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 $p ∈ H$

Let $H \leq G$ 
Let $H'= \{aHa^- \}$ be some congjugate Subgroup cnjugated by some $ a \in G$ where
$\{ah_1a^- , ah_2a^- .. \} \in H' $ for every $\{h_1 , h_2 ..\} \in H $


*

*$(ah_1a^-)(ah_2a^-) = ah_1h_2a^- = ah_3a^- \in H'$. This proves closure

*Associativity can be inhertied from group compsotion since they are all either subset or subgroup of the larger group $G$

*By Conjuagation of $a$ with $e \in H$ , $aea^- \in H' \Rightarrow e \in H'$. 

*For some $\{h ,h^- \}\in H$ there is $ \{ aha^-,ah^-a^-\} \in H'$ . Direct compostion of these two elements in $H'$. we have $$(aha^-)(ah^-a^-) = e $$ 
We have containment of inverse element in the group.

