# Centralizer of Group Element is Subgroup

## 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 1

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:

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

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

Let $x \in C_G \paren a$.

Then:

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

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.

$\blacksquare$

## Proof 2

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

We have that:

$\forall a \in G: e \circ a = a \circ e \implies e \in \map {C_G} a$

Thus $\map {C_G} a \ne \O$.

Let $x, y \in \map {C_G} a$.

$a \circ x \circ y^{-1} = x \circ y^{-1} \circ a$

so:

$x \circ y^{-1} \in\map {C_G} a$

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

$\blacksquare$