Conjugate of Subgroup is Subgroup

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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 1

Let $H \le G$.


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

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


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

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


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

$\blacksquare$


Proof 2

Let $*: G \times G / H \to G / H$ be the group action on the (left) coset space:

$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$

It is established in Action of Group on Coset Space is Group Action that $*$ is a group action.


Then from Stabilizer of Coset under Group Action on Coset Space:

$\Stab {a H} = a H a^{-1}$

where $\Stab {a H}$ the stabilizer of $a H$ under $*$.


The result follows from Stabilizer is Subgroup.

$\blacksquare$


Sources