Prove for each $a ∈ G, aHa^{-1}$ is a subgroup of G. Question:
Let $H$ be a subgroup of $G$. For any $a \in G$, let $aHa^{-1} = \{axa^{-1} : x \in H\}$; $aHa^{-1}$ is called a conjugate of $H$. Prove: For each $a \in G$, $aHa^{-1}$ is a subgroup of $G$.
I know in order to prove something is a subgroup it needs to be nonempty, closed under operation, contains it's identity, and inverse. 
I do believe to prove H contains it's inverse it goes as follows:
$aea^{-1} = aa^{-1}e = e$ for any $a \in G$. Therefore $e \in H$.
but after that I am at a loss.

One slick way to prove a subset $K$ of a group $G$ is a subgroup is to show:
1) The set $K$ is nonempty.
2) For every $x,y \in K, xy^{-1} \in K$.
(See if you can prove this criterion to be a subgroup!)
You've shown 1), since the identity $e \in K = aHa^{-1}$. Now let $x,y \in aHa^{-1}$. Then $x = aha^{-1}$ and $y = ah'a^{-1}$ for some $h,h' \in H$. But $y^{-1} = ah'^{-1}a^{-1}$ (check if you aren't familiar with this formula). Hence $xy^{-1} = aha^{-1}ah'^{-1}a^{-1} = ahh'^{-1}a^{-1}$. Since $H$ is a subgroup of $G$, $hh'^{-1} \in H$, so $xy^{-1} \in aHa^{-1}$, which completes the proof.
