Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

There is some facts about finite non abelian $p$-groups over the site. For example, when $n=3$: Nonabelian groups of order $p^3$. I have found the following problem in my very old works unsolved, claiming:

$G$ is a finite non-abelian $p$-group, $|G|=p^n$. Then $|G'|\neq p^{n-1}$

When $p=3$ we start with $Z(G)\neq 1$ to show that $|Z(G)|=|G'|=p$. But in above problem it seems to me that induction on all $p$-groups (finite and non abelian) with orders less than $n$ may be applicable. In fact for such a group, $|G|=p^n$, we have:$$|Z(G)|=p, p^2,... \mathrm{or}\ p^{n-2}$$Is my approach valid? Thanks.

share|cite|improve this question
up vote 8 down vote accepted

We can assume $n > 2$ because otherwise $G$ is abelian. In $p$-groups, there is a normal subgroup of every possible order. In particular, there is a normal subgroup $N$ of order $p^{n-2}$. Then $G/N$ is abelian and $N$ contains $G'$, proving the statement.

share|cite|improve this answer
Looks great. Very simple. The Frattini thing is occasionally useful ($G/G'$ cannot be any kind of cyclic), but for this specific index version, yours is much simpler. :-) – Jack Schmidt Sep 2 '12 at 18:37
+1 Painfully and beautifully simple. – DonAntonio Sep 3 '12 at 3:00

If $M$ is a maximal subgroup of the finite $p$-group $G$, then $M$ is normal in $G$ and the quotient is an abelian group of order $p$ ($M$ is properly contained in its normalizer, but it is maximal, so $G$ is its normalizer).

In particular, $G' \leq M$ for every maximal subgroup $M$. Now suppose $G/G'$ has order $p$, so that $G/G' = \langle zG' \rangle$ for some $z \in G$. Then if $z \in G'$, $G=G'=1$ (which is abelian, so irrelevant). Otherwise $z \notin G'$. If $\langle z \rangle = G$, then $G$ is abelian (so irrelevant). Hence $\langle z \rangle < G$ must be contained in some maximal subgroup $M$. But then both $z$ and $G'$ are contained in $M$, so $G = \langle z, G' \rangle \leq M < G$ is a contradiction.

In other words, if $G/\Phi(G)$ is cyclic, then $G$ is cyclic. In a $p$-group, $\Phi(G) = G^p G'$.

share|cite|improve this answer
So the induction in not a good way, however, I thought as you noted until $G/G'$ is cyclic but not further. Thanks. – Babak S. Sep 2 '12 at 18:54
So according to your answer we cannot have $Z(G)\leq M$ where $M$ is any maximal subgroup here. Right? – Babak S. Sep 2 '12 at 18:57
(Re 2nd comment:) I think you might have the logic backwards. If $H \leq M$ for all maximal subgroups $M$ and $H$ is normal in $G$, then $G/H$ cannot be cyclic without $G$ itself being cyclic. Of course $G/Z(G)$ cannot be cyclic (of order greater than 1) either, but this does not restrict the relationship between $Z(G)$ and $M$. – Jack Schmidt Sep 2 '12 at 22:20

Your Answer


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.