= Solution
Suppose $\widehat\Gamma\cong\mathbb Z_p$. This completion is an <abelian group>, so every finite quotient of $\Gamma$ is abelian. Consequently the quotient map to the <abelianization> $A=\Gamma^{\mathrm{ab}}$ induces
$$
\widehat\Gamma\cong\widehat A.
$$
The group $A$ is a finitely generated <abelian group>. If its free rank is zero, $\widehat A=A$ is finite. If its free rank is positive, $A$ and hence $\widehat A$ 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 $\widehat\Gamma\not\cong\mathbb Z_p$.
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