==== TABLE OF CONTENTS ====
An-infinite-finitely-presented-residually-finite-2-group//introduction.tex
An-infinite-finitely-presented-residually-finite-2-group//ring-phases.tex
An-infinite-finitely-presented-residually-finite-2-group//steinberg.tex
An-infinite-finitely-presented-residually-finite-2-group//shifts.tex
An-infinite-finitely-presented-residually-finite-2-group//long-entries.tex
An-infinite-finitely-presented-residually-finite-2-group//detection.tex
An-infinite-finitely-presented-residually-finite-2-group//stability.tex
An-infinite-finitely-presented-residually-finite-2-group//construction.tex
An-infinite-finitely-presented-residually-finite-2-group//references.tex
===========================

==== BEGIN An-infinite-finitely-presented-residually-finite-2-group//introduction.tex ====
\section{Introduction}\label{sec:introduction}

We prove the following statement.

\begin{theorem}[Main theorem]\label{thm:main}
There exists an infinite, residually finite group with an ordinary
finite presentation in which every element has finite 2-power order.
\end{theorem}

The rest of this section fixes terminology, surveys the relevant
history, and outlines the argument. Readers new to group theory may
skip straight to the short dictionary at the end of the section.

\subsection*{The terms of the statement}

A \emph{group} is a set $G$ with an associative binary operation, an
identity element $1$, and inverses. It is \emph{finitely generated}
if there is a finite set whose products (and inverses) give all of
$G$. An \emph{ordinary finite presentation} consists of finitely many
generators $s_1,\dots,s_m$ and finitely many relations $w_k=1$; the
group is the quotient of the free group on the generators by the
normal closure of the words $w_k$. A group is \emph{periodic} if
each of its elements has finite order. It is a \emph{2-group} if
each element has order $2^a$ for some integer $a\ge0$; the group
itself is allowed to be infinite. It is \emph{residually finite} if,
for every non-identity element $g$, there is a homomorphism from the
group to a finite group whose value at $g$ is non-identity. Finite
presentation is taken in the category of abstract discrete groups,
not in a variety of groups or a pro-2 category.

\subsection*{Why the statement is surprising}

Burnside \cite{Burnside1902} asked in 1902 whether finitely generated
groups of bounded exponent must be finite. Novikov and Adian
\cite{NovikovAdian1968} answered no, constructing infinite
finitely generated groups of small bounded exponent. Those groups
are notoriously hard to describe, and they cannot be residually
finite: Zel'manov's solution of the restricted Burnside problem for
2-groups \cite[Theorem~2]{Zelmanov1992} implies that a finitely
generated residually finite group whose element orders are bounded
by a fixed $2^a$ is finite. Theorem~\ref{thm:main} sits at the other
extreme: the group is infinite and periodic, and its finite quotients
are still plentiful enough to separate all of its elements. Its
element orders are necessarily unbounded; and, because every element
has 2-power order, Cauchy's theorem shows that every finite quotient
is itself a finite 2-group.

\subsection*{History}

Golod \cite[Example~2]{Golod1964}, using the dimension criterion of
Golod and Shafarevich \cite{GolodShafarevich1964}, produced the
first infinite finitely generated residually finite $p$-groups. His
mechanism anticipates ours in outline: nilpotence produces finite
$p$-power orders, while a grading separates elements in finite
quotients. Grigorchuk \cite{Grigorchuk1980} gave the first concrete
example, an infinite finitely generated residually finite 2-group
which is not finitely presented. Neither construction produces an
ordinary finite presentation of the whole group.

Whether a finitely presented infinite periodic group exists is
recorded as an open question in
\cite[p.~45]{OlshanskiiSapir2003} and again in
\cite[Section~2.2(IX)]{HaettelOsajda2025}. The companion paper
\cite{Companion} answers it affirmatively, by constructing an
infinite finitely presented periodic group as a Steinberg group over
a finitely presented graded algebra with a matrix-nil positive-degree
ideal. The companion's group is not known to be residually finite.
The present paper strengthens the companion in two ways: it proves
residual finiteness, and it produces a finite-index subgroup whose
every element has 2-power order. The companion's algebra is an
essential input, quoted precisely at the start of
Section~\ref{sec:ring-phases}.

