If $H$ is of finite index, Prove that, there is a subgroup $N$ of $H$ and of finite index in $G$ such that $aNa^{-1}= N$ for all $a$ in $G$. Can you find an upper bound for $[G:N]$?
Let, $a_1H,a_2H,..,a_mH$ is the complete list of left cosets of $H$ in $G$. Then, $\cup_{i=1}^{m} a_i H=G$ $\implies \cup_{i=1}^{m} a_i H a_j^{-1}=G\implies \cup_{i=1}^{m} b_i(a_j H a_j^{-1})=G$
for each $j=1,2,..m$ and $b_i\in G$ such that, $b_ia_j=a_i$ for $i=1,2,..,m$.
Now, we consider the subgroup, $N=\cap_{x\in G} xHx^{-1}=\cap_{i=1}^{m} a_iHa_i^{-1}$. Then, clearly $N$ is a normal subgroup of $G$ which is contained in $H$. Since, each of $a_iHa_i^{-1}$ is of finite index in $G$. Then, $\cap a_iHa_i^{-1}$ has also finite index in $G$. And we are done. Also, it's easy to see that, $[G:N]\le[G:a_1Ha_1^{-1}][G:a_2Ha_2^{-1}]..........[G:a_mHa_m^{-1}]$
Can anyone check it?
You proof works, but the end of the first paragraph (with the $b_j$'s) is superfluous. You also might want to spell out in more detail why the in the index of H in G is an upper bound on the index of an intersection of m+1 conjugates in an intersection of m of those m+1.
Finally, the simple proof is that H acts transitively on its left cosets in G. This provides a homomorphism from G to $S_{[G:H]}$. The kerner N is contained in every point stabilizer, one of which is H, and N has finite index since the image of the homomorphism is finite.