Let $G$ a simple group such that $|G|>n$ for every positive integer $n$. Show that $G$ has no proper subgroups of finite index.
An infinite simple group has no proper subgroup of finite index.
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group-theory
simple-groups
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1You are supposed to ask questions, not issue commands. – 2017-01-04
1 Answers
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If a group has a proper subgroup of index $k$, then it has a normal subgroup of index at most $k!$.
Let $H \le G$ have index $k > 1$. Consider the action of $G$ on the right cosets of $H$. This yields a homomorphism $\varphi : G \to S_{k}$. By the first isomorphism theorem, $\ker(\varphi)$ has index at most $\lvert S_{k} \rvert = k!$ in $G$.