-48)+2(-20)f 15 = 0.
In all three cases we have a single character of degree 15, which is therefore
necessarily rational. Hence on restriction to V(7) (writing ~15 for a
character of degree 15)
x15 1 Y(7) = Go+7 pz. (5.9)
Here u. is the identity character for V(7) and ,LL~ a character of degree 2,
necessarily rational since x15 is. But by the Schur result (3.13) a rational
character of degree 2 can represent a group whose g order is divisible by
at most 23* 3 = 24 and by no prime p * 5. But w = 15 or 5 and this conflicts
with the Schur result if V(7) is to be faithfully represented, as it
must be if G is simple.
For 225 S(7)‘s IN(7)1 = 112 = 16.7 we must have q = 2, w = 8 and
as the only degrees possible are if 5 - 6 = 0 the Stanton condition excludes
this. Again for 400 S(7)‘s, IN(7)[ = 63 = 9.7, we must have q = 3,
w = 3 and the degrees are all exchtded by the Stanton condition.
For 1800 S(7)‘s IN(7)1 = 14 = 2.7 we have q = 2, w = 1. The possible
degrees 1 + Sofa+ Sd2 = 0 are:
1+5-6 = 0,
l-16+15 = 0, (5.10)
l- 9f 8=0.
The first case may be excluded by Schur’s result since the rational character
~6 cannot faithfully represent a group whose order is divisible by 25.
Simple groups of order less than one million 155
Here N(7) has the following defining relations: a7 = 1, b2 = 1, bab = a-l.
The table of characters for N(7) is as follows:
c(x) 14 7 7 7 2
h(x) 1 2 2 2 7
X 1 a a2 a3 b
Al 1 1 1 1 1
Al. 1 1 1 1 -1
CL1 2 71 72 r/3 0
P2 2 q2 qa 71 0
P3 2 73 71 72 0
(5.11)
x is an element, h(x) the number of conjugates of x, c(x) the order of C,(x).
Here 171 = E+ c-l, 72 = e2+ .sm2, 73 = E3+.sm3, where E is a primitive 7th
root of unity, and ql+rj2+q3 = - 1
From the Brauer results (3.3) the other degrees in (5.10) correspond to the
following characters in G:
g 7 7 7
1 3600 3600 3600
1 a a2 a3 b
1 1 1 1 1
15 1 1 1
16 71 ~2 9-13
16 712 r/s rll
16 73 rll 772
g 7 7 7
1 3600 3600 3600
1 a a2 a3 b
1 1 1 1 1
8 1 1 1
9 71 72 73
9 rl2 r/3 71
9 73 r1 r/2
(5.12)
(5.13)
Since g = 25200 = 16.9.25.7, the characters in (5.12) or, 02, 8s are of
highest type for p = 2 and so vanish for b, a 2-singular element. All further
characters for G have degrees multiples of 7 and vanish for a, a2, and a3.