Prove the following.
Theorem. If \(A\) is a free abelian group of rank \(n\), then any subgroup \(B\) of \(A\) is a free abelian group of rank at most \(n\).
Proof. We can assume \(A = \mathbb{Z}^n\), the \(n\)-fold cartesian product of \(\mathbb{Z}\) with itself. Let \(\pi_i: \mathbb{Z}^n \to \mathbb{Z}\) be projection on the \(i\)th coordinate. Given \(m \leq n\), let \(B_m\) consist of all elements \(\mathbf{x}\) of \(B\) such that \(\pi_i(\mathbf{x}) = 0\) for \(i > m\). Then \(B_m\) is a subgroup of \(B\).
Consider the subgroup \(\pi_m(B_m)\) of \(\mathbb{Z}\). If this subgroup is nontrivial, choose \(\mathbf{x}_m \in B_m\) so that \(\pi_m(\mathbf{x}_m)\) is a generator of this subgroup. Otherwise, set \(\mathbf{x}_m = \mathbf{0}\).
Show \(\ts{\mathbf{x}_1, \ldots, \mathbf{x}_m}\) generates \(B_m\), for each \(m\).
Show the nonzero elements of \(\ts{\mathbf{x}_1, \ldots, \mathbf{x}_m}\) form a basis for \(B_m\), for each \(m\).
Show that \(B_n = B\) is free abelian with rank at most \(n\).
Solution
We prove (a) and (b) simultaneously by induction on \(m\).
For \(m=0\), \(B_0=0\), so there is nothing to prove. Assume the statements hold for \(m-1\), and take \(b\in B_m\). By construction, \(\pi_m(B_m)\) is either zero or the cyclic subgroup generated by \(\pi_m(x_m)\).
If \(\pi_m(B_m)=0\), then \(B_m=B_{m-1}\) and \(x_m=0\), so the induction hypothesis gives both assertions.
Otherwise there is an integer \(q\) such that \[ \pi_m(b)=q\pi_m(x_m). \] Then \[ b-qx_m\in B_m \] has zero \(m\)-th coordinate, hence belongs to \(B_{m-1}\). By induction it is generated by \(x_1,\dots,x_{m-1}\). Therefore \(b\) is generated by \(x_1,\dots,x_m\). This proves (a).
For linear independence, suppose \[ a_1x_1+\cdots+a_mx_m=0. \] If \(x_m\ne0\), then \(\pi_m(x_m)\ne0\), while \(\pi_m(x_i)=0\) for \(i<m\). Applying \(\pi_m\) gives \[ a_m\pi_m(x_m)=0, \] so \(a_m=0\). The induction hypothesis then gives \(a_i=0\) for every nonzero \(x_i\) with \(i<m\). If \(x_m=0\), the induction hypothesis applies immediately. Hence the nonzero members of \(\{x_1,\dots,x_m\}\) are linearly independent and, by (a), generate \(B_m\). They form a basis, proving (b).
Taking \(m=n\), we have \(B_n=B\). Its basis is a subset of the \(n\) elements \(x_1,\dots,x_n\), so \(B\) is free abelian of rank at most \(n\). This proves (c).