# If $[G:H]<\infty$, then $H$ contains a normal subgroup $N$ of $G$ such that $[G:N]<\infty$.

Let $G$ be a group and $H$ be a subgroup of $G$. I want to prove that if $[G:H]<\infty$, then $H$ contains a normal subgroup $N$ of $G$ such that $[G:N]<\infty$.

Professor gave me the following sketch of proof : Since $H\leq N_G(H)\leq G$, $[G:N_G(H)]< \infty$. Let $[G:N_G(H)]=t$. Then there exists $g_1, g_2, \cdots, g_t\in G$ such that $G=N_G(H)g_1\cup \cdots \cup N_G(H)g_t$. Then $N=\bigcap_{i=1}^{t}g_iHg_i^{-1}$ is a normal subgroup of G such that $[G:N]<\infty$.

But I cannot complete the proof because the last line does not clear for me. Please give me the detail. Thanks.

There are two claims in the last sentence: that $N$ is finite-index and that $N$ is normal. The first claim follows from Does the intersection of two finite index subgroups have finite index? and induction on $t$ (note that $gHg^{-1}$ is a subgroup of $G$ of index $[G:gHg^{-1}]=[G:H]$ for each $g\in G$).
For the second claim, choose $g\in G$ and $n\in N$. We wish to show that $gNg^{-1} = N$. We have: $$gNg^{-1} = g \left( \bigcap_i g_i H g_i^{-1}\right) g^{-1} = \bigcap_i (g g_i) H (g g_i)^{-1}$$ It would be easier if instead of what you had written, the $g_i$s are coset representatives on the other side, and so that's what I will assume: $G = g_1 N_G(H) \cup \dots \cup g_t N_G(H)$. Then for each $i$ we may factor $gg_i = g_{j(i)}m_i$ for some $j$ depending on $i$ and $m_i \in N_G(H)$. Then $$(g g_i) H (g g_i)^{-1} = g_{j(i)}m_i H m_i^{-1} g_{j(i)}^{-1} = g_{j(i)} H g_{j(i)}^{-1}$$ since $m_i$ normalizes $H$. Therefore: $$\bigcap_i (g g_i) H (g g_i)^{-1} = \bigcap_i g_{j(i)} H g_{j(i)}^{-1} \supseteq \bigcap_j g_j H g_j^{-1}$$ A priori, the last containment might not be an equality: the intersection over $i$ is equivalent to the intersection over only those $j$ of the form $j(i)$ (the function $j(i)$ depends on the choice of $g$), whereas the right-hand side intersects over possibly more $j$s, and so can result in something smaller.
But this does prove that $gNg^{-1} \supseteq N$ for all $g\in G$. In particular, $g^{-1}Ng \supseteq N$, but conjugating by $g$ shows $N \supseteq gNg^{-1}$. This completes the proof.
• Yes. The "real" reason is that $gGg^{-1} = G$, and $[G:H] = [gGg^{-1}:gHg^{-1}]$. But more hands on, suppose that $\{g_i\}_{i\in I}$ is a (possibly infinite) set of coset representatives for $G/H$. Then $\{gg_ig^{-1}\}_{i\in I}$ is a set of coset representatives for $G/gHg^{-1}$. – Theo Johnson-Freyd Mar 4 '14 at 2:39
There is an easy alternative for this proof (which upon inspection is probably just another way to say the same thing). By hypothesis there are finitely many cosets $gH$ in $G/H$. The group $G$ acts by left multiplication on this set of cosets, defining a group morphism $\rho:G\to S(G/H)$, the codomain being the finite group of permutations of the cosets. Then $\ker\rho$ is a normal subgroup (like any kernel) of finite index in$~G$ (the index is equal to the order of the image of$~\rho$, a subgroup of $S(G/H)$). Also $\ker\rho< H$ since its left multiplication by any element of$~\ker\rho$ must in particular stabilise the coset $eH$.