Show that the additive group of rationals \((\mathbb{Q}, +)\) is not finitely generated.
Show that \((\mathbb{Q}, +)\) is not a free abelian group.
Conclude that the statement “every torsion-free abelian group is free” is false if the hypothesis “finitely generated” is omitted.
Solution
1. (a) \((\QQ,+)\) is not finitely generated.
1.1. For \(a_1,\ldots,a_k\in\ZZ\) and \(b_1,\ldots,b_k\in\ZZ_{>0}\), put \(B=\prod_{i=1}^k b_i\). Then \(\left\langle \frac{a_1}{b_1},\ldots,\frac{a_k}{b_k}\right\rangle\subseteq\frac1B\ZZ\).
Every element of the subgroup is an integer combination \[x = \sum_{i=1}^k c_i \frac{a_i}{b_i} = \frac{1}{B} \sum_{i=1}^k c_i a_i \left(\prod_{j \ne i} b_j\right) \in \frac{1}{B}\ZZ.\]
1.2. For every \(B\in\ZZ_{>0}\), \(\frac1B\ZZ\ne\QQ\).
Let \(p\) be a prime with \(p>B\). If \(\frac1p=\frac mB\) with \(m\in\ZZ\), then \(B=mp\), so \(p\mid B\), which contradicts \(0<B<p\). Hence \(\frac1p\notin\frac1B\ZZ\).
1.3. Q.E.D.
Every finite subset of \(\QQ\) can be written as \(\left\{\frac{a_1}{b_1},\ldots,\frac{a_k}{b_k}\right\}\) with \(b_i>0\). By steps 1 and 2 the subgroup it generates is a proper subgroup of \(\QQ\).
2. (b) \((\QQ,+)\) is not a free abelian group.
2.1. \((\QQ,+)\) has no basis with exactly one element.
A free abelian group with a one-element basis is cyclic, hence finitely generated, which contradicts step 1.
2.2. No two distinct elements of \(\QQ\) are \(\ZZ\)-linearly independent.
Let \(x=\frac ab\) and \(y=\frac cd\) be distinct elements of a candidate basis, with \(a,b,c,d\in\ZZ\setminus\{0\}\); basis elements are nonzero. Then \[(b c) x - (a d) y = (b c) \left(\frac{a}{b}\right) - (a d) \left(\frac{c}{d}\right) = a c - a c = 0,\] and the coefficients \(bc\) and \(-ad\) are nonzero.
2.3. Q.E.D.
A basis of \((\QQ,+)\) is nonempty, since the empty set generates \(0\ne\QQ\). By step 1 it cannot have exactly one element, and by step 2 it cannot have two or more elements.
3. (c) \((\QQ,+)\) is a torsion-free abelian group that is not free.
For \(n \in \ZZ \setminus \{0\}\) and \(\frac{a}{b} \in \QQ\), \(n \cdot \frac{a}{b} = \frac{n a}{b} = 0\) implies \(n a = 0\), hence \(a = 0\) and \(\frac{a}{b} = 0\). So \((\QQ,+)\) is torsion-free. By step 2 it is not free, and by step 1 it is not finitely generated. Hence the statement “every torsion-free abelian group is free” is false.