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;
* $Z(G) \leq \phi(G)$.


1) 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 $G=M\langle g\rangle$ and $g^{p^{n}} \in M$. Looking at $G'=[G,G]$, I showed that $G'=[M,g].$ Is it true, under these assumptions, that $C\_{G}(G')=MZ(G)$ ? 


2) If the answer is no, is there some other assumption for which my thesis is true? 


3) 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



Centralizer of derived subgroup
To summarize the material in the comments so that the question does not appear as unanswered...


(1) The first question has a negative answer as stated. For example, take 
$$G = \langle x,y\mid x^8=y^8=[x,y]^2=[x,y,x]=[x,y,y]=1\rangle$$
a $2$-group of order 128 and class $2$. Take $M=\langle [x,y],x\rangle$, so that $G/M$ is of order $8$. Since $Z(G)=\langle x^2,y^2,[x,y]\rangle$, then $MZ(G) = \langle x^2,y,[x,y]\rangle\neq G = C\_G(G')$. 


Question (2) seems a bit harder to answer precisely.


For Question (3), [Derek Holt](https://mathoverflow.net/users/35840/derek-holt) provides the following:


**Proposition.** *Let $G$ be a metabelian $p$-group. The following are equivalent:*


1. $C\_G(G')=G'$.
2. *$G'$ is maximal among abelian subgroups of $G$.*
3. *$G'$ is maximal among abelian normal subgroups of $G$.*


*Proof.* (1)$\implies$(2) If $G'\subseteq A$ with $A$ abelian then $A\subseteq C\_G(G')$. Hence $G'=A$.


(2)$\implies$(3) Immediate.


(3)$\implies$(1) That $G'\subseteq C\_G(G')$ follows because $G$ is metabelian. And any subgroup of $G$ that contains $G'$ is normal; so if $x\in C\_G(G')$, then $\langle G',x\rangle$ is abelian and normal, so by maximality $x\in G'$. $\Box$


