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$\DeclareMathOperator\rk{rk}$ 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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2 Answers 2

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 ?

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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 is better left as a comment to Derek Holt's answer, 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

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