If $|G| \nmid m!$, then $\exists \ H \neq \{e\} $ that $H \leq K$ and $H \unlhd G$.

Question

Let $G$ be a Group and $K \leq G$ such as $[G:K]=m$.

Show that if $|G| \nmid m!$, then $\exists \ H \neq \{e\} $ that $H \leq K$ and $H \unlhd G$.

Any leads?

Answer 1

Let set $U=\{xK, x\in G\}$,
for each element $g \in G$, define $\sigma_g: U\to U$ so that $\sigma_g(xK)=(gx)K$. and $\sigma_{g_1}\sigma_{g_2}=\sigma_{g_1g_2}$ It is easy to show that $\{\sigma_g, g\in G\}$ forms a group $S$ which is a subgroup of $S_m$.

The mapping $\varphi: G\to S$ with $\varphi(g)=\sigma_g$ is a group homomorphim so that $|G|=|\ker(\varphi)||S|$.

Since $|G| \nmid m!$, it means $|\ker(\varphi)| \gt 1$ so that $\ker(\varphi) \unlhd G$