42 1. Sylow Theory
Proof. Write C#, C#, CD and Cj to denote the centralizers in G of i7, if,
D and J, respectively, and note that Cj =
CH^CK
and that C#, C#- C
CD-
Then
| J| |#K| - W^ and
lw|
and it follows that
m\ \n\\r \\EM\£M}\£]i rnG(H)mG(K)
mG(D) = \D\\CD\ -JJl ^ - = ™G(J) '
which establishes the inequality. If equality occurs, then \J\ = \HK\ and
\CD\

\CHCK\,
and thus J = HK and
CD
=
CHCK,
as required.
We mention that if i7, K C G are subgroups, then the subgroup J =
(iJ, K) is often called the join of H and if. Also, a collection C of sub-
groups of a group G is said to be a lattice of subgroups if it is closed under
intersections and joins.
1.44. Theorem. Given a finite group G, let C = C(G) be the collection
of subgroups of G for which the Chermak-Delgado measure is as large as
possible. Then
(a) C is a lattice of subgroups of G.
(b) IfH.Ke C, then (H,K) = HK.
(c) IfHeC, then CG(H) G C and CG(CG(H)) = H.
Proof. Let m be the maximum of the Chermak-Delgado measures of the
subgroups of G, and let H,K G C. Then rriG(H) = m = mciK), and
so by Lemma 1.43, we have
m2
= mG{H)rriG{K) rriG{T))mG{J), where
D = H H K and J = (H,K). But mG{D) m and mG(J) m by the
maximality of m, and thus
ITIG(D)
= m
7TIG(J),
and hence D and J lie
in £, as required. Also, since equality holds in Lemma 1.43, we know that
HK = J, and this proves (b). Finally, statement (c) follows because the
maximality of
ITIG(H)
forces equality in Lemma 1.42.
1.45. Corollary. Every finite group G contains a unique subgroup M, min-
imal with the property that
I71G(M)
is the maximum of the Chermak-Delgado
measures of the subgroups of G. Also M is abelian and M D Z(G).
Proof. Since C = C{G) is a lattice, the intersection of all of its members
lies in £, and this is the desired subgroup M. Also, since M G £, we know
that CG(M) G £, and thus M C
C G ( M ) ,
and it follows that M is abelian.
Finally, M = CG{CG(M)) D Z(G), and the proof is complete.
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