34 1. Sylow Theory

1.33. Theorem. Let \G\ = 24, and suppose ri2(G) 1 and 713(G) 1.

ThenG^S^

Proof. Since 123 exceeds 1, is congruent to 1 modulo 3 and divides 8, the

only possibility is that 77, 3 = 4, and thus \G : N\ = 4, where T V = Ns(P)

and P G Syl3(G). Now let K — corec(N), so that G/K is isomorphic to a

subgroup of £4 by Theorem 1.1. It suffices to show that K — 1 since that

will imply that G is isomorphic to a subgroup of S4, and this will complete

the proof since \G\ = 24 = |S4|.

Recall that K C T V = N

G

(P), where P G Syl3(G). Then P is a normal

Sylow 3-subgroup of KP, and so P is characteristic in KP. Since P is not

normal in G, however, we conclude that KP cannot be normal in G. But

KP/K is a Sylow 3-subgroup of G/K, and it follows that ns(G/K) 1. In

particular, G/K is not a 2-group, and so \K\ is not divisible by 3.

Since \G : N\ = 4, we see that |JV| = 6, and hence the only possibilities

are |K| = 1, as wanted, or \K\ = 2. Assuming now that \K\ = 2, we work

to obtain a contradiction. Since \G/K\ = 12, Theorem 1.31 applies, and

we deduce that either n^iG/K) — 1 or 77, 3 (G/K) — 1. Since we have seen

that ns(G/K) 1, however, we conclude that G/K has a unique Sylow

2-subgroup S/K. Then S/K G/ET, and so S G. Also |5| = 8 since

\KI = 2 and |5/if| = 4, and thus S is a normal Sylow 2-subgroup of G.

This contradicts the hypothesis that /12(G) 1. •

We mention that the group £4, of order 24, is not simple, and it follows

that no group of order

p3q

is simple, where p and q are primes. In fact,

S4 has a normal subgroup of order 12 and one of order 4. The subgroup

of order 12 is the alternating group A4 (see the discussion below) and the

normal subgroup of order 4 is the so-called Klein group consisting of the

identity and the three elements of order 2 in 64 that have no fixed points.

We can eliminate a quarter of all positive integers as possibilities for the

order of a simple group. The argument relies on the notion of odd and even

permutations, with which, we assume the reader is familiar.

Briefly, the facts are these. Every permutation of a finite set can be

written as a product of transpositions (2-cycles). A permutation that can

be written as a product of an even number of transpositions is even and

one that can be written as a product of an odd number of transpositions is

odd. It is a triviality that the even permutations in the symmetric group

Sn form a subgroup. (It is called the alternating group and is denoted

An.) It is also true, but much less trivial to prove, that no permutation can

be both even and odd, and since odd permutations certainly exist for n 2,

we see that An is proper in Sn in this case.