0
$\begingroup$

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.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – zibadawa timmy
    Commented Jul 3, 2016 at 16:41

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .