The number of subgroups conjugate to a given subgroup of a finite group Let $H$ and $K$ are subgroups of $G$ conjugate to each other.
$A$ is defined as 
$$A = \{a \in G  \mid aHa^{-1} = H \}$$ for all $a\in G$. 
Prove that $A$ is a subgroup of $G$ and prove that if $G$ is finite, then the number of subgroups that are conjugate to $H$ equals $|G|/|A|$.
So far I've proved $A$ is a subgroup of $G$ by showing that it is a nonempty set that is closed under the operation and contains the inverse:
When $a,\ b \in A$, then 
$$(ab)H(ab)^{-1} = abHb^{-1}a^{-1}=a(bHb^{-1})a^{-1} = aHa^{-1} = H$$ 
Therefore if $a,b \in A,\ ab\in A$ and the set is closed under the operation.
For $a\in A$ and $h \in H,\ aha^{-1} = k$ for some $k\in H$.
Then $a^{-1}aha^{-1}a = a^{-1}ka$ and $a^{-1}ka = h$ for $k \in H,\ h \in H$ and $a^{-1}Ha = H$. Therefore when $a\in A,\ a^{-1} \in A$. 
As the subset $A$ is closed under the operation and contains the inverse of its elements, $A$ is a subgroup of $G$.

If $aHa^{-1}=bHb^{-1}$, $b^{-1}aH(b^{-1}a)^{-1}=H$, so $b^{-1}a\in A$, which is equivalent to $aA=bA$.
Thus ths number of different $aHa^{-1}$ is same as the number of different cosets of $A$, which is same as $|G|/|A|$ by Lagrange's theorem in group theory.
