Integer abelian structure and rank by finite reduction

For $K\le {\mathbb{Z}}^n$, induct on $n$. When $n=0$, $K=0$ has the empty basis. For $n>0$, project to the last coordinate: its image is $d\mathbb{Z}$. If $d=0$, apply the induction hypothesis in ${\mathbb{Z}}^{n-1}$. If $d>0$, take $v\in K$ projecting to $d$. For each $x\in K$, write its last coordinate uniquely as $ad$; then $x-av$ lies in the projection kernel $K_0$. Thus $K=\mathbb{Z}v\oplus K_0$: the intersection is zero because $ad=0$ implies $a=0$. A basis of $K_0$ together with $v$ spans and is independent, so has at most $n$ members. Only one lift at each of at most $n$ stages is selected.

Choose a finite ordered generating list of $M$, giving $\pi :{\mathbb{Z}}^n\twoheadrightarrow M$. For any subgroup $L\le M$, its inverse image under $\pi$ is free with a finite basis by step 1.1. Images of that basis generate $L$, since every $l\in L$ has a preimage. In particular $K=\ker \pi$ has a finite basis, whose columns form an integer matrix $A$.

Row swaps, column swaps, sign changes, and adding an integer multiple of another row or column are invertible: undo the swap or sign, or subtract the same multiple. Column operations preserve the image subgroup, while a row operation carries it by an automorphism of ${\mathbb{Z}}^n$ and therefore induces an isomorphism of quotients. If the current rectangle is zero, stop. Otherwise move a nonzero entry to its top left and change its sign to obtain a positive pivot $d$.

If an entry in the pivot row is $a=qd+r$ with $0<r<d$, subtract $q$ times the pivot column and swap that column into the pivot position. The new positive pivot is $r<d$. The same procedure with rows treats the pivot column. If all row and column entries are divisible by $d$, clear them. If the remaining rectangle has an entry $b$ not divisible by $d$, add its row to the pivot row. The first pivot stays $d$, while the pivot row now contains $b$; column division again decreases the pivot. Each failed divisibility therefore strictly decreases a positive integer. Such descents terminate, since the attained pivots have a least member.

The terminal pivot divides the entire rectangle. Clear its row and column and repeat on the smaller rectangle. There are at most $\min (n,\mathrm{columns}(A))$ pivots. The resulting diagonal $d_1,\dots ,d_k>0$ presents the quotient as ${⨁}_{j=1}^k\mathbb{Z}\mathrm{/}{d}_j\mathbb{Z}\oplus {\mathbb{Z}}^{n-k}$: the coordinate quotient map is onto and its kernel is exactly the diagonal image. Delete unit summands, which are zero. A zero matrix has $k=0$; an empty matrix gives the same rule.

Each finite cyclic summand has exactly the $d_j$ residues $0,\dots ,d_j-1$. Their finite product is finite and torsion. A nonzero integer vector has infinite order, since a nonzero coordinate cannot be annihilated by a nonzero integer. Hence the displayed finite summand is exactly the torsion subgroup. In particular a finitely generated torsion group is finite, and a finitely generated torsion-free abelian group is free.

Let $V(M)=\mathrm{Hom}(M,(\mathbb{Q},+))$ with pointwise rational addition and scalar multiplication. The vector space laws follow pointwise from the field laws. A map to $\mathbb{Q}$ kills every finite-order element: $da=0$ implies $a=0$. Evaluation on the $r$ free generators identifies $V(M)$ with ${\mathbb{Q}}^r$: any assigned rational values extend by $\sum n_j{e}_j+t\mapsto \sum n_j{a}_j$, and this is the only extension. If a second decomposition has $r^{\mathrm{\prime }}$ free generators, the resulting two bases of $V(M)$ give $r\le r^{\mathrm{\prime }}$ and $r^{\mathrm{\prime }}\le r$. Thus $r=r^{\mathrm{\prime }}$, including $r=0$.

If $f:M\twoheadrightarrow N$ has finite kernel, every $u:M\to \mathbb{Q}$ kills that kernel by step 7.1. Define $\bar{u}(f(x))=u(x)$; different lifts differ by the kernel, so this is well-defined, additive and unique. Thus precomposition by $f$ is a rational-linear bijection $V(N)\to V(M)$. Transporting bases and applying the independence bound twice proves equal ranks.