Let $G$ be a group and $H\le G$ with $(G:H)<\infty$. Then, there is a subgroup $N$ of $H$ such that $N\unlhd G$ and $(G:N)<\infty$.
I have constructed such $N$: $$ N=\bigcap_{i=1}^m\ g_i H g_i^{-1}, $$ where $G=\bigcup_{i=1}^m g_i N_G(H)$ ($m$ conjugacy classes of $H$), and showed $N$ is normal.
What remains to prove is $(G:gHg^{-1})=(G:H)$, so that $$ (G:N)\le\prod_{i=1}^m\ (G:H)<\infty. $$ How can I prove that?