COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

152 Marshall Hall Jr.
easily found to be
15.59 = 885 S(17)‘s with JN(17) 1 = 8.7.17
168.59 = 9912 S(17)‘s with IN(17) 1 = 5.17.
If I iV(17) j = 5.17, then S(17) is in the center of its normalizer and by
the Burnside condition (9) b has a normal 17 complement. We may exclude
this. An S(2) of order 8 is necessarily the elementary Abelian group
because of (2) and (4) and so with 1 N(17) I = 8 * 7.17 we must have q = 2,
w = 28 for N(17). Thus the relations (3.2) reduce to
l+S,fo+S,fi = 0, S&Z, = -2(17), S,fi = l(l7) (5.4)
and fo, fi divide 2.885 = 2.3.5.59, J;: -C 8. No such divisors exist and so
we may exclude 885 S(17)‘s. We conclude that no simple group of this
order exists. This completes the elimination of all orders which are multiples
of 59.
EXAMPLE 2. Block separation.
g = 783216 = 16.27.49.37, 5292 S(37)‘s, q = 4, w = 1.
Degrees in Bo( 37) : l-189+2(112)-36 = 0.
Here Sofa = - 189, S,fl = S,fd= 1 12,@f3 = -36.
Here there are (37- 1)/4 = 9 37-conjugate characters of degree 189, and
in Bo(37) two characters of degree 112 and one of degree 36. The characters
of degree 189 and 112 are of 7-defect 1. For any 37-regular element
x we have by (3.12)
1 - x1*&) + XW) + x1?202(x) - x3644 = 0. (5.5)
The characters of degrees 189 and 112 belong to 7-blocks of defect 1,
those of degree 1 and 36 to 7-blocks of defect 2. Hence by block separation
as given in Lemma 2 in (11) of 9 3 we must have
1-~~~(1) = l-36 = -35 z 0 (mod 49)
But this is a conflict and so there is [no simple group of order g. (All
other possibilities were eliminated by more elementary arguments.)
EXAMPLE 3. The Stanton condition
g = 92400 = 16.3.25.7.11.
For p = 11 the only number of S(ll)‘s for which degrees satisfying
(3.2) exist is 210 S(ll)‘s. Here IN(11) 1 = 440 = 8.5.11.
For q = 2 the degrees 1 +&fO+&fi = 0 are 1+20-21 = 0.
For q = 5 degrees are 1 + SOfO+ &fi+ &f3+ c%fd+ S& = 0,
where &fO = 6, 50, or 105 and S& = 210, -175, 100 -21, 12, -10.
For q = 10 degrees are 1 + f SiJ;, = 0
i-1
with Sif;, = 210, -175, 100, -21, 12, -10.
For q = 5 or q = 10 there is in every instance a degree 10 or 12.
Simple groups of order less than one million 153
By the Stanton principle quoted in (8) since there is in every instance
a character in B,-,( 11) of degree less than 2p = 22 in g = p(1 +rp)qw
we must have w = 1. But here qw = 40 and so for q = 2, w = 20;
q = 5, w = 8 ; q = 10, w = 4. Hence the condition is violated and there
is no simple group of order g = 92400.
EXAMPLE 4. The Brauer method for groups divisible by p2.
g = 202800 = 16.3.25.169
Numbers of the form 1 + 13k dividing g are 1, 40, 300. By the Brodkey
argument (5) as 40-= 169 we cannot have 40 S(13)‘s. Suppose we have
300 S(13)‘s and IN(13)1 = 676 = 4.169. As 300 = 1+23.13, and
23 E$ 0 (13), if PO is an S(13) there must be a PI with [PO:POnP1] = 13.
Write POn PI = K. In the notation of (3.14)
300rK = (1+ 13bK)[G : No(K)]. (5.6)
As 1 -C 1 + 13bK -C 300 and as (1 + 13b,)/g; its only possible value is
40 = 1+3*13. Hence
Here (3.15) takes the form
300~ = 40[G : No(K)]
15rK = 2[G : No(K)].
299 = 13 C rKbK+a-169
K
23 = c rKb,+ 13a.
K
(5.7)
(5.8)
But l+bg13 = 40 and so bK = 3 in every case and from (5.7) it follows
that rK is even. Thus in (5.8) a is odd and 13a -= 23 and so a = 1 giving
23 = c 3rK+ 13 whence 3 I23- 13 = 10 which is false. We have reached
a conf&t and conclude that there is no simple group of order 202 800.
An alternate argument applicable here rests on showing that no group
contains exactly 40 S(13)‘s. From the writer’s result (14), since 40 is not
a prime power and 40 has no proper factorization 40 = (1 + 13r)(l+ 13s),
it follows that if there is a group with 40 S(13)‘s then there is a simple
group with 40 S(13)‘s. Since 40 -Z 169 there is no simple group G with
40 S(13)‘s if 13=1g, by the Brodkey argument (5). If g = 40*13qw,since
12.40 does not have a further factorization (13u+ 1)(13v- l), the Brauer-
Reynolds argument (11) shows that there is no simple group of such an
order, as 13- 1 is not a power of 2 and PSL2 (13) has 13 + 1 = 14 S(13)‘s.
Hence no group has 40 S(13)‘s. Thus it is impossible that H,(K) has
40 S(13)‘s in the above argument.
EXAMPLE 5. g = 25200 = 16.9.25.7.
For p = 7 the admissible Sylow numbers and orders of N(7) are as
follows :
11.