Conjugate of Set by Identity

Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $S \subseteq G$.

Then the conjugate of $S$ by $e$ is $S$:

$S^e = S$

Proof

\(\ds S^e\)

\(=\)

\(\ds \set {y \in G: \exists x \in S: y = e \circ x \circ e^{-1} }\)

Definition of Conjugate of Group Subset

\(\ds \)

\(=\)

\(\ds \set {y \in G: \exists x \in S: y = x}\)

Definition of Identity Element

\(\ds \)

\(=\)

\(\ds S\)

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 45$