Q1: Must the centralizer of a non-identity element of a group be abelian?

I challenge everyone by asking further about this question one step further.

The definition of centralizer is:

Let a be a fixed element of a group G. The centralizer of a in G, C(a), is the set of all elements in G that will commute with a. In symbols, $C(a)=\{g \in G∣ga=ag\}$.

We know if G is non-Abelian, then $C(1)=G$ with the identity $1$ still can be non-Abelian.

Q2: How about other $g' \in G$? Can $C(g')$ be non-Abelian, if $g' \neq 1$? Any explicit example please?


Q3: Is there a cyclic condition for $g′$ such that in order to have a non-Abelian $C(g′)$, we must have $(g′)^n=1$ for some integer $n$?

  • $\begingroup$ Let $G$ be non-abelian, but with nontrivial centre. Then the centraliser of a non-identity central element ... $\endgroup$ Jul 25, 2014 at 20:51
  • $\begingroup$ Take the quaternion group and $g' = -1$. $\endgroup$ Jul 25, 2014 at 20:54
  • $\begingroup$ But, assuming that $D_4$ and $H_8$ are the two nonabelian groups of order $8$, they both do work! $\endgroup$
    – Derek Holt
    Jul 25, 2014 at 21:06
  • $\begingroup$ Ok, so to summarize the above, is there a cyclic condition for $g'$ such that in order to have a non-Abelian $C(g')$, we must have $(g')^n=1$? $\endgroup$
    – wonderich
    Jul 25, 2014 at 21:13
  • $\begingroup$ @ Derek Holt, yes, $G=D_4$ and $G=Q_8$ with certain $g' \in G$ for $(g')^2=1$ works for $C(g')=G$. $\endgroup$
    – wonderich
    Jul 25, 2014 at 21:17

2 Answers 2


Certainly not in general. For example, let $G$ be a finite group which has a non-Abelian Sylow $p$-subgroup $P.$ Then $Z(P) \neq 1,$ and $C_{G}(z)$ is non-Abelian for each non-identity element of $Z(P).$

In fact, the structure of a finite group in which the centralizer of each non-identity element is Abelian is extremely restricted, and such groups were classified by M. Suzuki in the 1950s. This was an important step on the way to the classification of finite simple groups and, in particular, gave some pointers towards the proof by Feit and Thompson of the solvability of finite groups of odd order.


The centralizer of a central element is the entire group. So for your second question use a non-abelian group; $p$-groups will have a non-trival center. For your third question consider the group $G$ of invertible 2 by 2 matrices over a field $F$ of characteristic 0. The scalar element $2I_2\in G$ has a non-abelian centralizer and no power is the identity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.