Let $G$ be a group and $H\leq G$. Suppose $[G:H]$ is finite. Show that
there exists a normal subgroup $N \subseteq H$ in $G$ which is also of
finite index in $G$.
My idea was to use $$N := \bigcap_{g \in G} gHg^{-1}$$ It is clear that $N$ is normal in $G$ and that $H$ contains $N$. Is it true that this specific $N$ is of finite index? How would I show it?
Best Answer
Thanks to the hint of user1952009 I managed to solve the exercise with proof. Let $$X := \{ xH : x \in G\}$$ and consider the group action $$\begin{cases} G \times X\to X\\ (g,xH) \mapsto gxH\end{cases}$$ Naturally we can associate with each group action a bijection $$\lambda_g:\begin{cases}X\to X\\ xH \mapsto gxH\end{cases}$$ for $g \in G$ and a homomorphism $$\lambda:\begin{cases}G\to S_X\\ g \mapsto \lambda_g\end{cases}$$ where $S_x$ denotes the symmetric group of $X$. Now we show $$\ker \lambda = \bigcap_{g \in G}gHg^{-1}$$ Let $y \in \bigcap_{g \in G}gHg^{-1}$. Thus we can write $y = xh_x x^{-1}$ with $x \in G$ and $h_x \in H$. Therefore for any $x \in G$ we have $$\lambda_y(xH) = yxH = xh_xx^{-1}xH = xh_xH = xH$$ Conversly, if $y \in \ker \lambda$ we have $\lambda_y = \operatorname{id}_X$. Hence $$yxH = xH$$ for all $xH \in X$. Therefore $$yxh_1 = xh_2 \quad \Leftrightarrow \quad y = xh_2h_1^{-1}x^{-1} \in xHx^{-1}$$ for all $x \in G$. Thus $y \in\bigcap_{g \in G}gHg^{-1}$. By the isomorphism theorem we have $$G/\ker \lambda = G/\bigcap_{g \in G}gHg^{-1} \cong \operatorname{Im} \lambda \leq S_X$$ And since $[G:H]$ is finite, $X$ is finite of cardinality $[G:H]!$. Thus we conclude that $[G:\bigcap_{g \in G}gHg^{-1}]$ is also finite.