Normalized root words and reflective completion $\widehat G=\varprojlim G/P_n(G)$ with exact power filtration

Proof. For a subset $T\subseteq G$, let $N$ be the normal closure of $\{[t,s]:t\in T,\ s\in S\}$. In $G/N$, the image of each $t\in T$ commutes with the images of all elements of $S$, and hence is central. The normal closure of the image of $T$ is therefore central. It follows that

The reverse inclusion used here holds because each displayed commutator belongs to the subgroup on the left, which is normal. Starting with the normal closure of $S$, induction gives the first assertion. If $\gamma_{c+1}(G)=1$, then $\gamma_c(G)$ is central. The subgroup generated by its displayed generators is consequently normal, so normal and ordinary generation agree. ◻

Proof. For weight one the assertion is immediate. Suppose $c\geq2$ and write $t=[u,s]$, where $u$ has weight $c-1$ and $s$ is either $x$ or $y$. The group $G/Z(G)$ has class at most $c-1$. Induction shows that $u(a,b^q)$ and $u(a,b)^{q^{d'}}$ have the same image in $G/Z(G)$, where $d'$ counts the occurrences of $y$ in $u$. Elements differing by a central factor have equal commutators with every fixed element.

The commutator $[u(a,b),s(a,b)]$ belongs to the central subgroup $\gamma_c(G)$. For a central commutator one has

Indeed, the identity $[v_1v_2,z]=v_1[v_2,z]v_1^{-1}[v_1,z]$ gives $[v^m,z]=[v,z]^m$ by induction. Applying this with the variables interchanged and using $[v,z]^{-1}=[z,v]$ gives the second exponent. If $s=x$, formula (11) gives exponent $q^{d'}$; if $s=y$, it gives $q^{d'}q=q^{d'+1}$. This proves (10). A commutator of weight at least two involving only $a$ is trivial, proving the last assertion. ◻

Proof. By Lemma 3, the central subgroup $T=\gamma_c(N)$ is generated by the values of weight-$c$ simple commutators in $a,b$. For a generator $t$ with $d\geq1$ occurrences of $b$, Lemma 4 gives

Generators with $d=0$ are trivial. The identity has the root $1$. If $z_i^q=\varphi(e_i)$ with $z_i\in T$, centrality gives

Induction under products and inverses now gives (12) for every $e\in T$. ◻

Proof of Theorem 2. Let $F=F(x,y)$ and $N_c=F/\gamma_{c+1}(F)$. The images of $x,y$ will be denoted by the same letters. We construct a word whose power equation holds in $N_c$ and whose normalization $W(x,1)=x$ holds in $F$ itself.

For $c=1$, take $W_{1,q}=xy$. Suppose $c\geq2$ and the word $v=W_{c-1,q}$ has been constructed. The canonical map $N_c\to N_{c-1}$ has kernel $\gamma_c(N_c)$. Thus, writing $v$ also for its value in $N_c$,

The substitution $x\mapsto x$, $y\mapsto y^q$ induces an endomorphism $\varphi$ of $N_c$, since homomorphisms preserve lower-central terms. Lemma 5 supplies $z\in\gamma_c(N_c)$ with $z^q=\varphi(e)$.

Surjective homomorphisms map each lower-central term onto the corresponding term. Choose $t\in\gamma_c(F)$ mapping to $z$, and put

The element $z$ is central in $N_c$, so

Since $c\geq2$, substitution $y=1$ kills $\gamma_c(F)$: its image lies in the corresponding lower-central term of the cyclic group generated by $x$. Hence $t(x,1)=1$, and (16) gives $W_{c,q}(x,1)=x$ in $F$. Finally every pair $a,b$ in a group of class at most $c$ defines a homomorphism $N_c\to G$. Evaluation transfers both identities to $G$. ◻

Proof. First let $z=t^{p^{r+k}}$. Set $b=w(a,t^{p^k})$. The first identity in (5) gives $b^p=a^p z$. In the quotient $G/P_k(G)$, the second argument of $w$ becomes $1$, so the normalization gives $bP_k(G)=aP_k(G)$.

