A lower bound on the size of maximal abelian subgroups

Abstract

Let G be a p-group for some prime p. Let n be the positive integer so that |G:Z(G)| = pn. Suppose A is a maximal abelian subgroup of G. Let pl = max \|Z(CG (g)):Z(G)| : g ∈ G Z(G)\, pb = max \|cl(g)| : g ∈ G Z(G) \, and pa = |A:Z(G)|. Then we show that a n/(b+l).

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