Exercise 4.2.8 (If $|G:H| = n$, there is some normal subgroup $K$ of $G$ with $K \leq H$ and $|G : K | \leq n!$)

Proof. Let $G$ acts on the set $A$ of left cosets of $H$, and let ${\pi }_H$ be the corresponding representation. By Theorem 3,

Put $K=\ker <!--\; FUNCTION\; APPLICATION\; -->({\pi }_H)$ ( $K$ is the core of $H$ in $G$). Then $K\le H$ and $K⊴G$ (all is said in the proof of Theorem 3).

Moreover, $G∕K=G∕\ker <!--\; FUNCTION\; APPLICATION\; -->({\pi }_H)\simeq {\pi }_H(G)\le S_A$, thus $G∕K$ is isomorphic to a subgroup of $S_A$, where the index of $H$ in $G$ is $|A|=n$, so $S_A\simeq S_n$. This shows that $G∕K$ is isomorphic to a subgroup of $S_n$, so $|G:K|\le n!$.

If $H$ has a finite index $n$ then there is a normal subgroup $K$ of $G$ with $K\le H$ and $|G:K|\le n!$. □