Subgroups of finitely generated free abelian groups and of free groups

For free abelian groups, let \[ H\le \ZZ^n. \] Since \(\ZZ\) is a PID, every submodule of a free \(\ZZ\)-module is free. Hence \[ H\cong\ZZ^r \] for some \(0\le r\le n\).

One can see this inductively without invoking the full structure theorem. Project \(H\) to the first coordinate. Its image is \(d\ZZ\) for some \(d\ge0\). The kernel lies in \(\ZZ^{n-1}\) and is free by induction. If \(d>0\), choose \(h\in H\) projecting to \(d\); then \[ H=\ker(\pi|_H)\oplus\ZZ h. \] Thus every subgroup of a finitely generated free abelian group is again finitely generated free abelian.

For free groups, the Nielsen–Schreier theorem says that every subgroup of a free group is free. A geometric proof realizes \[ F_r=\pi_1\left(\bigvee_{i=1}^r S^1\right). \] A subgroup \(H\le F_r\) corresponds to a connected covering graph \(X_H\) of the bouquet. Since the fundamental group of every connected graph is free, \[ H\cong\pi_1(X_H) \] is free.

Unlike the abelian case, \(H\) need not be finitely generated even when \(F_r\) is. For example, \[ [F_2,F_2]\le F_2 \] is free of countably infinite rank. If \(H\) has finite index \(d\) in \(F_r\), then the Schreier formula gives \[ \operatorname{rank}H=1+d(r-1), \] so finite-index subgroups are finitely generated.