Let \(G\) be a finitely generated abelian group in which no element (except 0) has finite order. Show that \(G\) is a free abelian group.
Solution
1. There are integers \(r\ge0\) and \(d_1,\ldots,d_t>1\) such that\[ G\cong \ZZ^r\oplus \ZZ/d_1\ZZ\oplus\cdots\oplus\ZZ/d_t\ZZ. \]
This is the structure theorem for finitely generated abelian groups, applied to the finitely generated abelian group \(G\).
2. One must have \(t=0\).
If \(t>0\), then the element corresponding to \[ (0,\ldots,0,1\bmod d_1,0,\ldots,0) \] is nonzero and has finite order \(d_1\). This contradicts the hypothesis that no nonzero element of \(G\) has finite order.
3. Hence \(G\cong\ZZ^r\), so \(G\) is free abelian.
By 1 and 2 the finite cyclic summands are absent, leaving precisely the free abelian group \(\ZZ^r\).