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Let $G$ be a finite group and $H ≤ G$ with index $k > 1$. If $|G|$ does not divide $k!$, show that there is a normal subgroup $N$ of $G$, different from$\{e\}$ and $G$.

P.D. I have been trying to solve this problem, and I have found similar problems on the site. I think that maybe I could take some ideas from those solutions but I still do not understand what role does the fact that $|G|$ does not divide $k!$ plays on this problem as a comparison to the question that I linked above where it is not specified.

If anyone can give me a hint or advice I would be very grateful.

Shaun
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AdrinMI49
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  • If $|G|$ does not divide $k!$, then a morphism from $G$ into a group of order $k!$ cannot be one-to-one. – Arturo Magidin Jan 21 '21 at 05:25
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    I am completely with you on this, but I still do not know how to use this information to my advantage. – AdrinMI49 Jan 21 '21 at 05:35
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    This problem is usually stated in a slightly different way: if a finite simple group $G$ has a subgroup of index $k>1$, then $|G|$ divides $k!$. That statement is logically equivalent to what you are trying to prove - do you see that? You might find it less confusing to try and solve the problem ion that alternative formulation. – Derek Holt Jan 21 '21 at 07:54
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    @adrinmajestuosso: if you can get a homomorphism into a group of order $k!$, that will give you a nontrivial kernel. – Arturo Magidin Jan 21 '21 at 16:28
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    Once you get that a homomorphism from $G$ to $S_k$ exists (do you?), then its kernel can't be trivial (which assumption would this contradict?). In order to rule out also the case of the kernel to be the whole $G$, you need to know how such a kernel is made in term of $H$, so as to use your second assumption ($[G:H]>1$). For further reference on this latter point, see e.g. here: https://math.stackexchange.com/a/3631336/870827 –  Jan 22 '21 at 15:25

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Hint: Let $G$ act on the set of cosets $G/H$. This gives a morphism into $S_k$. (The kernel of that morphism will be normal.)