Could this identity have an application? Let $H$ be a subgroup of $G$.
While trying to show another thing, I observed that $$C_G(C_G(C_G(H)))= C_G(H)$$
Does this identity have any application?
 A: I don't know about applications, but perhaps I can offer a bit of perspective.

Definition 0. Let $P$ denote a poset. Then by an orthomap on $P$, I mean an order-reversing mapping $P \rightarrow P$
  denoted $x \mapsto x^\perp$ satisfying either and therefore both of
  the following.
  
  
*
  
*$x^{\perp\perp} \geq x$
  
*$x^\perp \geq y \iff x \leq y^\perp$.
  

Exercise: show these are equivalent.

Proposition. Every orthomap satisfies $x^{\perp\perp\perp} \;= x^\perp.$

Proof. We know that LHS $\geq$ RHS, by substituting $x^\perp$ for $x$ in (1). On the other hand, we know that LHS $\leq$ RHS by applying the function $x \mapsto x^\perp$ to both sides of (1) and noting that it is order-reversing.
Orthomaps arise everywhere, in the following way. Suppose $X$ is a set and $R$ is symmetric relation on $X$. Then we get an orthomap $\mathcal{P}(X) \rightarrow \mathcal{P}(X)$ as follows. Given $A \subseteq X$, we define $A^\perp$ to be the unique subset of $X$ such that for all $x \in X$, the following are equivalent.


*

*$x \in A^\perp$

*for all $a \in A$, we have that $R(a,x)$.


Exercise: show this is an orthomap.
This gives many examples.


*

*Suppose $X$ is a group, and let $R$ denote the binary relation on $X$ defined by $R(x,y) \iff xy=yx.$ Then $A^\perp$ is the centralizer of $A$.

*Consider the case where $X$ is a commutative ring and $R$ is given by $R(x,y) \iff xy=0$. Then $A^\perp$ is the annihilator of $A$.

*Let $X$ denote an inner product space and suppose $R$ is given by $R(x,y) \iff \langle x,y\rangle = 0$. Then $A^\perp$ is the orthogonal complement of $A$.
Interestingly, we always get a notion of "closure operator" from any orthomap. Let me be a little more precise:

Definition 1. Let $P$ denote a poset. Then by a closure operator on $P$, I mean an order-preserving mapping $P \rightarrow P$
  denoted $x \mapsto \mathrm{cl}(x)$ satisfying both of the following conditions.



*

*$\mathrm{cl}(x) \geq x$

*$\mathrm{cl}(\mathrm{cl}(x)) = \mathrm{cl}(x)$


Exercise. Suppose $P$ is a poset equipped with an orthomap. Show that the function $\mathrm{cl} : P \rightarrow P$ defined by $\mathrm{cl}(x) = x^{\perp \perp}$ is a closure operator.

Definition 2. Suppose $P$ is a poset equipped with a closure operator. Then we call $k \in P$ closed iff one - and therefore all
  three - of the following holds.
  
  
*
  
*$k=\mathrm{cl}(p)$ for some $p \in P$.
  
*$\mathrm{cl}(k) = k$.
  
*$\mathrm{cl}(k) \leq k$.
  

In this enriched language, we can state our earlier proposition differently: it is just saying that $x^\perp$ is always closed.
