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.
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