Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 2 / b

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 $A\cong \mathbb{Z}$.

Suppose $\hat{\Gamma }\cong {\mathbb{Z}}_p$. This completion is an abelian group, so every finite quotient of $\Gamma$ is abelian. Consequently the quotient map to the abelianization $A={\Gamma }^{\mathrm{ab}}$ inducesThe group $A$ is a finitely generated abelian group. If its free rank is zero, $\hat{A}=A$ is finite. If its free rank is positive, $A$ and hence $\hat{A}$ have a nontrivial quotient $\mathbb{Z}/q\mathbb{Z}$ for every sufficiently chosen prime $q$. But ${\mathbb{Z}}_p$ has no nontrivial finite quotient of order coprime to $p$. Both cases are impossible, so $\hat{\Gamma }\ncong {\mathbb{Z}}_p$.