# If $G$ is simple and $|G|=\infty$ then $G$ doesn't have a non-trivial subgroup of finite index

Attempt

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?

• A counterexample: $G=\mathbb{Z}$ – Lee Mosher Mar 25 '18 at 16:35
• Aside from the trivial example ($G$ is a subgroup of itself after all), what if $G=\mathbb Z \times H$ where $H$ is finite? – lulu Mar 25 '18 at 16:35
• My apologies...$G$ is said to be simple,hence my question... – 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.
• 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? – Γιάννης Παπαβασιλείου Mar 25 '18 at 16:39