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?
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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 $n$?
Best Answer
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.