Centralizer of Group Element is Subgroup/Proof 2

Proof
Let $\struct {G, \circ}$ be a group.

We have that:
 * $\forall a \in G: e \circ a = a \circ e \implies e \in \map {C_G} a$

Thus $\map {C_G} a \ne \O$.

Let $x, y \in \map {C_G} a$.

Then from Commutation with Group Elements implies Commuation with Product with Inverse:
 * $a \circ x \circ y^{-1} = x \circ y^{-1} \circ a$

so:
 * $x \circ y^{-1} \in\map {C_G} a$

The result follows by the One-Step Subgroup Test, the result follows.