Problems IE 37

and this is a contradiction since £5 has a normal subgroup of index 2: the

alternating group A5.

We close this section with one further result which, though somewhat

more general than those that we have already proved, is still only a very

special case of Burnside's

paqb-theorem.

1.36. Theorem. Suppose that \G\ =

paq,

where p and q are primes and

a 0. Then G is not simple.

Proof. We can certainly assume that p ^ q and that np 1, and thus

rip = q. Now choose distinct Sylow p-subgroups S and T of G such that

\S fl T\ is as large as possible, and write D — S fl T.

If D — 1, then every pair of distinct Sylow p-subgroups of G intersect

trivially, and so the Sylow ^-subgroups of G account for a total of

q(pa

— 1)

nonidentity elements of G. All of these elements, of course, have orders

divisible by p, and this leaves room for at most \G\ —

q(pq

— 1) = q elements

with order not divisible by p. It follows that if Q G Syl^(G), then Q is exactly

the set of these elements, and in particular Q is unique, and so Q G and

G is not simple, as required.

We can now assume that D 1, and we let N =

~NG{D).

Since D S

and D T and we know that "normalizers grow" in p-groups, we can

conclude that N D S D and N H T D.

Next, we show that N is not a p-group. Otherwise, by the Sylow D-

theorem, we can write N C R G Sylp(G). Then RnS ^ N C\S D, and

so by the choice of D, we see that the Sylow subgroups R and S cannot

be distinct. In other words, S — R and similarly, T = R. But this is a

contradiction since S ^ T.

It follows that q divides |iV|, and so if we choose Q G Sy\q(N), we have

IQI = q. Now S is a p-group and Q is a g-group, and thus S D Q = 1

and we have \SQ\ = \S\\Q\ =

paq

= \G\. Then SQ = G, and hence if

g G G is arbitrary, we can write g = xy, where x G S and y G Q. Then

Sg

=

Sxy

—

Sy

3

Dy

— D, where the second equality holds since x G S and

the last equality follows because y G Q C N =

NQ(D).

We see, therefore,

that D is contained in every conjugate of S in G. Then D is contained in

every Sylow p-subgroup of G, and we have 1 D C Op(G). Thus Op(G) is

a nonidentity proper normal subgroup G, and this completes the proof that

G is not simple. •

Problems IE

1E.1. Let \G\ —

p2q2,

where p q are primes. Prove that nq{G) = 1 unless

\G\ = 36.