Prove that a conjugate of a subgroup is a subgroup 
Let $G$ be a group and $H$ be a subgroup of $G$ and $a\in G$ fixed, then
$$H^{a}=aHa^{-1}=\{aHa^{-1} \colon h\in H\}$$
is a subgroup of $G$.

my attempt
Identity
$aha^{-1} \in H^a$
$aea^{-1} \in H^a\ \ \ \ $ Since $e ∈ H$
$aa^{-1} \in H^a$
Therefore $aa^{-1} = e ∈ H^{a}$
Closed under the operation of $G$
Let $p,q ∈ H^{a}$ and $x,y ∈ H$
$p = axa^{-1}$ q = $aya^{-1}$
as $H$ is a group
$x*y ∈ H$
$p*q = a(xy)a^{-1} ∈ H$
Inverse
let p ∈ $H^{a}$ and $x ∈ H$
$p = axa^{-1}$
$p^{-1} = ax^{-1}a^{-1}$
Then, $p^{-1} ∈ H^a$ for all $p ∈ H$

Use subgroup criterion:

$a 1 a^-1 = 1 \in H'$, so H' is not empty.

Say $\{h_1, h_2\} \in H$, so $\{a h_1 a^{-1}, a h_2 a^{-1}\} \in H'$. We show that $a h_1 a^{-1} (a h_2 a^{-1})^{-1} \in H'$:

$a h_1 a^{-1} (a h_2 a^{-1})^{-1} = a h_1 a^{-1} a h_2^{-1}a^{-1} = a h_1 h_2^{-1}a^{-1} \in H'$.

