1G 41
Hint. In fact, if x G N is chosen so that \P D
is as small as possible,
then P acts faithfully on the P-orbit containing x.
Suppose that A C G is an abelian subgroup, and that the index n = \G : A\
is relatively small. Can we conclude that G has a normal abelian subgroup
T V such that the index \G : N\ is under control? Yes, certainly; we know
that \G : corec(A)| n!, and of course, corec(A) is normal, and since this
subgroup is contained in A, it is abelian.
A bound much better than \G : N\ n\ is available if the abelian
subgroup A happens to be a Sylow subgroup of G. Assuming that A is a
Sylow p-subgroup, we can take N = Op(G), and then Corollary 1.39 tells
us that \G : N\
Surprisingly, it is not necessary to assume that A
is a Sylow subgroup; there always exists an abelian normal subgroup N
with \G : N\
Although this remarkable theorem of A. Chermak and
A. Delgado does not depend on Sylow theory, it seems appropriate to include
it here since it can be viewed as a generalization of Corollary 1.39.
1.41. Theorem (Chermak-Delgado). Let G be a finite group. Then G has
a characteristic abelian subgroup N such that \G : N\ \G :
for every
abelian subgroup A C G.
As we shall see, the key idea in the proof of the Chermak-Delgado the-
orem is very elementary. But it is powerful, and it yields even more than
we stated in Theorem 1.41. The trick is to define the integer
for arbitrary subgroups H of a finite group £?, and to consider
subgroups H C G where this Chermak-Delgado measure is maximized.
We begin with a trivial observation.
1.42. Lemma. Let H C G, where G is a finite group, and write C =
mG{C) and if equality occurs, then H =
Proof. Since H C
C G ( C ) ,
we see that
mG{C) = |C||C
(C)| |C||ff| = mG(H),
and the result follows.
Next, we consider two subgroups simultaneously.
1.43. Lemma. Let H and K be subgroups of a finite group G, and write
D = HnKandJ={H,K). Then
mG{H)mG{K) mG{D)mG{J)
and if equality holds, then J = HK and
Previous Page Next Page