Centralizer of Group Element is Subgroup

Theorem
Let $$\left({G, \circ}\right)$$ be a group and let $$a \in G$$.

Then $$C_G \left({a}\right)$$, the centralizer of $$a$$ in $$G$$, is a subgroup of $$G$$.

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


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

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.