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Upon learning that all $p$-groups of order $p^5$ are metabelian I wanted to investigate the case of order $p^4$. A group $G$ is metabelian if $G'$, its commutator subgroup, is abelian; equivalently, $G$ is metabelian if and only if there is an abelian normal subgroup $A$ such that the quotient group $G/A$ is abelian.

A natural choice of abelian normal subgroup is the center, $Z(G)$. If $|G:Z(G)|=1$, $G$ is abelian and is therefore metabelian. $G$ is also abelian if $|G:Z(G)|=p$ because if $G/Z(G)$ is cyclic then $G$ is abelian. When $|G:Z(G)|=p^2$, $G/Z(G)$ is abelian and therefore $G$ is metabelian.

I haven't been able to work out the last case, where $|G:Z(G)|=p^3$ (i.e., $|Z(G)|=p$). I tried showing $G'$ is abelian by showing $G' \le Z(G)$ since we know $G'$, being normal in $G$, intersects the center nontrivially.

How can I finish this proof that all $p$-groups of order $p^4$ are metabelian, or is there a counterexample?

Alex Petzke
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1 Answers1

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Note that if $|Z(G)| = p$ then $G/Z(G)$ need not be abelian. Example: Semidihedral group of order 16.

Instead note that a $p$-group has normal subgroups of every order $p^k$. So take one of order $p^2$. Both the quotient and the group itself have order $p^2$ and are thus abelian. This completes the proof.

mathperson314
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