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 Chermak-Delgado 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)\
|A|2,
where the last inequality holds because A is
abelian. Then
IC-.ll
2 | G | 2 | G | 2
-
| G | | G | | G |
=\C-M\
^ ^ \A\*-mG(M) |M| |C
G
(M)| " |M|
| U
''
as required.
Another application of the Chermak-Delgado 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 Chermak-Delgado subgroup of G and observe that
ITIG(M)
mdH) \G\. But the Chermak-Delgado 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 maximum-measure lattice C(G) and
that |G:Z(G)| = |G:
A\2.
1G.2. Let C{G) be the maximum-measure 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 |iJ||C3(.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
.
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