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.
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