The word problem for finitely generated abelian groups is solvable

First, every subgroup $H\le {\mathbb{Z}}^n$ is finitely generated, by induction on $n$. For $n=0$ it is zero. For $n>0$, projection on the last coordinate has image $d\mathbb{Z}$ by [L2], and its kernel is a subgroup of ${\mathbb{Z}}^{n-1}$, finitely generated by induction. If $d=0$, these kernel generators generate $H$. Otherwise fix one $h\in H$ with last coordinate $d$; every $v\in H$ has last coordinate $qd$ for some integer $q$, so $v-qh$ lies in the kernel. The kernel generators together with $h$ therefore generate $H$. This finite induction uses only finitely many choices of witnesses.

Now use the given finite presentation, and let $e(w)\in {\mathbb{Z}}^n$ be the vector of exponent sums of a word $w$. Let $M\in M_{n\times m}(\mathbb{Z})$ have columns $e(r_j)$, and let $L=M{\mathbb{Z}}^m$. Put $F=F(x_1,\dots ,x_n)$ and let $N$ be the normal closure of the relators. Since $F\mathrm{/}N=A$ is abelian, the kernel of the exponent-sum map $e:F\to {\mathbb{Z}}^n$, namely the commutator subgroup by [L1], is contained in $N$. Also $e(N)=L$: every element of a normal closure is a finite product of conjugates of relators and their inverses, whose exponent vectors are sums of signed columns of $M$, and every such integer combination occurs. Hence $w\in N$ if and only if $e(w)\in L$. For the reverse implication, choose $v\in N$ with $e(v)=e(w)$; then $w{v}^{-1}\in \ker e\subseteq N$.

Compute suitable $U,V,D$ by enumerating integer square matrices $U$ of size $n$ and $V$ of size $m$, with all entries bounded in absolute value by $b=0,1,2,\dots$. Each bounded stage is finite. Check whether $\det U,\det V\in \{1,-1\}$ and whether $UMV$ is a rectangular diagonal matrix with its nonzero diagonal entries first and each dividing the next. Determinants, matrix multiplication and integer divisibility are computable by finitely many integer operations; for a zero-dimensional square matrix the determinant is $1$. The search terminates by [L3], since the required pair has a finite entry bound. Integer matrices with determinant $\pm 1$ have integer inverses by the adjugate formula, so every accepted pair has the required invertibility.

Choose a finite generating list for $A$. Commutativity gives a surjection ${\mathbb{Z}}^n\to A$ sending standard basis vectors to these generators. Its kernel has a finite generating list $h_1,\dots ,h_t$ by step 1.1. A finite presentation of $A$ is obtained by imposing the finitely many pairwise commutator relators and the words $x_1^{(h_j{)}_1}\cdots x_n^{(h_j{)}_n}$: the commutator relators make the quotient ${\mathbb{Z}}^n$, and the remaining relators make its quotient by precisely the kernel. Thus every finitely generated abelian group admits a finite presentation.

Write the nonzero diagonal entries of $D$ as $d_1,\dots ,d_s$. Because $V{\mathbb{Z}}^m={\mathbb{Z}}^m$, we have $UL=D{\mathbb{Z}}^m$. Compute $y=Ue(w)$. Then $e(w)\in L$ if and only if $d_i$ divides $y_i$ for $1\le i\le s$ and $y_i=0$ for $s<i\le n$: these are exactly the coordinate conditions for the image of a rectangular diagonal matrix. The conditions include the zero matrix, empty relator list and empty generating list. Each is a finite integer check, and step 1.2 identifies the outcome with identity of $w$ in $A$.

The preprocessing in step 1.3 terminates, and exponent sums and the tests in step 2.2 terminate for every input word. Therefore the word problem for every finite presentation of an abelian group is solvable; step 2.1 supplies such presentations for every finitely generated abelian group. For another fixed finite generating set, fix words in the presentation generators for its finitely many generators, substitute these into the input word, and apply the same algorithm. No infinite choice or effective recognition of whether an arbitrary presentation is abelian is required.