IE 33

(Another example of this phenomenon is that all automorphisms of the

symmetric group Sn are inner for all integers n 1 except n — 6.)

1.32. Theorem. Let \G\ —

p3q,

where p and q are primes. Then G has

either a normal Sylow p-subgroup or a normal Sylow q-subgroup, except when

\G\ = 24.

Proof. As in the previous proof, we assume that np 1 and that nq 1,

and we conclude that p ^ q and that np p and nq q. Continuing to

reason as we did previously, we have nq q = np p. Since nq divides

p3,

we are left this time with two possibilities: either nq =

p3

or nq =

p2.

Suppose that nq =

p3.

Then the

p3

subgroups of order q contain a total of

p3(q—

1) elements of order #, and thus G contains at most |G| —

p3(q

— 1) =

p3

elements that do not have order q. It follows that if P G Sylp(G), then P is

exactly the set of elements of G having order different from g, and thus P

is unique, contradicting the assumption that np 1.

We conclude, therefore, that nq =

p2,

and thus

p2

= 1 mod q. In other

words, q divides

p2

— 1 = (p + l)(p—l). Since q is prime, we see that q must

divide either p+1 or p — 1, and so in either case, we have q p + 1. But

we know from the first paragraph of the proof that p q, and so the only

possibility is that q = p + 1. Now p and q are primes that differ by 1, and

so one of them must be even. We deduce that p = 2 and q = 3, and thus

|G| - 24. •

Of course, the previous proof does not tell us that there actually is an

exception of order 24; it only permits the possibility of such an exception.

But an exception really does exist. Consider 54, the symmetric group on

four symbols, which, of course, has order 4! = 24. By examining the possible

cycle structures for elements of 54, it is easy to count that there are exactly

nine elements of order 2, eight elements of order 3 and six elements of order

4. (As a check, note that 9+8+6 = 23, so that together with the identity, we

have accounted for all 24 elements.) By the Sylow D-theorem, every element

of 2-power order in 54 lies in some Sylow 2-subgroup. Since each Sylow 2-

subgroup has order 8, and yet there are more than eight elements of 2-power

order, it follows that 54 must have more than one Sylow 2-subgroup. Similar

reasoning shows that there must be more that one Sylow 3-subgroup. (In

fact, it is easy to see that n2(54) — 3 and that 723(54) = 4.)

There is still more that we can say about this situation. We know

that among groups of order

p3q,

only groups of order 24 can fail to have a

nontrivial normal Sylow subgroup, and we know that at least one group of

order 24 actually does fail to have such a Sylow subgroup. The remaining

question is whether or not there are any other groups of order 24 that can

occur as exceptions. The answer is "no".