32 1. Sylow Theory

1.30. Theorem. Let \G\ — pq, where q p are primes. Then G has a

normal Sylow p-subgroup. Also, G is cyclic unless q divides p — 1.

Proof. Since np = 1 mod p, we see that if np 1, we have np p q. This

is not possible, however, since np must divide q. We conclude that np = 1,

as wanted. Also, since every element of G of order p generates a subgroup

of order p, and there is only one of these, there are exactly p — 1 elements

of order p in G.

Now if nq 1, then p = nq = 1 mod g, and thus q divides p — 1.

Otherwise ng = 1, and we see that there are exactly q — 1 elements of order

q in G. Together with the identity, we have now accounted for p + q — 1

elements of G. But p + q — 1 2p pq = |G|, and so G must have some

element g that does not have order either 1, g or p. Since o(g) divides

\G\ — PQ, the only possibility is that o(g) = pg, and thus G = (pg) is

cyclic. •

1.31. Theorem, iet |G| =

p2q,

where p and q are primes. Then G has

either a normal Sylow p-subgroup or a normal Sylow q-subgroup.

Proof. We assume that np 1 and nq 1, and we note that p ^ q. Since

np = 1 mod p, it follows that np p and similarly nq q.

Now np must divide g, and hence np — q and we have nq q = np p.

But n^ divides

p2,

and we see that the only possibility is that nq =

p2,

which

means that G has

p2

subgroups of order q.

If Q\ 7^ Q2 are subgroups of order g, then since q is prime, it follows

via Lagrange's theorem that Qi C\ Q2 — 1. The

p2

subgroups of order q

in G, therefore, have no nonidentity elements in common, and it follows

that G has at least

p2(q

— 1) elements of order q. (Actually, since every

element of order q in G must lie in some subgroup of order g, it follows

that G has exactly

p2{q

— 1) elements of order g, but we shall not need that

fact.) The number of elements of G that do not have order q is then at

most |G| —

p2(q

— 1) =

p2.

If P G Sylp(G), then, of course, no element

of P has order q, and since \P\ =

p2,

it follows that P is exactly equal

to the set of elements of G that do not have order q. In particular, P is

uniquely determined, and this is a contradiction since we are assuming that

np

1. •

By the previous two theorems, we see that if p 7^ q are primes, then

a group of order pq or of order

p2q

must either have a normal Sylow p-

subgroup or a normal Sylow ^-subgroup. This almost works for groups of

order

p3q

too, but there is an exception. This situation is endemic in finite

group theory: some general fact holds with a small number of exceptions.