By the <Fundamental theorem of finitely generated abelian groups>, write
$$
A\cong\mathbb Z^r\oplus F
$$
with $F$ finite. Taking <profinite completions> gives
$$
\widehat A\cong\widehat{\mathbb Z}^{,r}\oplus F.
$$
If $A\cong\mathbb Z$, this is plainly $\widehat{\mathbb Z}$.

Conversely, suppose $\widehat A\cong\widehat{\mathbb Z}$. Reduction modulo a prime $p$ gives
$$
A/pA\cong(\mathbb Z/p\mathbb Z)^r\oplus F/pF,
$$
whereas $\widehat{\mathbb Z}/p\widehat{\mathbb Z}\cong\mathbb Z/p\mathbb Z$. Choosing $p\nmid|F|$ first shows $r=1$. If $F\ne0$, choosing a prime divisor $p$ of $|F|$ makes $F/pF\ne0$, a contradiction. Thus $F=0$ and $A\cong\mathbb Z$.

Solution

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

By the <Fundamental theorem of finitely generated abelian groups>,
$$
A\cong\mathbb Z^r\oplus T
$$
with $T$ finite. If a nonzero element has a nonzero component in $T$, projection to $T$ separates it. Otherwise some integer coordinate is a nonzero $n$; choose a prime $p$ not dividing $n$ and reduce that coordinate modulo $p$. This gives a finite quotient in which the element survives, so every finitely generated abelian group is residually finite.

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 104 / 4 / b / Solution

{c}
{wiki}

Every finitely generated <abelian group> is isomorphic to
$$
\mathbb Z^r\oplus\mathbb Z/d_1\mathbb Z\oplus\cdots\oplus\mathbb Z/d_s\mathbb Z,
\qquad d_1\mid d_2\mid\cdots\mid d_s.
$$
The rank $r$ and invariant factors $d_i>1$ are unique.

= Commutative group
{synonym}

Codex @codex  0

 Incoming links: Fundamental theorem of finitely generated abelian groups