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Suppose $G$ is a finite group such that $\frac{G}{Z(G)}\cong Z_p\times Z_p\times Z_p$.
Is it true all centralizers of G are abelian?

m a s
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  • I guess this is not true. Suppose we have found some nonabelian $G$ such that $G / Z(G) \cong (Z/pZ)^3$. Then $C(Z(G)) = G$ which is nonabelian. The difficulty seems to be in finding such $G$ for each $p$. – Alex Vong Dec 09 '18 at 11:20

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As mentioned in the comments, if $G$ was non-abelian, then for $z\in Z(G)$, $C(z)=G$ would be a contradiction. But of course, if $G$ is abelian, then $G/Z(G)$ is trivial. So your claim is equivalent to: there exists no finite group $G$ with $G/Z(G)=Z_p^3$.

That's probably a hint this isn't true. Since $G$ is nilpotent, we can restrict attention to finite $p$-groups. Then we can get a counterexample for every $p$:

Let $H$ be the elementary abelian group of order $p^3$, generated by $x,y,z$. Let $\alpha,\beta\in Aut(H)$ be two order $p$ elements: \begin{align*} \alpha(x) = xy && \alpha(y)=y && \alpha(z)=z\\ \beta(x)=xz && \beta(y)=y && \beta(z)=z \end{align*} Then if $G$ is the group of order $p^5$, given by the semidirect product $H\rtimes\langle\alpha,\beta\rangle$, we have $Z(G)=\langle y,z\rangle$ and $G/Z(G)$ is the elementary abelian group generated by the images of $x$, $\alpha$, and $\beta$.

Hempelicious
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