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Let $G$ be a finite group and suppose $H$ is a subgroup of $G$ having index $n$. Show there is a normal subgroup $K$ of $G$ with $K\subset H$ and such that the order of $K$ divides $n!$ . any suggestion? Thanks.

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    $\begingroup$ This is false. Perhaps you mean the index of $K$ divides $n!$? $\endgroup$ – Chris Eagle Jun 23 '13 at 15:42
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    $\begingroup$ @ChrisEagle Well, there's always the trivial subgroup. $\endgroup$ – Alexander Gruber Jun 23 '13 at 15:47
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This has surely been asked before (with the correction order $\to$ index noted by @ChrisEagle). Anyway, consider the permutation representation of $G$ on the cosets of $H$, $$ g \mapsto (H a \mapsto H a g). $$ (That is, an element $g \in G$ is mapped to the permutation $Ha \mapsto Hag$ of the cosets of $H$.)

This gives you a homomorphism $\phi$ of $G$ to the symmetric group $S_{n}$. If the image is $T$, then the first isomorphism theorem tells you that $$ G / \ker(\phi) \cong T \le S_{n}. $$ By Lagrange's theorem, the order of $T$ divides the order $n!$ of $S_{n}$, and then $\lvert T \rvert = \lvert G : \ker(\phi) \rvert$.

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Hint. Let $G$ act on the left cosets of $H$ in $G$ by left multiplication, and consider the kernel of this action.

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Consider the action of $G$ on the right cosets of $H$ (by right multiplication); it gives you a morphism $\phi:G \to S_n$ so that the cardinality of $G/Ker(\phi)$ divides $n!$. That kernel is a normal subspace of $G$ contained in $H$ .

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