Describe the subgroups of \(\ZZ^2\).
Solution
Every subgroup \(H\le\ZZ^2\) is a free abelian group of rank \(0\), \(1\), or \(2\).
1. Rank \(0\) gives only the zero subgroup.
2. Every rank-\(1\) subgroup is infinite cyclic.
Since \(\ZZ\) is a PID, every submodule of the free module \(\ZZ^2\) is free. A rank-\(1\) subgroup therefore has the form \[ H=\ZZ(a,b) \] for some nonzero vector \((a,b)\in\ZZ^2\).
3. Every rank-\(2\) subgroup has finite index and admits a Smith normal form.
Choose a basis \(v_1,v_2\) of \(H\) and form the \(2\times2\) integer matrix whose columns are \(v_1,v_2\). Smith normal form gives unimodular matrices \(U,V\in GL_2(\ZZ)\) such that \[ UAV=\begin{pmatrix}d_1&0\\0&d_2\end{pmatrix}, \qquad d_1,d_2>0, \quad d_1\mid d_2. \] Thus, after an automorphism of \(\ZZ^2\), \[ H=d_1\ZZ\oplus d_2\ZZ. \] In particular, \[ [\ZZ^2:H]=d_1d_2<\infty. \]
Equivalently, every subgroup is either \(0\), cyclic, or a rank-\(2\) lattice of finite index in \(\ZZ^2\).