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?

  • $\begingroup$ The inner $C_G$ is with reference to an element of $G$ (namely $g$), the outer one to a subgroup of $G$ (namely $C_G(g)$): how would you define this latter?. $\endgroup$
    – user750041
    Mar 8, 2020 at 20:50
  • 4
    $\begingroup$ $g \in C_G(g) \Rightarrow C_G(C_G(g)) \le C_G(g) \Rightarrow C_G(C_G(g)) \le Z(C_G(g))$ which is abelian. $\endgroup$
    – Derek Holt
    Mar 8, 2020 at 21:01
  • $\begingroup$ Since centralizer of an element is a subgroup we can define the centralizer of its centralizer right? $\endgroup$
    – Rick
    Mar 8, 2020 at 21:02
  • $\begingroup$ @Derek Why is the last implication true? $\endgroup$
    – Rick
    Mar 8, 2020 at 21:03
  • 1
    $\begingroup$ @user750041 the centralizer of a subgroup is defined as the set of all elements of the group which commute with each element of the subgroup $\endgroup$
    – Rick
    Mar 8, 2020 at 21:06

2 Answers 2


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.


Let $H\le G$ and $g_1,g_2 \in C_G(H)$; then:

$$\forall h \in H, g_1g_2=g_1(hh^{-1})g_2=(g_1h)(h^{-1}g_2)=(hg_1)(g_2h^{-1})=h(g_1g_2)h^{-1} \tag 1$$

But $g_1g_2=g_2^{-1}(g_2g_1)g_2$, thence $(1)$ reads:

$$\forall h \in H, g_1g_2=h(g_1g_2)h^{-1}=h(g_2^{-1}(g_2g_1)g_2)h^{-1}=(hg_2^{-1})(g_2g_1)(hg_2^{-1})^{-1} \tag 2$$

Now, if $H=C_G(g)$, then $C_G(H) \le H^{(*)}$, so $\exists \bar h \in H \mid g_2=\bar h$; therefore $(2)$ implies:

$$g_1g_2=(\bar hg_2^{-1})(g_2g_1)(\bar hg_2^{-1})^{-1}=g_2g_1 \tag 3$$

Since $g_1,g_2$ are arbitrary in $C_G(C_G(g))$, this latter is abelian.

$^{(*)}$ In fact, let $H=C_G(g)$ and $\tilde g \in C_G(H)$; thence, $\tilde gh=h\tilde g, \forall h \in H$. Now, by definition of centralizer of $g$, it is $g \in H$; suppose, by contrapositive, $\tilde g \notin H$; thence, $\tilde gg \notin H \Rightarrow \tilde ggg\ne g\tilde gg \Rightarrow$ ($\tilde g$ commutes with every $h \in H$, and $g \in H$) $g\tilde gg \ne g\tilde gg$: contradiction. So, $\tilde g \in C_G(H) \Rightarrow \tilde g\in H$, whence $C_G(H)\le H$.


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