1G 41

Hint. In fact, if x G N is chosen so that \P D

Px\

is as small as possible,

then P acts faithfully on the P-orbit containing x.

1G

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\

n2.

Surprisingly, it is not necessary to assume that A

is a Sylow subgroup; there always exists an abelian normal subgroup N

with \G : N\

n2.

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 :

A\2

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

T71G(H)

=

\H\\CG(H)\

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 =

CG(H).

Then

IMG(H)

mG{C) and if equality occurs, then H =

CG(C).

Proof. Since H C

C G ( C ) ,

we see that

mG{C) = |C||C

G

(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

CG(D)

=

CG(H)CG(K).