The conclusion holds for $z=1$ by taking $b=a$. If it holds for $z$, apply the first step to its chosen root $b$ to absorb a further factor $t^{p^{r+k}}$. The coset condition is preserved because $P_k(G)$ is a subgroup. Inverse factors are absorbed by replacing $t$ with $t^{-1}$. This proves the assertion for the subgroup generated by all the indicated powers, namely $P_{r+k}(G)$. ◻

Proof. Choose $a_0=1$. We construct $a_i\in G$ such that

The error condition holds at $i=0$. Given $a_i$, apply (19) to $e_i$, with $n=r+k+i$ and $m=n+1$, to obtain $t_i\in P_{r+k+i}(G)$ satisfying

Lemma 6, with $k+i$ in place of $k$, gives $a_{i+1}$ such that

Then $e_{i+1}=\eta_G(t_i)^{-1}e_i$, proving the induction.

Products of (21b) give

Define $h_n=a_nP_n(G)$. Equation (24) makes $(h_n)$ a compatible sequence, hence an element $h\in\widehat G$. Its $k$th coordinate is trivial, so $h\in K_k(G)$. Since (21a) implies $e_n\in K_n(G)$, the $n$th coordinate of $h^p$ is $\pi_n(z)$ for every $n$. Therefore $h^p=z$.

Iterating this assertion proves (20). At the induction step, take a $p$th root in $K_{rn}(G)$ of an element of $K_{r(n+1)}(G)$, and then take its $p^n$-th root. ◻

Proof. Every element of $Q_n(G)$ has $p^n$-th power $1$, so $P_n(\widehat G)\leq K_n(G)$. Conversely, let $z\in K_n(G)$. Apply (19) at level $m=n+rn$ to obtain $g\in P_n(G)$ with

Lemma 7 writes $e=h^{p^n}$. The homomorphism $\eta_G$ sends $P_n(G)$ into $P_n(\widehat G)$, and therefore

 ◻

Proof. The product of the discrete groups $Q_n(G)$ is complete and Hausdorff. The compatibility equations in (3) define closed subsets, so $\widehat G$ is a closed, hence complete, subgroup. A finite set of coordinates is controlled by its largest index. Representing that coordinate by an element of $G$ proves density of $\eta_G(G)$. Finally,

which proves the pullback assertion.

If $G$ is complete, its image under this uniformity-inducing map is complete, hence closed in the Hausdorff limit. Density then makes $\eta_G$ surjective. Conversely, a surjective uniformity-inducing map onto a complete space makes its domain complete. For (27b), the pullback uniformity separates points exactly when the map to the Hausdorff limit is injective. ◻

Proof. A homomorphism $f:G\to H$ induces quotient maps

which commute with the transitions. Coordinatewise application gives $\widehat f:\widehat G\to\widehat H$ and $\widehat f\eta_G=\eta_H f$. Lemma 9 makes $\eta_H$ an isomorphism. Thus

is an extension.

For uniqueness, let $u,v:\widehat G\to H$ satisfy $u\eta_G=v\eta_G$. Fix $z\in\widehat G$ and $n\geq0$. Choose $g\in G$ representing $\pi_n(z)$. Proposition 8 gives

Homomorphisms preserve generated power subgroups. Hence the images of $u(z)$ and $u(\eta_G(g))$ in $Q_n(H)$ agree, as do the corresponding images under $v$. The assumed equality on $\eta_G(G)$ gives equality of the $n$th coordinates of $\eta_H(u(z))$ and $\eta_H(v(z))$. This holds for every $n$, and $\eta_H$ is injective, so $u(z)=v(z)$. ◻

Proof of Theorem 1. Theorem 2, with $q=p$, puts every nilpotent group in $\mathcal C_p$. For fixed $r,w$, identities (5) are preserved by quotient maps, products and subgroups. They therefore hold in every $Q_n(G)$, in their product, and in the compatible-sequence subgroup $\widehat G$. Thus $\widehat G\in\mathcal C_p$.

Proposition 8 identifies the two uniformity bases. Lemma 9 consequently makes $\widehat G$ complete and Hausdorff also for its intrinsic uniformity, so it lies in $\mathcal D_p$.

The coordinatewise maps $\widehat f$ preserve identities and composition. They define the asserted functor, and $\widehat f\eta_G=\eta_H f$ is naturality of its unit. Proposition 10 proves the bijections (7). These bijections are natural: postcomposition commutes with restriction, and precomposition is compatible with the displayed naturality square. Equivalently, uniqueness of the extensions identifies the corresponding composites. This proves the adjunction. ◻