1C 17
divides |iV|, we deduce that
23
divides \N\. Since also
73
divides \N\, we
have \G :N\ 8.
We can use what we have established to show that a group G of order
21,952 cannot be simple. Indeed, if 717(G) = 1, then G has a normal sub-
group of order
73,
and so is not simple. Otherwise, our subgroup N has
index at most 8, and we see that \G\ does not divide \G : N\\. By the n\-
theorem (Corollary 1.3), therefore, G cannot be simple if N G. Finally,
if N = G then D G and G is not simple in this case too.
In the last case, where D £?, we see that D is contained in all Sylow
7-subgroups of G, and thus D is the intersection of every two distinct Sylow
7-subgroups of G. In most situations, however, Theorem 1.16 can be used
to prove only the existence of some pair of distinct Sylow subgroups with
a "large" intersection; it does not usually follow that every such pair has a
large intersection.
To prove Theorem 1.16, we need the following.
1.18. Lemma. Let P G Sylp(G), where G is a finite group, and suppose
that Q is a p-subgroup
O / N G ( P ) .
Then Q C P .
Proof. We apply Sylow theory in the group T V =
N G ( P ) .
Clearly, P is
a Sylow p-subgroup of iV, and since P N, we deduce that P is the only
Sylow p-subgroup of N. By the Sylow D-theorem, however, the p-subgroup
Q of N must be contained in some Sylow p-subgroup. The only possibility
is Q C P, as required.
An alternative method of proof for Lemma 1.18 is to observe that since
Q Cj
NG?(P),
it follows that QP = PQ. Then QP is a subgroup, and it is
easy to see that it is a p-subgroup that contains the Sylow p-subgroup P. It
follows that P = QP 2 Q, as wanted.
Proof of Theorem 1.16. Let S act on the set Sylp(G) by conjugation.
One orbit is the set {5}, of size 1, and so if we can show that all other orbits
have size divisible by \S : S n T|, it will follow that np(G) = |Sylp(G)| = 1
mod |5 : S H T|, as wanted. Let O be an arbitrary 5-orbit in Sylp(G) other
than {S} and let P G (9, so that P ^ S. By the fundamental counting prin-
ciple, \0\ = \S : Q\, where Q is the stabilizer of P in S under conjugation.
Then Q C NG(P), and so Q C P by Lemma 1.18. But also Q C 5, and thus
|Q| ^ |5 fl P| |5 n T|, where the latter inequality is a consequence of the
fact that \S H T\ is as large as possible among intersections of two distinct
Sylow p-subgroups of G. It follows that \0\ = \S : Q\ \S : S n T\. But
since the integers \0\ and \S : S Pi T| are powers of p and |C?| \S : 5 Pi T|,
we conclude that |(9| is a multiple of I*! ? : S D T\. This completes the
proof.
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