Tor is always a torsion group

First show that \(\operatorname{Tor}(A,B)\) is torsion. Express \(A\) as the directed union of its finitely generated subgroups \(A_\alpha\). By Exercise 4, \[ \operatorname{Tor}(A,B) \cong\varinjlim_\alpha\operatorname{Tor}(A_\alpha,B). \] For each finitely generated \(A_\alpha\), write \[ A_\alpha\cong\mathbb Z^r\oplus\bigoplus_{j=1}^s\mathbb Z_{n_j}. \] Then \[ \operatorname{Tor}(A_\alpha,B) \cong\bigoplus_{j=1}^s B[n_j], \] which is a torsion group. Every element of the direct limit comes from one such torsion group, so it has finite order. Hence \(\operatorname{Tor}(A,B)\) is torsion.

Now suppose \(A\) and \(B\) each contain an element of order exactly \(n\). Let \[ C\cong\mathbb Z_n \] be the cyclic subgroup of \(A\) generated by such an element. From \[ 0\to C\to A\to A/C\to0 \] the long exact Tor sequence begins \[ \operatorname{Tor}_2(A/C,B) \longrightarrow \operatorname{Tor}_1(C,B) \longrightarrow \operatorname{Tor}_1(A,B). \] Over \(\mathbb Z\), every abelian group has projective dimension at most \(1\), so \(\operatorname{Tor}_2=0\). Hence \[ \operatorname{Tor}(C,B)\hookrightarrow\operatorname{Tor}(A,B). \] But \[ \operatorname{Tor}(\mathbb Z_n,B)\cong B[n]. \] Since \(B\) contains an element of order exactly \(n\), \(B[n]\) contains an element of order \(n\). Its image in \(\operatorname{Tor}(A,B)\) still has order \(n\), proving the claim.