# Prove that there is a subgroup $K$ : $k! | [H:K]$ [duplicate]

Consider a group $G$, and $H \in G$ - subgroup with $[G:H] = k$, then prove that there is exists normal subgroup $K$ in $H$ such that $k! | [G:K]$?

Actually I have no ideas. Any hints?

• @DietrichBurde edited – openspace Sep 20 '17 at 20:37
• Is it not the other way around, i.e., $[G:K]\mid k!$? – Dietrich Burde Sep 20 '17 at 20:40