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