Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$\DeclareMathOperator\rk{rk}$ 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.

share|improve this question

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 $n/3$.

share|improve this answer

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 ?

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.