Deep derived relations preserve finite mixed-coefficient quotients

Each $P^{i+1}L$ is a subgroup of $P^{i}L$. An automorphism of $L$ preserves $P^{i}L$ inductively, and induces an automorphism of its abelianization and of the displayed tensor group; their maps commute with $a\mapsto a\otimes 1$. Its kernel is therefore preserved, proving that all $P^{i}L$ are characteristic and in particular normal. Also $L^{(i)}\subseteq P^{i}L$ by induction: the assertion is equality at zero, and commutators of elements of $L^{(i)}$ lie in $[P^{i}L,P^{i}L]$, which maps to zero under abelianization and hence lies in $P^{i+1}L$. Derived subgroups are descending by their finite commutator definition. Consequently $N\subseteq G^{(r)}\subseteq G^{(i)}\subseteq P^{i}G$ for every $i\le r$.

We prove $P^{i}H=q(P^{i}G)$ through index $r$. It holds at zero. Suppose it holds at $i<r$ and put $A=P^{i}G$. The restriction $q:A\to P^{i}H$ is onto with kernel $N$. Moreover $N\subseteq G^{(r)}\subseteq G^{(i+1)}=[G^{(i)},G^{(i)}]\subseteq [A,A]\mathrm{.}$ The abelianization of $A\mathrm{/}N$ is $A\mathrm{/}[A,A]$: expanding commutators shows $[A\mathrm{/}N,A\mathrm{/}N]=[A,A]\mathrm{/}N$, since every finite commutator word in the quotient lifts by finitely many individual lifts. Thus the induced abelianization map is an isomorphism. Tensoring this isomorphism with $R_i$ is an isomorphism too, because tensoring its inverse gives an inverse on every generator $a\otimes u$; exactness of tensor is not being assumed. The two maps to these isomorphic tensor groups commute with $q$. Their kernels consequently satisfy $P^{i+1}H=q(P^{i+1}G)$: a fixed element of the former has a lift in $A$, and its tensor value is zero by the isomorphism, hence that lift is in $P^{i+1}G$. This proves the finite induction.

The kernel of each restricted $q:P^{i}G\to P^{i}H$ is $N$, already contained in $P^{i}G$ by step 1.1, proving the first displayed isomorphism. The preimage of $P^{i}H=q(P^{i}G)$ is $P^{i}G$, since it already contains $N$; this proves the second. Similarly the preimage in $P^{i}G$ of $P^{i+1}H$ is $P^{i+1}G$, giving the successive-quotient isomorphisms. For ordinary derived subgroups, expanding commutators and lifting each finite commutator word proves inductively $q(G^{(i)})=H^{(i)}$. Since $N\subseteq G^{(r)}$, the same preimage argument gives $G\mathrm{/}{G}^{(r)}\cong H\mathrm{/}{H}^{(r)}$. None of these conclusions follows merely from characteristicness: their actual kernels and preimages have been identified.

When all $R_i=\mathbb{Q}$, each quotient $P^{i}G\mathrm{/}{P}^{i+1}G$ embeds, by the kernel definition, in the additive group $(P^{i}G\mathrm{/}[P^{i}G,P^{i}G]){\otimes }_{\mathbb{Z}}\mathbb{Q}$. That group is a rational vector space by scalar multiplication on the second tensor coordinate, so its additive group is torsion-free abelian: multiplication by a nonzero integer has inverse multiplication by its reciprocal. The finite normal series $G\mathrm{/}{P}^{r}G⊵P^{1}G\mathrm{/}{P}^{r}G⊵\cdots ⊵P^{r}G\mathrm{/}{P}^{r}G=1$ therefore has torsion-free abelian factors and is precisely a PTFA series, with trivial factors permitted. For $r=0$ the quotient is trivial and the series is empty. For other ring lists steps 1.1–3.1 are unchanged, but a positive-characteristic tensor group can have additive torsion, so this last PTFA argument is intentionally not asserted. Every lift above concerned one finite word or one individual quotient element; no infinite set of representatives or vector-space basis was selected. This proves the full natural quotient assertions without a choice principle.