Question is to prove that :
Let $G$ be a group with a proper subgroup $H$ of finite index. Show that $G$ has a proper normal subgroup of finite index.
what i have tried is :
Assuming $|G/H|=n< \infty$ let $G$ act on set of left cosets of $H$ in $G$ and this would give me
$\eta :G\rightarrow S_n$ a homomorphism..
As $Ker(\eta)\unlhd G$ we do have a normal subgroup of $G$.
But, this question is asked before introducing group actions.
So, another solution without group actions is what i am trying to get.
I would like somebody to check my solution and let me know if there is any possible solution with out using group actions...
we assume $G$ to be infinite group.
This is because in any group $Z(G)\unlhd G$ and the index of $Z(G)$ would be finite in "finite groups"..