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

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

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?

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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 $n$?

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

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 $Z(P).$
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.