 Solution

By the Fundamental theorem of finitely generated abelian groups, writewith $F$ finite. Taking profinite completions givesIf $A\cong \mathbb{Z}$, this is plainly $\hat{\mathbb{Z}}$.

Conversely, suppose $\hat{A}\cong \hat{\mathbb{Z}}$. Reduction modulo a prime $p$ giveswhereas $\hat{\mathbb{Z}}/p\hat{\mathbb{Z}}\cong \mathbb{Z}/p\mathbb{Z}$. Choosing $p\nmid |F|$ first shows $r=1$. If $F\ne 0$, choosing a prime divisor $p$ of $|F|$ makes $F/pF\ne 0$, a contradiction. Thus $F=0$ and $A\cong \mathbb{Z}$.