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This is a 'Harder' problem 40 from Abstract Algebra(1996) by Herstein. I'm just not able to figure out how to do this. even though I found a very similar post. Following is a verbatim statement of the question.

If $G$ is a finite group, $H$ a subgroup of $G$ such that $n \nmid i_G(H)!$, where $n=|G|$, prove that there is a normal subgroup $N \neq (e)$ of $G$ contained in $H$.

P.S. I've been stuck on this for about a week, and now I'm throwing in the towel, so I'd really appreciate a solution, but I humbly implore you to give me hints instead so that I can kill this problem (sort of) on my own, although frankly, I've given up hope.

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  • $\begingroup$ What is $i_G(H)$? My guess from the rest of the question: the index of $H$ in $G$. In which case I have a solution. If it's something else, I'm not so sure. $\endgroup$ – David A. Craven Aug 17 at 14:28
  • $\begingroup$ I'm missing something maybe. Isn't a proof of what you need in the similar post you mention? $\endgroup$ – 1123581321 Aug 17 at 14:32
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Suppose that $H$ has index $n$ in $G$. The action on the (right, say) cosets of $H$ induces a homomorphism $\phi:G\to S_n$, and the kernel of this map, the core of $H$ in $G$, is the largest normal subgroup of $G$ contained in $H$. Thus the core is non-trivial if and only if the subgroup $N$ you require exists, so let $N$ denote this core. Since $G/N$ is isomorphic to a subgroup of $S_n$, $|G/N|\mid n!$. But $|G|\nmid n!$, and therefore $|N|>1$.

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  • $\begingroup$ The presence of a factorial suggests that this is the way to do it... $\endgroup$ – David A. Craven Aug 17 at 17:46
  • $\begingroup$ I've now looked in Herstein, and Problem 38 is what I am using. It's just that his notation is all over the place. So $i_G(H)$ is normally written $|G:H|$, and $A(S)$ is usually written $\mathrm{Sym}(S)$. I just noted that $\mathrm{Sym}(S)$ is naturally isomorphic to $S_n$. And you don't need isomorphism theorems. If $N=1$ then $G$ is a subgroup of $A(S)$, and that's all you are trying to prove. $\endgroup$ – David A. Craven Aug 17 at 17:53
  • $\begingroup$ Alright, I understand most of the proof now, and it'squite clever, I still don't get how you arrived at it. Just one thing though, how do we know $G/N$ is isomorphic to a subgroup of $S_n$ ? by Cayley's theorem ? $\endgroup$ – codesPliff Aug 18 at 11:39
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    $\begingroup$ @codesPliff It's best to assume that $N=1$ and derive a contradiction, if you don't have isomorphism theorems. Then the action of $G$ on the right cosets of $H$ is faithful, i.e., no element of $G$ acts trivially on the cosets of $H$. Thus it forms a subgroup of $A(S)$. This has order $n!$, so Lagrange gives you $|G|\mid n!$. This is a standard argument in group theory, then if $G$ has a subgroup of index $n$, then it has a normal subgroup of index at most $n!$. $\endgroup$ – David A. Craven Aug 18 at 12:31

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