Centralizer of Group Element is Subgroup/Proof 1

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:

Thus $C_G \left({a}\right)$ is closed under $\circ$.

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

Then:

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

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