Centralizer of Group Element is Subgroup/Proof 1

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

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

Thus $C_G \paren a \ne \O$.

Let $x, y \in C_G \paren a$.

Then:

Thus $C_G \paren a$ is closed under $\circ$.

Let $x \in C_G \paren a$.

Then:

So:
 * $x \in C_G \paren a \implies x^{-1} \in C_G \paren a$

Thus, by the Two-Step Subgroup Test, the result follows.