Proof of Conjugate Subgroup Isomorphism Let $G$ be a group, and let $H$ be a subgroup of $G$. Prove that if $a$ is an element of $G$, then the subset $aHa^{-1} = \{g ∈ G | g = aha^{-1} \text{ for some } h \in H\}$ is a subgroup of $G$ that is isomorphic to $H$. 
Proof:
Let $G$ be a group and $H$ is a subgroup of $G$. Suppose $a$ is any element of $G$. Now $ϕ: H \rightarrow aHa^{-1}$ be defined by $ϕ(h) = aha^{-1}.$  So $$ϕ(h_1 h_2) = ah_1h_2a^{-1} = ah_1a^{-1} ah_2a^{-1} = ϕ(h_1) ϕ(h_2)  $$
Thus, ϕ is a homomorphism.
(one-to-one)
Now
$$\begin{eqnarray*}h \in \operatorname{ker}(ϕ) &\text{if and only if}&ϕ(h) = 0\\
&\text{if and only if}&aha^{-1} = 0\\
&\text{if and only if}&h = 0\\
\end{eqnarray*}
$$
Hence, $\operatorname{ker}(ϕ) = \{0\}$. Thus $ϕ$ is one-to-one.
(onto)
Let $y \in aHa^{-1}$. Then $y = aha^{-1}$ for some $h \in H$.  As $h \in H$, $ϕ(h) = aha^{-1} = y$.  Thus $ϕ$ is onto.  
Hence, $ϕ$ is an isomorphism. 
How do I show that $aHa^{-1}$ is a subgroup of $G$?
 A: Let me show you how I would write each step for the subgroup verification.
(identity)
$aa^{-1}=1\in aHa^{-1}$, so $aHa^{-1}$ has identity.
(inverses)
If $x\in H$ then $x^{-1}\in H$, so $$\left(axa^{-1}\right)^{-1}=ax^{-1}a^{-1}\in aHa^{-1}.$$
(associativity)
Associativity is trivially inherited from the supergroup.
(closure)
Take $axa^{-1}\in aHa^{-1}$ and $aya^{-1}\in aHa^{-1}$ with $a,y\in H$.  $$axa^{-1}aya^{-1}=axya^{-1}$$ By closure of $H$, $xy\in H$, so $axya^{-1}\in aHa^{-1}$.

Alternatively, you could compose your homomorphism $\phi$ with the zero homomorphism $x\mapsto 1$.  The image of $\phi$ is the kernel of this homomorphism, and you have proven that $\phi$ is onto.
A: First, we need to establish the existence of inverses:
Consider any element $aha^{-1} \in aHa^{-1}$. Then $(aha^{-1})^{-1} = ah^{-1}a^{-1}$.  Certainly, this is in $aHa^{-1}$ since $h^{-1} \in H$.
Next, we check that it is closed under multiplication:
Consider a second element $ah'a^{-1} \in aHa^{-1}$.  Then $ah'a^{-1}aha^{-1} = ah'ha^{-1}$.  Certainly, this is in $aHa^{-1}$ since $h'h \in H$.
I'll let you check that the identity is also contained in our subgroup.
A: It suffices to check: $e \in aHa^{-1}$ and given arbitrary elements $x, y \in aHa^{-1}$, we have $xy^{-1} \in aHa^{-1}$. 
The first condition follows from the fact that $H$ is a subgroup and so is the the second: let $x = ala^{-1}$ and $y = aka^{-1}$, then $xy^{-1} = a(lk^{-1})a^{-1} = aha^{-1} \in aHa^{-1}$. 
