# Centralizer of derived subgroup

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

• For (1), no; say $G$ is the nonabelian group of order $p^3$ and exponent $p$, generated by $x$ and $y$; the center equals the commutator, $\langle [x,y]\rangle$; take $M=\langle [x,y],y\rangle$, so $G/M\cong C_p$. Then $MZ(G)=M$ because $Z(G)=[G,G]\subseteq M$, but $C_G(G')=C_G(Z(G))=G$. – Arturo Magidin Feb 16 '14 at 23:33
• The second and third questions are too imprecise. To get helpful answers you need to ask more directed questions. – Derek Holt Feb 17 '14 at 8:40
• @MarcoRuscitti: If you are "really" looking for a counterexample with more conditions, then you need to state those conditions. Otherwise, you are wasting everyone's time, including yours. State exactly what you want, with what conditions, rather than pose a question and then complain that the answers are not really what you want. Don't just add it in the comments, edit your question and put everything you want in the question. – Arturo Magidin Feb 17 '14 at 15:41
• (1) is still false; take $G = \langle x,y \mid x^8=y^8=[x,y]^2=[x,y,x]=[x,y,y]=1\rangle$. So $Z(G)=\langle x^2,y^2,[x,y]\rangle$, and $[G,G]=\langle x,y\rangle$. Take $M=\langle x,[x,y]\rangle$ so that $G/M\cong C_4$. Since $[G,G]\subseteq Z(G)$ (being class $2$), then $C_G(G')=G$, but $MZ(G)=\langle x,y^2,[x,y]\rangle\neq G$. More generally, take any nonregular group of class two with central quotient isomorphic to $C_{2^n}\times C_{2^n}$, and pick $M$ to the be the pullback of one of the cyclic factors; then $MZ(G)=M\neq G$, but $C_G(G')=G$. – Arturo Magidin Feb 17 '14 at 16:48
• In Qn 3, for a metabelian $p$-group $G$, the following three conditions are equivalent: (i) $C_G(G')=G'$; (ii) $G'$ is maximal as an abelian subgroup of $G$; (iii) $G'$ is maximal as a normal abelian subgroup of $G$. – Derek Holt Feb 17 '14 at 18:07

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 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 $x\in G'$. $\Box$