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

Hx~1.

Now g induces an

odd permutation on O, and hence gx induces an even permutation and we

have gx G H. Then g G

Hx~l,

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

and

3«52

all exceed 60, we see that the only possibilities that are

products of at least three primes are n =

33

and n =

32-5,

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 =

22«32,

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 =

24«3,

we get a violation of