Present a <finitely generated abelian group> $G$ as the <cokernel> of an integer matrix $A:\mathbb Z^m\to\mathbb Z^n$. Changing $A$ by the invertible row and column operations used in its <Smith normal form> does not change the isomorphism type of the cokernel. If
$$
UAV=\operatorname{diag}(d_1,\ldots,d_s,0,\ldots,0),
$$
then
$$
\boxed{G\cong\mathbb Z^r\oplus\mathbb Z/d_1\mathbb Z\oplus\cdots\oplus\mathbb Z/d_s\mathbb Z},
\qquad d_1\mid\cdots\mid d_s,
$$
where factors with $d_i=1$ may be omitted. This is the <Fundamental theorem of finitely generated abelian groups>; the <rank of an abelian group> $r$ and invariant factors $d_i>1$ are unique.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 11G / b / Solution

{wiki}

A finitely generated abelian group has a finite <generating set>. Equivalently, it is the <cokernel> of a homomorphism between finitely generated free abelian groups.

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 Incoming links: Finitely generated abelian group