# Centralizer of Group Element is Subgroup/Proof 1

## Theorem

Let $\struct {G, \circ}$ be a group and let $a \in G$.

Then $\map {C_G} a$, the centralizer of $a$ in $G$, is a subgroup of $G$.

## 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:

 $\displaystyle x \circ a$ $=$ $\displaystyle a \circ x$ $\displaystyle y \circ a$ $=$ $\displaystyle a \circ y$ $\displaystyle \leadsto \ \$ $\displaystyle x \circ y \circ a$ $=$ $\displaystyle x \circ a \circ y$ $\displaystyle$ $=$ $\displaystyle a \circ x \circ y$ $\displaystyle \leadsto \ \$ $\displaystyle x \circ y$ $\in$ $\displaystyle C_G \paren a$

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

Let $x \in C_G \paren a$.

Then:

 $\displaystyle x \circ a$ $=$ $\displaystyle a \circ x$ $\displaystyle \leadsto \ \$ $\displaystyle x^{-1} \circ x \circ a \circ x^{-1}$ $=$ $\displaystyle x^{-1} \circ a \circ x \circ x^{-1}$ $\displaystyle \leadsto \ \$ $\displaystyle a \circ x^{-1}$ $=$ $\displaystyle x^{-1} \circ a$

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.

$\blacksquare$