Center of finite metabelian p-groups $\DeclareMathOperator\rk{rk}$
Let $G$ be a finite metabelian $p$-group, i.e. the commutator subgroup $G'$ of $G$ is abelian. Then I ask myself under which conditions does the following hold:
$$\tag{$*$} \rk(G' \cap Z(G)) \le \rk(G/G')$$
where $Z(G)$ is the centre of $G$. I have constructed examples where $(*)$ does not hold, but in most cases it does hold. Do you know of any results in the literature? How high would you guess the percentage of $p$-groups satisfying $(**)$ in a numerical analysis?
Thanks a lot.
 A: I think it could be reasonably conjectured that asymptotically almost all $p$-groups are nilpotent of class 2 (and hence metabelian), and satisfy $(*)$, with ${\mathrm{rk}}(G' \cap Z(G)) \approx \frac{1}{2}{\mathrm{rk}}(G/G')$, but it would not be easy to prove this.
In support of this conjecture, it is  proved in
G. HIGMAN, 'Enumerating $p$-groups. I, Inequalities', Proc. London Math. Soc. (3) 10 (1960), 24-30
that the number of isomorphism classes of $p$-group of order $p^n$ is $p^{An^3}$, where $A \ge 2/27 - o(1)$. He also proved an upper bound on $A$, which was later improved in
C. SIMS, 'Enumerating $p$-groups', Proc. London Math. Soc. (3) 15 (1965), 151-66
to $A \le 2/27 + O(n^{-1/3})$.
Higman obtained his lower bound by estimating the number of $p$-groups of order $p^n$ in which $G/G'$ and $G' = Z(G)$ are both elementary abelian, with $|G/G'| =p^r$, $|G'|=p^s$ with $r$ about $2n/3$ and $s$ about $n/3$.
Certainly, among the groups of order $p^n$ in which $G/G'$ and $G' = Z(G)$ are both elementary abelian, you find the largest number of distinct isomorphism types when $|G/G'| =p^r$, $|G'|=p^s$ with $r$ about $2n/3$ and $s$ about $n/3$.
A: Let E be the elementary abelian group of order p^n. Then its Schur multiplier has rank n(n-1}/2 (Issai Schur). Therefore, the representation group of E (that group is special) does not satisfies ($*$). It is possible to construct infinite set of such examples of arbitrary exponents. Therefore, assertion that (*) fulfilled in the most cases, is sinceless (in any case, I do not know what it means).
