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 $p ∈ H$
 A: *

*You should write sentences and paragraphs, not just lists of equations.

*Why is $pq = a(xy)a^{-1}$? Explain. (This is easy, so it will be a short explanation. But it should be explained in a class at this level.)

*Don't let $x,y \in H$. Instead say: Let $p, q \in H^a$. Then there are $x,y \in H$ such that $p = axa^{-1}$ and $q = aya^{-1}$. Therefore...

*Again, for "Inverse", don't say let $p \in H^a$ and $x \in H$. Instead, say: Let $p \in H^a$. Then there is an $x \in H$ such that $p = axa^{-1}$. Therefore...

*Why is $p^{-1} = ax^{-1}a^{-1}$? Explain.

*Why is $p^{-1} \in H^a$? Explain. (You have said $p^{-1} = ax^{-1}a^{-1}$. So? Explain why this implies $p^{-1} \in H^a$. Hint: $x^{-1} \in$...)


"Identity" is a mess. You wrote: $H^a = aha^{-1}$. No. No! $H^a$ is a subset of $G$. $aha^{-1}$ is a single element of $G$. They are not the same kind of thing: one is a set, the other is an element. They are not equal. Are you trying to say that $H^a$ is the set of $aha^{-1}$, for all $h \in H$? If that's what you want to say, then say that. Although I don't know why you would want to say that. If what you want to say is something else, then say that instead.
In the next line you write $H^a = aea^{-1}$ since $e \in H$. No!! Are you trying to say that $H^a$ includes the element $aea^{-1}$? There is a correct way to say that. You seem to know what that correct way is: you wrote $e \in H$. And indeed, you can write $aea^{-1} \in H^a$.
Well, I'm sorry to be critical of how you wrote this. You clearly understand the main mathematical ideas. I hope that you will write your proof more carefully. 
A: 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.

A: 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 $a h_1 a^{-1} (a h_2 a^{-1})^{-1} \in H'$:

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

