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The canonical image of $\Gamma$ is dense in its profinite completion $\hat{\Gamma }$. Therefore every $x\in \hat{\Gamma }$ is a limit of a net $({\delta }_{\lambda })$ in $\Gamma$. Continuity of conjugation givesso every element of ${\mathrm{Cl}}_{\hat{\Gamma }}(\gamma )$ lies in the closure of ${\mathrm{Cl}}_{\Gamma }(\gamma )$. The reverse inclusion follows because the larger conjugacy class contains the smaller one and is closed by part b(iii). Hence

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 $A\cong \mathbb{Z}$.

For every open normal subgroup $U◃G$, the compositehas finite image and therefore factors uniquely through the profinite completion $\hat{\Gamma }$. These factor maps are compatible as $U$ varies. The universal property of an inverse limit consequently produces a continuous homomorphismwith $\hat{f}\iota =f$. It is unique because $\iota (\Gamma )$ is dense in $\hat{\Gamma }$ and two continuous maps into the Hausdorff group $G$ that agree on a dense subset agree everywhere. This is the universal property of profinite completion.