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 $|G:Z(G)| = p^n$. Suppose $A$ is a maximal abelian subgroup of $G$. Let $$p^l = {\rm max} {|Z(C_G (g)):Z(G)| : g \in G \setminus Z(G)},$$ $$p^b = {\rm max} {|cl(g)| : g \in G \setminus Z(G) },$$ and $p^a = |A:Z(G)|$. Then we show that $a \ge n/(b+l)$.