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How to prove that if $G$ is a group with a subgroup $H$ of index $n$, then $G$ has a normal subgroup $K\subset H$ whose index in $G$ divides $n!$
I'm trying to prove the following: "If $G$ is an infinite group and $H < G$ satisfies $(G:H) = r$, then there exists a normal subgroup $K$ of $G$ contained in $H$ such that $(G:K) \leq r!$". The equality is trivial if we if we find $K \triangleleft G$ such that there exists an injective action of $K$ on the cosets of $H$. However, I can't think of any such action, considering $K < H$.