Show that if \(G\) is an infinite simple group, then \(G\) cannot have a proper subgroup of finite index.
Solution
Suppose \(H<G\) has finite index \(n\). The action of \(G\) on the left cosets \(G/H\) gives a homomorphism \[ \rho:G\to S_n. \] Its kernel is the core \[ \ker\rho=\bigcap_{g\in G}gHg^{-1}, \] which is a normal subgroup of \(G\) contained in \(H\).
If \(G\) is simple, then \(\ker\rho\) is either \(1\) or \(G\). Since \(H\) is proper, \(\ker\rho\ne G\), so \(\ker\rho=1\). Hence \(\rho\) is injective and embeds \(G\) into the finite group \(S_n\), forcing \(G\) to be finite. This contradicts the hypothesis that \(G\) is infinite.
Therefore an infinite simple group has no proper subgroup of finite index.