I would like to ask about the structure of non-abelian groups ($p$-groups) in which the centralizer of any non-central element is cyclic. Is there any classification about these groups?

  • $\begingroup$ As stated, all abelian groups have this property, so you should probably insist that your group is nonabelian. $\endgroup$ – Derek Holt Oct 19 '15 at 8:53
  • $\begingroup$ See mathoverflow.net/questions/128841/…. $\endgroup$ – darko Oct 19 '15 at 9:19
  • $\begingroup$ Dear Darko,mathoverflow.net/questions/128841/… is about Classification of groups in which the centralizer of every non-identity element is cyclic $\endgroup$ – Takjk Oct 19 '15 at 9:49

Let $G$ have this property, and $P \in {\rm Syl}_p(G)$ for some prime $p$.

If $P$ is nonabelian, then it must have a unique subgroup of order $p$. For, if not, let $P_1$ and $P_2$ be two distinct subgroups of order $p$, with $P_1 \le Z(P)$. So $Q = \langle P_1, P_2 \rangle \cong C_p \times C_p$. Then if $Q \not\le Z(P)$, an element of $P_2$ has noncyclic centralizer, whereas if $Q \le Z(P)$ then any element in $P \setminus Z(P)$ has noncyclic centralizer. There is a result that a $p$-group with a unique subgroup of order $p$ is cyclic or (generalized) quaternion with $p=2$, so if $P$ is nonabelian, then it must be generalized quaternion.

On the other hand, if $P$ is abelian then it must be cyclic, snce otherwise with would have a subgroup $Q= C_p \times C_p$ which would lead again to acontradiction in cases $Q \le Z(G)$ and $Q \not\le Z(G)$.

So in any case $P$ is cyclic or generalized quaternion with $p=2$.

If all Sylow $p$-subgroups are cyclic, then $G$ is metacyclic. That does not completely answer the question in that case, since not all metacyclic groups have the property, but it does give you a lot of information about $G$.

Otherwise $G$ has (generalized) quaternion Sylow $2$-subgroup and there is classification of such groups which says that $G/O(G)$has a unique element of order $g$, where $O(G)$ denotes the largest normal subgroup of $G$ of odd order. If you factor that out, then you get a group with dihedral Sylow $2$-subgroups which (using the fact that Sylow $p$-subgroups are cyclic for odd $p$) has to be either a $2$-group or else $G/O(G) = {\rm SL}(2,q)$ for some prime $q$. It is not hard to see that the condition in the question imples $O(G)=1$. The groups ${\rm SL}(2,q)$ with $q$ prime do indeed have the required property.


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