\subsection*{The group and its finite quotients}

The group in Theorem~\ref{thm:main} arises as follows. The companion
supplies a unital associative $\F_2$-algebra
\[
 R=\F_2\oplus I,\qquad I=\bigoplus_{d\ge1}R_d,
\]
finitely presented, graded by nonnegative integers, with every $R_d$
finite-dimensional, and with the strong nilpotence property that
every finite square matrix with entries in $I$ is nilpotent. We
consider the Steinberg group
\[
 \Gamma=\St_{12}(R),
\]
generated by symbols $x_{ij}(a)$, $a\in R$, $i\ne j$, subject to
the algebraic relations stated in Section~\ref{sec:steinberg}. Its
matrix image is the elementary matrix group $\E_{12}(R)$ generated
by the elementary matrices $e_{ij}(a)$, and $\Gamma$ has an
ordinary finite presentation by \cite[Theorem~4.1]{Companion} (see
also \cite[Theorem~3]{KrsticMcCool}).

The finite quotients of $\Gamma$ come from degree truncations of
$R$. For $d\ge1$ put
\[
 R_{\ge d}=\bigoplus_{e\ge d}R_e,\qquad B_d=R/R_{\ge d}.
\]
Each $B_d$ is finite: it is a direct sum of finitely many
finite-dimensional $\F_2$-vector spaces. Moreover
$B_d=\F_2\oplus I_d$ with $I_d^d=0$, so the companion's nilpotence
hypothesis still applies, and we obtain finite quotients
\[
 \Gamma\longrightarrow\St_{12}(B_d).
\]
The heart of the paper is the following detection statement: for
every non-identity $g\in\Gamma$, some $d\ge1$ makes the image of
$g$ in $\St_{12}(B_d)$ non-identity.

\subsection*{How detection works}

Detection is proved in two nested layers. The outer layer,
Section~\ref{sec:stability}, is a stability theorem: adjoining
coordinates does not enlarge the kernel. Concretely, the standard
map $\St_{12}(R)\to\St(R)$ to the stable Steinberg group is
injective. This lets detection be carried out in the stable group
$\St(R)$, where the proof may freely use additional coordinates.

The inner layer, Sections~\ref{sec:shifts}--\ref{sec:detection},
separates points of the stable group. Put
\[
 \E(R)=\varinjlim_n\E_n(R),\qquad
 J(R)=\ker\bigl(\St(R)\longrightarrow\E(R)\bigr).
\]
The kernel $J(R)$ is central in $\St(R)$
(Lemma~\ref{lem:stable-center}). Where $\E(R)$ is concerned,
separation is immediate: $g\notin J(R)$ already has a non-identity
matrix image over $R$, which some finite truncation sees. The real
content is detection inside the central kernel $J(R)$.

That is done by two cyclic-shift computations. Let $C_b$ be the
$b\times b$ cyclic permutation matrix over $\F_2$. Applying the
graded substitution $t\mapsto C_b$ to a root word for $z\in J(R)$
produces an element $W_z(C_b)\in J(R)$. The two computations are:

\begin{itemize}\itemsep0.2em
\item \emph{Power of two.} If $b$ is a power of two and large
  enough, then $C_b$ is conjugate in $\GL_b(\F_2)$ to an upper
  triangular matrix with all diagonal entries $1$. Triangular
  evaluation and coordinate permutations then give $W_z(C_b)=z^b$
  whenever $b$ is large enough that the wrap of $C_b$ can be
  removed (Lemma~\ref{lem:wrap}); on the other hand
  Lemma~\ref{lem:cyclic-vanishing} gives $W_z(C_b)=1$ whenever $b$
  is large enough. Hence $z^{2^a}=1$ for some $a\ge0$.
\item \emph{Odd power.} If $L$ is odd and $z$ is missed by every
  finite truncation, then a representative of $z$ with very long
  entries exists, and the \emph{phases} of long words in $R$ let us
  transport root endpoints to coordinates labeled by their phases.
  This gives $W_z(C_L)=z^L$ (Proposition~\ref{prop:odd-power}).
  Combined with the vanishing $W_z(C_L)=1$ for large enough odd $L$
  (Lemma~\ref{lem:cyclic-vanishing}), we obtain $z^L=1$.
\end{itemize}

Since $2^a$ and $L$ are coprime, $z=1$. That completes detection of
$J(R)$, hence of $\St(R)$, hence (by stability) of $\Gamma$.

\subsection*{The 2-group}

Once $\Gamma$ is residually finite, we pass to
\[
 G=\ker\bigl(\St_{12}(R)\longrightarrow\St_{12}(\F_2)\bigr).
\]
This is finite index in $\Gamma$ because $\St_{12}(\F_2)$ is finite
(the case $d=1$ of the finite-quotient construction above). It is
infinite and finitely presented by the Reidemeister--Schreier
theorem \cite[Chapter~II, Section~4]{LyndonSchupp}, and it inherits
residual finiteness from $\Gamma$. Every element has finite 2-power
order. Indeed, if $g\in G$ has matrix image $\Id+A$ with $A\in
M_{12}(I)$ and $A^r=0$, choose $a$ with $2^a\ge r$. In
characteristic two,
\[
 (\Id+A)^{2^a}=\Id+A^{2^a}=\Id,
\]
so $g^{2^a}\in J_{12}(R)$. Its stable image lies in $J(R)$ and
reduces to $1$ in $\St(\F_2)$, so
Proposition~\ref{prop:relative-torsion} gives an integer $b\ge0$
with $g^{2^b}=1$ stably. Injectivity of
$\St_{12}(R)\to\St(R)$ then gives $g^{2^{a+b}}=1$ in $\Gamma$, hence
in $G$.

