# Centralizer of centralizer of an element is abelian

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?

• 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?.
– user750041
Mar 8, 2020 at 20:50
• $g \in C_G(g) \Rightarrow C_G(C_G(g)) \le C_G(g) \Rightarrow C_G(C_G(g)) \le Z(C_G(g))$ which is abelian. Mar 8, 2020 at 21:01
• Since centralizer of an element is a subgroup we can define the centralizer of its centralizer right?
– Rick
Mar 8, 2020 at 21:02
• @Derek Why is the last implication true?
– Rick
Mar 8, 2020 at 21:03
• @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
– Rick
Mar 8, 2020 at 21:06

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 $$C_G(H)\le H$$.