# p-groups with special property on its centralizers

Thanks for any help or comment.

Suppose $G$ is a finite non-abelian p-group. Suppose $G$ has a proper non-abelian subgroup $M$ such that for every non-central element $x\in M$, $C_G(x)\subseteq M$. Is someone have any comment, theorem or reference about the structure of these group.

If $M$ is abelian then AC-groups are example of such groups and if $G$ is center less then groups with cc-subgroups are examples of these groups but in my consideration $G$ has center and $M$ is non-abelian.

• The condition implies that $C_G(M)=Z(M)$, and there are some results concerning $p$-groups that satisfy this for certain or all non-abelian subgroups. See this paper, for example. Those don't cover your specific (in some sense more general) question, though maybe it can lead you in good directions. Jul 3, 2016 at 16:41

We claim that the class of $G$ is not bounded. Let $G$ be a $3$-group of maximal class and order $3^n>3^4$ containing no abelian subgroup of index $3$. Then $G$ has a metacyclic subgroup $M$ of index $3$ such that $|M'|=3$. It follows that $M$ is minimal nonabelian so that $|M:\text{Z}(M)| =3^2$. If $x\in M-\text{Z}(M)$, then $\text{C}_G(x)=\langle x,\text{Z}(M)\rangle$ is a subgroup of $M$ of index $3$. It is known that $|G|$ is not bounded so the same holds for $\text{cl}(G)$. But it is not known if the derived length of $G$ is bounded.