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Prove that if the order of the group $G$ is bigger than $n!$ and $H < G$ is a subgroup with $|G:H| <n$, then $G$ cannot be a simple group.

We got the hint that we should represent $G$ on the right cosets of $H$ with right multiplication, but I can't really begin.

Any help appreciated.

Atvin
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  • See also https://math.stackexchange.com/questions/88719/how-to-prove-that-if-g-is-a-group-with-a-subgroup-h-of-index-n-then-g-h – lhf Feb 24 '18 at 20:40

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Hint: find a normal subgroup of $G$ by considering the kernel of a homomorphism from $G$ to somewhere. What is the "somewhere" that might be useful, bearing in mind that $n! < |G|$?

Second hint: you already know you need to make $G$ act on the set of cosets of $H$. This action induces a homomorphism into a symmetric group.

Patrick Stevens
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    I am guessing that somewhere is $S_n$? :) – Atvin Feb 24 '18 at 20:36
  • @Atvin yes, and groups, like people, are known by their actions :) – operatorerror Feb 24 '18 at 20:37
  • Correct. But which incarnation of $S_n$? (I added another bit of hint.) – Patrick Stevens Feb 24 '18 at 20:37
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    So, the action of $G$ on the right cosets of $H$ is a homomorphism from $G$ to $S_n$. Since $|G| > n!$, this can't be injective. Then we get, that the Kernel is a normal subgroup? – Atvin Feb 24 '18 at 20:41
  • The kernel is a normal subgroup; you need to check that it's not $G$ and it's not ${e}$. It's not ${e}$, because the homomorphism is not injective. You need to prove that it's not $G$, though. – Patrick Stevens Feb 24 '18 at 20:57