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.


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.

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    $\begingroup$ 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. :-) $\endgroup$ – Jack Schmidt Sep 2 '12 at 18:37

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, $\Phi(G) = G^p G'$.

  • $\begingroup$ So the induction in not a good way, however, I thought as you noted until $G/G'$ is cyclic but not further. Thanks. $\endgroup$ – Mikasa Sep 2 '12 at 18:54
  • $\begingroup$ So according to your answer we cannot have $Z(G)\leq M$ where $M$ is any maximal subgroup here. Right? $\endgroup$ – Mikasa Sep 2 '12 at 18:57
  • $\begingroup$ (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 $M$. $\endgroup$ – Jack Schmidt Sep 2 '12 at 22:20
  • $\begingroup$ @JackSchmidt: What is the meaning of $\Phi(G)$? $\endgroup$ – Bumblebee Nov 10 '20 at 6:25

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