# Is it true all centralizer of G are abelian?

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?

• 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 9 '18 at 11:20

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 $$\beta$$.