A lower bound on the size of maximal abelian subgroups

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).