36 1. Sylow Theory

the n!-theorem by taking P G Syl2(G), so that \G : P\ = 3. This completes

the proof that 60 is the smallest possible order of a nonabelian simple group.

To prove that no number between 60 and 168 can be the order of a

simple group, it is convenient to have two more nonsimplicity results of the

type we have been discussing: if G has order

p2q2

or order pqr, where p, q

and r are primes, then G cannot be simple. (We leave the proofs of these

to the exercises at the end of this section.) Using these two facts and the

other results that we have established, all but about a dozen of the numbers

between 60 and 168 can be eliminated, and most of those are easily dealt

with by ad hoc methods. For example, we need to consider the numbers

84 =

22-3-7,

140 =

22-5-7

and 156 =

22-3-13,

and these are easily eliminated

by the observation that if p is the largest prime divisor, then np — 1 since

no divisor of n exceeding 1 is congruent to 1 modulo p.

The number 72 =

23«32

can be eliminated by observing that if G is

simple and |G| = 72, then n% 1, and therefore the only possibility is

ri3 = 4. Then \G : N\ = 4, where N is the normalizer of a Sylow 3-

subgroup, and this violates the n!-theorem. More interesting is the case

where \G\ = 144 =

24-32.

This case can be handled using Theorem 1.16,

with an argument analogous to the one we used in the discussion following

Corollary 1.17, where we showed that groups of order

26«73

are not simple.

Another number that requires some work is 132 =

22«3'11.

If a group of this

order is simple, then nu — 12, and hence there are at least 120 elements of

order 11, leaving at most 12 other elements. It follows that ns = 4, and this

yields a contradiction using the n!-theorem.

The number between 60 and 168 that seems to be the most difficult to

eliminate is 120 =

23«3«5.

One (rather tricky) approach is this. If G is simple

of order 120, it is easy to see that n^(G) = 6. Then G acts (nontrivially)

on the six right cosets of the normalizer of a Sylow 5-subgroup, and this

determines a nontrivial homomorphism from G into the symmetric group

SQ. Since G is simple, this homomorphism is injective, and thus there is an

isomorphic copy H of G with H C SQ. But G has no normal subgroup of

index 2, and hence by Lemma 1.34, no element of G can act oddly on the

six cosets, and thus H actually lies in the alternating group AQ, and since

\AQ\ — 360, we see that 1^6 : H\ — 3. Recall, however, that the alternating

groups An are simple when n 5, and so by the n!-theorem, AQ cannot

have a subgroup of index 3. This contradiction shows that 120 is not the

order of a simple group.

An alternative approach for eliminating 120 is to quote Problem 1C.4,

which asserts that if \G\ = 120, then there exists a subgroup H C G with

l | G : i J | 5 . IfGi s simple, this yields an isomorphism of G into the

symmetric group 55. Since IS5I = 120 = |G|, we conclude that G = S$,