Let $H<G$ such that $H$ has finite index in the infinite group $G$. Then $G$ contains a normal subgroup of finite index contained in $H$.
Can I create a subgroup of index $2$ in $G$ using elements of $H$?
Every subgroup of finite index contained in an infinite group $G$ contains a normal subgroup of $G$.
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There are infinite groups with finite index subgroups but no index $2$ subgroup. For instance, $\mathbb{Z}_3$. – Captain Lama Apr 24 '16 at 18:01
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Hint: say $[G:H] = n$, i.e. the set of left cosets $G/H$ has $n$ elements. Define an action of $G$ on $G/H$ by left translation. This induces a homomorphism from $G$ into the symmetric group on $G/H$ (which is isomorphic to $S_{n}$). What can you say about the kernel of this homomorphism?
Alex Wertheim
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I haven't learned about group actions or kernels yet (next two chapters) and don't want to jump the gun. Is there a more elementary way of proving this? The question comes from M.A. Armstrong's book Groups & Symmetry, question 15.11. – Jacopo Stifani Apr 25 '16 at 08:18