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.

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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 $n/3$.

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Thanks you very much first, Derek. I tried to run a numerical analysis of that question on the computer program MAGMA. For groups of order \le p^7, the statement is true. But this can be verified rather easily on foot as well. Unfortunately, MAGMA only has groups of order equal or less to p^7 (when p\ge 5). Do you know of any other computer programs who can run such an analysis for groups of higher order for p\ge 5 ? – Tobias Bembom Aug 12 '11 at 15:57
A full classification of groups of order $p^n$ has been completed only for $n \le 7$, and the current belief seems to be that it would not be feasible to do this for $n=8$. So you would need to find a different approach to the problem! – Derek Holt Aug 12 '11 at 19:40
This answer should have been left as a comment to Derek Holt's answer (and we will convert it to such), but it seems that since you created a new account with the same name, that was not possible for you. I've merged your two accounts, but unless you register or find a way to keep your browser cookie, you will continue to create new accounts. – S. Carnahan Aug 13 '11 at 4:13

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).

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If you do not know what it means, then you cannot answer the question can you? I do not know what "sinceless" means, so I am not sure what you are trying to say. But what the assertion means is that, if $a(n)$ denotes the number of isomorphism classes of $p$-groups of order at most $n$, and $b(n)$ is the number of those that satisfy $(*)$, then $b(n)/a(n) > 1/2$ for all sufficiently large $n$. (We could let $p$ be a fixed prime number here, and conjecture that $\lim_{n \to \infty} b(n)/a(n)=1$ for every prime $p$.) The existence of infinitely many examples does not contradict that conjecture. – Derek Holt Dec 24 '15 at 17:01
What is not correct in -2? It follows from this and other estimates of my posts that the person which estimates, is awfully non-competent since correct assertions he estimates negatively. Therefore, I ask to erase all my posts snce I intend to stop participate in this portal. – yakov yesterday