Coordinate boxes and word balls have matching size

A word of length at most n has layer-i free exponents at most $C\max (1,n{)}^i$ by F2. Choose b>=1 such that $b^i\ge C$ for all finitely many i. Then for $n\ge 1$ every such integer exponent is at most $\lfloor (bn{)}^i\rfloor$ in absolute value; its residues are allowed in Q. Thus $B_S(n)\subseteq Q(bn)$.

We prove $Q(R)\subseteq B_S(KR)$ for $R\ge 1$ by induction on class. For G=1 the empty tuple represents only 1. For class one, if $g=\prod u_j^{a_j}\prod v_j^{b_j}$ is in Q(R), repeating fixed S-words for the u_j costs at most $R\sum \mathrm{\mid }{u}_j{\mathrm{\mid }}_S$, and the finite residues cost at most $\sum (d_j-1)\mathrm{\mid }{v}_j{\mathrm{\mid }}_S$. Since $R\ge 1$ this is at most KR.

For class $c\ge 2$ put $H={\gamma }_c(G)$. The coordinate system truncated before layer c is a mixed coordinate system of G/H: for i<c the subgroup H lies in ${\gamma }_{i+1}$, so those quotient factors are unchanged. For g in Q(R), its image lies in the quotient box. By induction it has a word of length at most K_0 R in the quotient generators; lift its letters to an S-word w with the same length. The element $h=w^{-1}g$ lies in H. An explicit weighted word for it is the reversed inverse S-word for w followed by the ordered coordinate word for g. In weight i this has at most $\lambda R^i$ letters for a fixed lambda: w contributes only O(R) weight-one letters, the free powers contribute O(R^i), and finite residues contribute fixed bounded counts.

Apply F2 to that weighted word. Since h is in H, only last-layer coordinates remain; their free exponents are O(R^c), with bounded residues. The corresponding coordinate generators are a finite generating set of the abelian H, so $\mathrm{\mid }h{\mathrm{\mid }}_H\le K_1{R}^c$ after increasing K_1. F3 gives $\mathrm{\mid }h{\mathrm{\mid }}_S\le K_{2}R+K_2\le 2K_{2}R$. If H is finite its fixed ambient diameter gives the same conclusion. Hence $\mathrm{\mid }g{\mathrm{\mid }}_S=\mathrm{\mid }wh{\mathrm{\mid }}_S\le (K_0+2K_2)R$, completing the induction. Taking K>=1 and a=1/K gives $Q(an)\subseteq B_S(n)$ whenever a$n\ge 1$.

By uniqueness, every permitted tuple represents a different element. A free weight-i coordinate has exactly $2\lfloor R^i\rfloor +1$ possible values, and a residue coordinate has d possible values, independently. This proves the product formula, including the empty product 1. For $R\ge 1$, $R^i\le 2\lfloor R^i\rfloor +1\le 3R^i$. Thus with $T=\prod d_{ij}$ and $h=\sum r_i$, $T{R}^{D(G)}\le \mathrm{\mid }Q(R)\mathrm{\mid }\le T{3}^h{R}^{D(G)}$. Combining the two box inclusions gives positive upper and lower multiples of $n^{D(G)}$ for all sufficiently large n. If all ranks vanish, Q(R) has the constant size T and the same argument gives degree zero.