Show that a direct limit \(\varinjlim G_\alpha\) of torsionfree abelian groups \(G_\alpha\) is torsionfree. More generally, show that any finitely generated subgroup of \(\varinjlim G_\alpha\) is realized as a subgroup of some \(G_\alpha\).
Solution
Let \(G=\varinjlim G_\alpha\).
First suppose \(x\in G\) and \(mx=0\) for some nonzero integer \(m\). Represent \(x\) by \(g_\alpha\in G_\alpha\). The equality \(mx=0\) means that for some \(\beta\ge\alpha\) the image \(g_\beta\) satisfies \[ mg_\beta=0. \] Since \(G_\beta\) is torsionfree, \(g_\beta=0\), hence \(x=0\) in the direct limit. Thus \(G\) is torsionfree.
For the stronger statement, let \(H\le G\) be generated by \(x_1,\ldots,x_r\). Choose one stage \(G_\alpha\) containing representatives \(g_1,\ldots,g_r\). Let \[ \phi:\mathbb Z^r\to H, \qquad e_i\mapsto x_i. \] Its kernel \(K\) is a subgroup of the finitely generated free abelian group \(\mathbb Z^r\), hence is finitely generated. Choose generators \(k_1,\ldots,k_s\) of \(K\).
Each relation \(k_j\) becomes zero at some later stage. By directedness there is a single \(\beta\ge\alpha\) at which all these relations vanish. Let \(H_\beta\le G_\beta\) be the subgroup generated by the images of the \(g_i\). Every relation among these images is certainly a relation in the direct limit, so its relation lattice is contained in \(K\); conversely, by construction all generators of \(K\) already vanish in \(G_\beta\). Hence the two relation lattices coincide. Therefore \[ H_\beta\cong\mathbb Z^r/K\cong H. \] Thus every finitely generated subgroup of the direct limit is realized as a subgroup of some stage \(G_\beta\).