Centralizer of Group Element is Subgroup

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

Then from Commutation with Group Elements implies Commuation with Product with Inverse:

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

so:

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

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

$\blacksquare$


Sources

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