Properties of centralization in group theory Recall Let $C\subseteq G$ where $G$ is a group. If $x\in G$ for any $c\in C,$ $xc=cx,$ then $x$ centralizes $C$
$$C_G(C) := \{x \in G : \text{ for any } x\in C, xc=cx \}$$
Clearly,
$$C_G(C)=\bigcap_{c\in C}C_G(c)$$

Prove that
a) $C\subseteq C_G(C_G(C))$
b) If $C\subseteq D$ then $C_G(D)\subseteq C_G(C)$
c) $C_G(C_G(C_G(C)))=C_G(C)$

My Proof-trying:
a) Let $c\in C$. We will show $c\in C_G(C_G(C)).$ We need to show for any $x\in C_G(C),$ $xc=cx.$
Note that $x\in C_G(C)$ $\iff$ for any $c\in C$ $xc=cx$ by definition.
Hence $c\in C_G(C_G(C)).$ Therefore $C\subseteq C_G(C_G(C)).$
b) Assume $C\subseteq D.$ Let $x\in C_G(D).$ Then for any $d\in D$, we have $dx=xd.$ Since $C\subseteq D$, then for any $c\in C$, we have $cx=xc.$ So we are done.
c)I couldn't show it.
Can you check my proof, if there is a false, can you edit? Can you help for c)? Thanks...
 A: The proofs for a) and b) are OK.
For c), first, apply b) to the inclusion a) to obtain one inclusion.
Second, to obtain the reverse inclusion, apply a)  to $C_G(C)$.
A: 
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).
