# Herstein exercise: A subgroup of a finite group G such that $|G| \nmid i_G(H)!$ must contain a non-trivial normal subgroup.

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.

• 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. – David A. Craven Aug 17 at 14:28
• I'm missing something maybe. Isn't a proof of what you need in the similar post you mention? – 1123581321 Aug 17 at 14:32

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 $$|N|>1$$.
• 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. – David A. Craven Aug 17 at 17:53
• 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 ? – codesPliff Aug 18 at 11:39
• @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 $n!$. – David A. Craven Aug 18 at 12:31