Conjugate of Subgroup is Subgroup/Proof 1

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Theorem

Let $G$ be a group.

Let $H \le G$ be a subgroup of $G$.


Then the conjugate of $H$ by $a$ is a subgroup of $G$:

$\forall H \le G, a \in G: H^a \le G$


Proof

Let $H \le G$.


First, we show that $x, y \in H^a \implies x \circ y \in H^a$:

\(\ds x, y\) \(\in\) \(\ds H^a\)
\(\ds \leadsto \ \ \) \(\ds a x a^{-1}, a y a^{-1}\) \(\in\) \(\ds H\) Definition of Conjugate of Group Subset
\(\ds \leadsto \ \ \) \(\ds \paren {a x a^{-1} } \paren {a y a^{-1} }\) \(\in\) \(\ds H\) Group Axiom $\text G 0$: Closure
\(\ds \leadsto \ \ \) \(\ds a \paren {x y} a^{-1}\) \(\in\) \(\ds H\)
\(\ds \leadsto \ \ \) \(\ds x y\) \(\in\) \(\ds H^a\) Definition of Conjugate of Group Subset


Next, we show that $x \in H^a \implies x^{-1} \in H^a$:

\(\ds x\) \(\in\) \(\ds H^a\)
\(\ds \leadsto \ \ \) \(\ds a x a^{-1}\) \(\in\) \(\ds H\) Definition of Conjugate of Group Subset
\(\ds \leadsto \ \ \) \(\ds \paren {a x a^{-1} }^{-1} = a x^{-1} a^{-1}\) \(\in\) \(\ds H\) Power of Conjugate equals Conjugate of Power
\(\ds \leadsto \ \ \) \(\ds x^{-1}\) \(\in\) \(\ds H^a\) Definition of Conjugate of Group Subset


Thus by the Two-Step Subgroup Test, $H^a \le G$.

$\blacksquare$


Sources

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