Conjugate of Subgroup is Subgroup

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 $\text G 0$: 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.

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

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

The result follows from Stabilizer is Subgroup.

$\blacksquare$