# If $H\leq G$ is of finite index, then, we have a normal subgroup $N\leq G$ of finite index.

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"..

Thank you.

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The kernel is the intersection of all conjugates of $H$, and you can check directly hat this is a normal subgroup of finite index, contained in $H$. –  Martin Brandenburg Sep 15 '13 at 8:33
@MartinBrandenburg : Yes, That was another guess of me but was not so confident to put that as i have not checked it.. Now i wil check it out.. Thank you :) –  Praphulla Koushik Sep 15 '13 at 8:34
"But, this question is asked before introducing group actions." This might be one of those exercises that want you to invent a concept you'll learn about later, for motivation. Otherwise, what kinds of things have been introduced thus far? –  anon Sep 15 '13 at 8:37
It does not seem to be as an exercise for motivation (at least for me)... not many things are done before this.. just normal subgroups, symmetric groups, isomorphisms ... that's it... But Mr.Martin's answer suits well for my doubt :) –  Praphulla Koushik Sep 15 '13 at 8:41
If you add in the assumption that your group is finitely generated, then there are finitely many subgroups of a given finite index. You can then intersect them all to get a characteristic subgroup of finite index. –  user1729 Sep 15 '13 at 10:32
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Your solution is correct provided you add the easy verification that the stabiliser of the coset $eH$ in the action of $G$ on $G/H$ is $H$, so that $\ker\eta\leq H$ is a proper subgroup. If you don't want to mention group actions, you can define the morphism $\eta:G\to \operatorname{Sym}(G/H)$ explicitly. Identifying the stabilisers of the other cosets as conjugates of $H$, and $\ker\eta$ as their intersection, you get the argument in the comment by Martin Brandenberg.