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. •