Centralizer of Group Element is Subgroup/Proof 1
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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
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.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 37.1$. Some important general examples of subgroups