In all questions suppose $G$ metabelian p-group such that
G is not regular ( so $cl(G) \geq p$ ), G is not a wreath product;
$Z(G) \leq \phi(G)$.
Let $M$ normal abelian subgroup of $G$ such that $\frac{G}{M} \cong C_{p^{n}}$ with $n \geq 2$. So it exists an element $g \in G - M$ such that $G=M<g>$ and $g^{p^{n}} \in M$. Looking at $G'=[G,G]$ i showed that $G'=[M,g].$ Is it true, under these assumptions, that $C_{G}(G')=MZ(G)$ ?
If the answer is no, is there some other assumption for which my thesis is true?
In every metabelian p-group G, since G' is abelian, we have that $G' \leq C_{G}(G')$. Are there suitable assumptions for which $G' = C_{G}(G')$? I know that this is true when $G'$ is maximal (but this means G cyclic) and when $G'$ is maximal over normal abelian subgroups.
I edited my post since it was not clear, i apologize for this fact, and i'm grateful for your attention to my problem.
Best regards
Marco, PhD student