Centralizer of centralizer of an element is abelian

Question

Is $C_G(C_G(g)))$ abelian for a group $G$?

I suppose this statement is false but cannot come with an example.

Can I get some hints, please?

There are examples where the centralizer itself is not abelian but how do I extend to centralizer of a centralizer?

Answer 1

Proposition Let $H,K$ subgroups of a group $G$, then the following hold.
$(a)$ If $H \leq K$ then $C_G(K) \leq C_G(H)$.
$(b)$ $H \leq C_G(C_G(H))$.
$(c)$ $C_G(H)=C_G(C_G(C_G(H)))$.
$(d)$ If $H$ is abelian, then $C_G(C_G(H)) \subseteq C_G(H)$.
$(e)$ If $H$ is abelian then $Z(C_G(H))=C_G(C_G(H))$, in particular $C_G(C_G(H))$ is abelian.
Conversely, if $Z(C_G(H))=C_G(C_G(H))$, then $H$ is abelian.

Proof $(a)$ is obvious.
$(b)$ Let $h \in H$, and $x \in C_G(H)$, then $xh=hx$ by definition, hence $h$ centralizes $C_G(H)$.
$(c)$ Replacing $H$ by $C_G(H)$ in (b) we obtain $C_G(C_G(H)) \subseteq C_G(C_G(C_G(H)))$. But applying (a) to (b) yields the reverse inclusion: $C_G(C_G(C_G(H))) \subseteq C_G(C_G(H))$.
$(d)$ If $H$ is abelian, then obviously $H \subseteq C_G(H)$. Hence, by (a) we are done.
$(e)$ Observe that in general $Z(H)=H \cap C_G(H)$. If $H$ happens to be abelian, then, by applying (d) we have $Z(C_G(H))=C_G(H) \cap C_G(C_G(H))=C_G(C_G(H)).$ The converse statement follows from (b).

Remark Since $C_G(g)=C_G(\langle g \rangle)$, the above proves your question.

Note (added January 2024) From $(c)$ above we see that the centralizers are the same at each odd iteration, while in $(b)$ equality does hold necessarily (e.g. take $H$ a proper subgroup of an abelian group $G$). However, if $H$ is a member of the so-called Chermak-Delgado lattice, then $H=C_G(C_G(H))$, see chapter $1G$, p. 41 in Finite Group Theory, by M.I. Isaacs.