Finite normal quotients preserve lower-central ranks

Surjectivity gives $q({\gamma }_{1}G)=G\mathrm{/}F$. If $q({\gamma }_{i}G)={\gamma }_i(G\mathrm{/}F)$, then $q([g,u])=[q(g),q(u)]$ shows that the images of the generators of ${\gamma }_{i+1}G$ generate exactly ${\gamma }_{i+1}(G\mathrm{/}F)$. This proves equality for every $i$ and gives a surjection on each factor. Both source and target factors are finitely generated abelian, since a quotient of a finite generating list is finite and the quotient group is nilpotent.

Its kernel in layer $i$ consists of $x{\gamma }_{i+1}$ with $x\in {\gamma }_i\cap F{\gamma }_{i+1}$. Write $x=fy$, $f\in F,y\in {\gamma }_{i+1}$. Then $f=x{y}^{-1}\in F\cap {\gamma }_i$, and $x{\gamma }_{i+1}=f{\gamma }_{i+1}$. Conversely every such $f$ maps to the identity. The kernel is therefore the image of $F\cap {\gamma }_i$, a finite set. Finite-kernel rank preservation gives equal ranks layer by layer. Summing them with weights $i$ or $1$ proves equality of $D$ and $h$.

For $H={\gamma }_c$ and $i<c$, $H\le {\gamma }_{i+1}$, so the map ${\gamma }_i\mathrm{/}{\gamma }_{i+1}\to ({\gamma }_i\mathrm{/}H)\mathrm{/}({\gamma }_{i+1}\mathrm{/}H)$ is bijective: the kernel is zero and every coset lifts. In layer $c$ the quotient factor is trivial, and all later factors of both groups are trivial. Thus its dimension loses exactly $c{r}_c$. For $c=1$ the quotient is $G\mathrm{/}G=1$ and this says $0=D(G)-r_1$. For $G=1$, the first assertions are equality of empty sums.