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

Question

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?

Answer 1

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.