Integral coordinates from a central cyclic refinement

The center is a subgroup of torsion-free $G$, so is torsion-free. Every upper-central factor is torsion-free and abelian; it is finitely generated as a quotient of a finitely generated subgroup. Hence it has a finite ordered free basis. Refine it by the spans of its successive basis vectors, omitting zero factors. Lifting to $G$ gives a finite central series: for each lifted intermediate subgroup the commutators with $G$ lie in the previous upper-center subgroup, hence in the previous refined subgroup. Normality follows from this containment. Each new nontrivial factor is infinite cyclic.

Reverse the series and choose one generator lift $u_j$ per cyclic factor. For $g\in H_0$, there is a unique integer $a_1$ with $g{H}_1=u_1^{a_1}{H}_1$. Then $u_1^{-a_1}g\in H_1$. Iterate: at stage $j$ remove $u_j^{a_j}$ on the left. The last remainder is in $H_m=1$, giving $g=u_1^{a_1}\cdots u_m^{a_m}$. All selections are finite; the exponents are uniquely determined, without choices.

If two products are equal, their images in $H_0\mathrm{/}{H}_1$ force equality of their first exponents, since that factor is infinite cyclic. Cancel those first powers and repeat in $H_1\mathrm{/}{H}_2$, obtaining equality of every exponent. The zero tuple represents $1$; when $G=1$, $m=0$ and the single empty tuple represents its identity. Thus the product map is bijective.