Let $H\leq G$ with $[G:H]=n<\infty$.We consider the action of $G$ on $G/H$ and we get a homomorphism $\phi:G\to S_n$.But $H\not= G$ so $ker\phi \not=G\Rightarrow ker\phi \lhd G$ and $[G:ker\phi]=m|n!$

Any ideas on how to proceed from here?

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    $\begingroup$ A counterexample: $G=\mathbb{Z}$ $\endgroup$ – Lee Mosher Mar 25 '18 at 16:35
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    $\begingroup$ Aside from the trivial example ($G$ is a subgroup of itself after all), what if $G=\mathbb Z \times H$ where $H$ is finite? $\endgroup$ – lulu Mar 25 '18 at 16:35
  • $\begingroup$ My apologies...$G$ is said to be simple,hence my question... $\endgroup$ – Γιάννης Παπαβασιλείου Mar 25 '18 at 16:41

The statement you're trying to prove is actually false. For example, you can consider the infinite group $G=\mathbb{Z}$ and its finite index subgroup $H=2\mathbb{Z}$.

Edit: Indeed, if you require that $G$ is an infinite simple group, then your argument is correct (i.e., it does provide a contradiction), by constructing something that would have to be a non-trivial normal subgroup.

  • $\begingroup$ Yes ,thanks.I should had add that $G$ is simple(I didn't see it).But if $G$ is simple then we can embed $G$ into $S_n$,which is a contradiction,right? $\endgroup$ – Γιάννης Παπαβασιλείου Mar 25 '18 at 16:39
  • $\begingroup$ @giannispapav: Yes, that's a valid argument. I've updated my answer. $\endgroup$ – Zev Chonoles Mar 25 '18 at 16:42

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