Conjugate Groups 
$H$ and $K$ are conjugates of a group $G$ with $a \in G$, where $aHa^-1= \{aha^{-1} : h \in H\}= K$.
Prove that the set $A= \{a \in G : aHa^{-1}=H\}$ is a subgroup of $G$.

For $a, b \in A$, $aHa^{-1} = H$ and $bHb^{-1}=H$. And, $(ab)H((ab)^{-1}) = abHb^{-1}a^{-1} = a(bHb^{-1})a^{-1} = aHa^{-1}$. So, $aHa^{-1}$ is contained within $G$, and $A$ is a subgroup of $G$.
Is the proof organized?
 A: Let $G$ be a group and $H$ a subgroup of $G$. We want to prove that the set $A=\{a\in G\,|\, aHa^{-1}=H\}$ is also a subgroup of $G$.
How do we show any subset of $G$ is a subgroup? There are three steps: the subset should be non-empty, the subset should be closed under products, and the subset should be closed under inverses. What do I mean by closure under products? I mean that if $a$ and $b$ are in the subset, then $ab$ is in the subset. What do I mean by closure under inverse? I mean that if $a$ is in the subset, then $a^{-1}$ is in the subset.
So now let's look at our particular subset $A$. We need to show three things: $A$ is not the empty set, $A$ is closed under products, and $A$ is closed under inverses. The important thing to remember throughout this is what does it mean to be an element of $A$ ? An element $x\in G$ is an element of this subset $A$, by definition, if $xHx^{-1}=H$.
Now, to see that $A$ is not empty, notice that $eHe^{-1}=eHe=H$ where $e$ is the identity element in $G$. The equality on the line above exactly means that $e\in A$, so $A$ is not the empty set.
Next we show that $A$ is closed under products. Take $a, b\in A$. Then $aHa^{-1}=H$ and $bHb^{-1}=H$ as you noted in your original post. So $$(ab)H(ab)^{-1}=abHb^{-1}a^{-1}=a(bHb^{-1})a^{-1}=aHa^{-1}=H$$ and this means that $ab\in A$.
Lasty, we show that $A$ is closed under inverses. Take $a\in A$. Then $aHa^{-1}=H$. Multiplying this equation on the right by $a$ gives $aH=Ha$; multipling this equation on the left by $a^{-1}$ gives $H=a^{-1}Ha$. But recall that $a=(a^{-1})^{-1}$. So this last equation can be rewritten: $$H=a^{-1}H(a^{-1})^{-1}$$ which means that $a^{-1}\in A$.
Thus, $A$ is non-empty, closed under products, and closed under inverses. So we conclude that $A$ is a subgroup of $G$.
