The order of an element \(a\) of an abelian group \(G\) is the smallest positive integer \(m\) such that \(ma = 0\), if such exists; otherwise, the order of \(a\) is said to be infinite. The order of \(a\) thus equals the order of the subgroup generated by \(a\).
Show the elements of finite order in \(G\) form a subgroup of \(G\), called its torsion subgroup.
Show that if \(G\) is free abelian, it has no elements of finite order.
Show the additive group of rationals has no elements of finite order, but is not free abelian. [Hint: If \(\ts{a_\alpha}\) is a basis, express \(\frac{1}{2}a_\alpha\) in terms of this basis.]
Solution
Let \[ T=\{g\in G:g\text{ has finite order}\}. \] Certainly \(0\in T\). If \(a,b\in T\), choose positive integers \(m,n\) with \(ma=0\) and \(nb=0\). Then \[ mn(a-b)=n(ma)-m(nb)=0, \] so \(a-b\in T\). Hence \(T\) is a subgroup.
Let \(G\) be free abelian with basis \(\{e_\alpha\}\), and suppose \(g\ne0\). Write \[ g=n_1e_{\alpha_1}+\cdots+n_re_{\alpha_r} \] with some \(n_i\ne0\). If \(mg=0\) for a positive integer \(m\), uniqueness of coordinates gives \(mn_i=0\) in \(\mathbb Z\), impossible. Thus every nonzero element has infinite order.
The additive group \(\mathbb Q\) is torsion-free: if \(mq=0\) with \(m>0\), then \(q=0\). Suppose nevertheless that \(\mathbb Q\) were free abelian with a nonempty basis \(\{a_\alpha\}\). Choose a basis element \(a=a_\alpha\). Since \(a/2\in\mathbb Q\), write \[ \frac a2=n_1a_{\alpha_1}+\cdots+n_ra_{\alpha_r} \] with integers \(n_i\). Multiplying by \(2\), \[ a=2n_1a_{\alpha_1}+\cdots+2n_ra_{\alpha_r}. \] Uniqueness of basis coordinates says the coefficient of \(a\) on the right must be \(1\), but every coefficient on the right is even. Contradiction. Therefore \(\mathbb Q\) is torsion-free but not free abelian.