Centralizer of Group Element is Subgroup/Proof 1

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$


Sources