Centralizer of Group Element is Subgroup/Proof 2
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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:
- $\forall a \in G: e \circ a = a \circ e \implies e \in \map {C_G} a$
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
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Exercise $(11)$