Uniqueness of finite-index subgroups forces infinite cyclic abelianization

If a finitely generated group $S$ has exactly one subgroup of each finite index, every such subgroup is normal. The quotients of prime order give surjections of its finitely generated abelianization onto $C_p$ for all primes, forcing positive free rank. A free rank of at least two gives two distinct index-$p$ subgroups. Positive finite torsion together with the free rank-one factor also gives two index-$p$ subgroups for a suitable prime. Hence $S_{\mathrm{ab}}\cong \mathbb{Z}$.

// Target: algebra.bigb

The epimorphism $S\to S_{\mathrm{ab}}\cong \mathbb{Z}$ has preimage of $n\mathbb{Z}$ as the unique index-$n$ subgroup. Thus every finite-index subgroup contains the commutator subgroup. Residual finiteness makes their intersection trivial, forcing the commutator subgroup to vanish and $S\cong \mathbb{Z}$.

// Target: geometry-and-topology.bigb