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I am very bad with problems involving divisibility of orders and such. If anyone can give me some help with the following problem, it will be very much appreciated:

Prove that every subroup $H$ of $G$ of index $n$ must contain a normal subgroup $N \unlhd G$ such that $[G : N]$ divides $n!$ (this is a factorial and not an exclamation mark, just in case anyone gets confused as I did).

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2 Answers 2

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With $H$ come $n$ cosets of the form $gH$. The group $G$ operates on the set of these cosets by left multiplication. The kernel of this operation is a normal subgroup $N$ of $G$. We have $N<H$ because $gH=H$ implies $g\in H$. We have $[G:N]|n!$ because $G/N<S_n$.

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Amazingly simple! Thank you very much! –  user44069 Nov 11 '12 at 23:14

Consider the action of $G$ on the left cosets of $H$ by left multiplication. This gives rise to a homomorphism $$\varphi: G \to S_n$$ and if an element $x$ is in the kernel, $xH = H$, so $x \in H$. Hence, $\ker\varphi \subseteq H$. By the first isomorphism theorem, $G/\ker\phi \cong \varphi(G)$, so $[G: \ker\phi]$ divides $n!$.

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