MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$\DeclareMathOperator\rk{rk}$ 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|cite|improve this question

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|cite|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 ? – Tobias Bembom Aug 12 '11 at 15:57
1  
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).

share|cite|improve this answer
    
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

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.