IE 35
More generally, suppose that G is an arbitrary group that acts on some
finite set 0, and assume that there is at least one element a;G(? that induces
an odd permutation on fi. It should be clear that the elements of G that
induce even permutations on Q form a subgroup, which we call H. Since
x 0 H, we see that H G, and we claim that \G : H\ 2. To see this, it
suffices to show that the set G H is contained in a single right coset of
H. We show, in fact, that if g G G H, then g G
Now g induces an
odd permutation on O, and hence gx induces an even permutation and we
have gx G H. Then g G
as claimed. Since a subgroup of index 2 is
automatically normal, we have established the following useful fact.
1.34. Lemma. Let G act on a finite set ft and suppose that some element
of G acts "oddly". Then G has a normal subgroup of index 2.
1.35. Theorem. Suppose that \G\ = 2n, where n is odd. Then G has a
normal subgroup of index 2.
Proof. By Cauchy's theorem, we can find an element t G G of order 2. In
the regular action of G on itself by right multiplication, t has no fixed points,
and so the permutation induced by t consists entirely of 2-cycles, and there
are \G\/2 = n of them. Since n is odd, the permutation induced by t is odd,
and the result follows by Lemma 1.34.
We have stated that 60 is the smallest number £hat occurs as the order
of a nonabelian simple group. In fact, using what we have now proved, this
is not very hard to see. First, we can eliminate all powers of primes since if
G is a simple p-group, then because 1 Z(G) G, we see that Z(G) = G
and G is abelian. Also, by Theorem 1.30, we can eliminate all products of
two primes, and thus we can assume that |G| factors as a product of at least
three primes, counting multiplicity.
Suppose now that \G\ 60 and write n \G\. If n is odd, then, since
34, 32«7
all exceed 60, we see that the only possibilities that are
products of at least three primes are n =
and n =
and we have seen
that neither of these numbers can be the order of a simple group.
We are left with the cases where n is even. By Theorem 1.35, we can
suppose that n is divisible by 4, and we write n 4m, where m 15. We
have eliminated the cases where m is prime or is a power of 2, and also
the cases where m = 2p, where p is an odd prime. The the only surviving
possibilities for m are 9 and 12, and so we need to show that there is no
simple group of order 36 or of order 48.
If G is simple of order 36 =
we see that \G : P\ 4, where
P G Syl3(G), and yet \G : P\ does not divide 4! = 24. This violates the
n!-theorem, Corollary 1.3. Similarly, if IGI = 48 =
we get a violation of
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