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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 $G=M<g>$ and $g^{p^{n}} \in M$. Looking at $G'=[G,G]$ i showed that $G'=[M,g].$ Is it true, under these assumptions, that $C_{G}(G')=MZ(G)$ ?

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

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

Say $G$ a $p$-group, $M$ normal abelian subgroup of G such that $$\frac{G}{M} \cong C_{p^{n}}.$$

It exists $g \in G - M $ such that $G=M\langle g\rangle$.

I showed that the derived subgroup is $G' = [M,g]$.

1. Is it true that $C_{G}(G')=MZ(G)$ ?

2. When is this assertion true?

3. In a metabelian $p$-group, $G'$ is abelian, so it's contained in its centralizer, right? When is it self centralizing?

Best regards

Marco, PhD student.

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 $G=M<g>$ and $g^{p^{n}} \in M$. Looking at $G'=[G,G]$ i showed that $G'=[M,g].$ Is it true, under these assumptions, that $C_{G}(G')=MZ(G)$ ?

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

Say G a p-group, M normal abelian subgroup of G such that $\frac{G}{M} \cong C_{p^{n}}.$

It exists $g \in G - M $ such that $G=M<g>.$

I showed that the derived subgroup is $G' = [M,g].$

1. Is it true that $C_{G}(G')=MZ(G) ?$

2. When is this assertion true?

3. In a metabelian p-group G' is abelian, so it's contained in its centralizer, right? When is it self centralizing?

Best regards

Marco, PhD student.

Say $G$ a $p$-group, $M$ normal abelian subgroup of G such that $$\frac{G}{M} \cong C_{p^{n}}.$$

It exists $g \in G - M $ such that $G=M\langle g\rangle$.

I showed that the derived subgroup is $G' = [M,g]$.

1. Is it true that $C_{G}(G')=MZ(G)$ ?

2. When is this assertion true?

3. In a metabelian $p$-group, $G'$ is abelian, so it's contained in its centralizer, right? When is it self centralizing?

Best regards

Marco, PhD student.