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Let $G$ be a group and $H$ a subgroup with finite index $n$. Prove that $G$ has a normal subgroup $N$ such that $N\subseteq H$and $|G: N| \le n!$.

csasba
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  • This was on an old qualification exam and also given to me without any context other than being a group theory problem. I will try to edit it when I get a chance. – csasba Apr 02 '18 at 05:23
  • -2? Stack Exchange is ruthless. – csasba Jul 01 '20 at 05:50

2 Answers2

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$G$ acts on $G/H$ by left multiplication and this action is sujective, you deduce a morphism $f:G\rightarrow S_n$ which is not trivial, take $N$ the kernel of $f$.

Tsemo Aristide
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You don't show any self work so I can't know what you've tried and what you know, so I'll give the answer and you find out whether it is true and its properties:

$$N=\bigcap_{g\in G} g^{-1}Hg$$

and it gets even better, since in fact $\;N\lhd G\;$ .

DonAntonio
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  • I was clueless and didn't know what direction to take. I had been staring at the problem for too long. – csasba Apr 01 '18 at 00:20
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    Where did you find the problem, and what things had been talked about in the book / class recently? (For instance, this is an early proposition in Isaacs' Finite Group Theory; I do think that if you haven't seen the proof, it can be tricky to come up with). That's usually a good way to include context, since "naked" problem statements are generally not well-received. – pjs36 Apr 01 '18 at 01:07
  • It's on a past exam. I am studying for my qualification exam. I stated the problem as it appears in the exam. I left out, "N has finite index," but I thought that was obvious from the inequality. – csasba Apr 01 '18 at 01:57