Problems 1G 43
We shall refer to the subgroup M of Corollary 1.39 as the Chermak
Delgado subgroup of G. It is, of course, characteristic in G.
Proof of Theorem 1.41. Since the ChermakDelgado subgroup M is char
acteristic and abelian, it suffices to show that \G : M\ \G :
A\2
for all
abelian subgroups A C G. By the definition of M, we have
ITIG(M)
rriG(A) =
\A\\CG(A)\
A2,
where the last inequality holds because A is
abelian. Then
IC.ll
2  G  2  G  2

 G   G   G 
=\CM\
^ • ^ \A\*mG(M) M C
G
(M) " M
 U
''
as required.
Another application of the ChermakDelgado subgroup is the following.
1.46. Corollary. Let H be a subgroup of a finite group G, and assume that
\H\\CQ{H)\
\G\. Then G is not a nonabelian simple group.
Proof. Let M be the ChermakDelgado subgroup of G and observe that
ITIG(M)
mdH) \G\. But the ChermakDelgado measure of the identity
subgroup is equal to G, and it follows that M 1. Since M is abelian and
normal in G, the result follows. •
Problems 1G
1G.1. Let A C G , where A is abelian, and assume that there does not exist
a characteristic abelian subgroup N of G such that G:A/"G:^4
2
. Show
that A = CG(A) is a member of the maximummeasure lattice C(G) and
that G:Z(G) = G:
A\2.
1G.2. Let C{G) be the maximummeasure lattice as in Theorem 1.44, and
suppose that H G C(G) and that H G. Show that there exists a normal
subgroup M of G such that H C M G.
Hint. Observe that all conjugates of H in G lie in C{G) and thus the product
of any two of them is a subgroup. If H is not contained in a proper normal
subgroup of G, show that it is possible to write G = HK, where K G
and K contains a conjugate of H. Deduce a contradiction from this.
1G.3. Let G be simple and suppose that H C G and that iJC3(.ff") =
G. Show that H = 1 or H = G.
Hint. If G is nonabelian, show that H G C(G).
1G.4. Let A C G, where A is abelian and G is nonabelian. Show that there
exists a normal abelian subgroup N of G such that  G : i V   G : A 
2
.