Centralizer of Group Element is Subgroup/Proof 2

Proof
Let $\left({G, \circ}\right)$ be a group.

We have that:
 * $\forall a \in G: e \circ a = a \circ e \implies e \in C_G \left({a}\right)$

Thus $C_G \left({a}\right) \ne \varnothing$.

Let $x, y \in C_G \left({a}\right)$.

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:
 * $x \circ y^{-1} \in C_G \left({a}\right)$

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