\subsection*{A dictionary for readers new to the subject}

Readers comfortable with the terms above may skip this list.

\begin{itemize}\itemsep0.2em
\item \textbf{Group.} A set with an associative multiplication, an
  identity, and inverses.
\item \textbf{Generator.} An element whose products (and their
  inverses) produce the whole group.
\item \textbf{Relation.} An equation between two products of
  generators, usually written $w=1$ for a word $w$.
\item \textbf{Finite presentation.} Finitely many generators and
  finitely many relations.
\item \textbf{Order.} The smallest $n\ge1$ with $g^n=1$, if it
  exists.
\item \textbf{Periodic.} Every element has finite order.
\item \textbf{$p$-group.} Every element has order a power of the
  prime $p$; here $p=2$.
\item \textbf{Residually finite.} For every non-identity $g$, some
  homomorphism to a finite group sends $g$ to something
  non-identity.
\item \textbf{Matrix and Steinberg groups.} $\E_n(R)$ is generated
  by the elementary matrices $e_{ij}(a)$ (identity plus $a$ in one
  off-diagonal position). $\St_n(R)$ is generated by formal symbols
  $x_{ij}(a)$, subject to the same algebraic relations; it covers
  $\E_n(R)$ with kernel $J_n(R)$.
\end{itemize}

For a reader encountering these objects for the first time, it may
help to picture $\E_n(R)$ as the group generated by very small
matrix tweaks, and $\St_n(R)$ as its well-behaved formal lift:
symbols $x_{ij}(a)$ know only the algebraic structure of these
tweaks, not the particular matrices they happen to be.

\subsection*{How to read the rest}

Section~\ref{sec:ring-phases} records the companion algebra's word
presentation and extracts the phases of long words.
Section~\ref{sec:steinberg} introduces the Steinberg group and the
faithful-strip lemmas that the rest of the paper uses without
comment. Section~\ref{sec:shifts} proves the power-of-two
computation and the vanishing of cyclic shifts.
Section~\ref{sec:long-entries} controls opposite roots for long
entries. Section~\ref{sec:detection} combines both computations
into detection by finite degree truncations.
Section~\ref{sec:stability} transfers detection to fixed rank $12$.
Section~\ref{sec:construction} constructs the finite quotients
explicitly and proves Theorem~\ref{thm:main}.

All arguments establishing the phase, stabilization, and detection
claims are included below. From the companion we quote only the
algebra construction and the finite-rank elementary facts needed to
apply it, with hypotheses stated at the points of use.
==== END An-infinite-finitely-presented-residually-finite-2-group//introduction.tex ====

==== BEGIN An-infinite-finitely-presented-residually-finite-2-group//ring-phases.tex ====
\section{The algebra and phases of long words}\label{sec:ring-phases}

The work in this paper depends on a specific graded algebra $R$. It
has two features. Its positive-degree part behaves like a nilpotent
ideal: every finite matrix over that part is nilpotent. And its
longest words carry consistent positions modulo arbitrarily large
odd integers. The first feature produces finite torsion orders. The
second is what allows finite quotients to detect elements of the
Steinberg group that matrices alone cannot see. This section records
both. The algebra itself is taken from the companion
\cite{Companion}; only the phase construction is proved here.

Throughout, all algebras are unital and associative over $\F_2$. An
ideal $I$ of such an algebra is \emph{matrix-nil} if every square
matrix of every finite size with entries in $I$ is nilpotent.

\begin{proposition}[The companion algebra]\label{prop:ring-input}
There is an infinite finitely presented unital associative algebra
\[
 R=\bigoplus_{d\geq0}R_d=\F_2\oplus I,
 \qquad R_0=\F_2,
 \qquad I=\bigoplus_{d\geq1}R_d,
\]
such that every homogeneous component $R_d$ is finite-dimensional
and the ideal $I$ is matrix-nil. The algebra has the word
presentation and hierarchy specified below in
Proposition~\ref{prop:hierarchy-input}.
\end{proposition}

This is the algebra constructed in \cite[Theorem~3.7]{Companion}. In
a presentation by words, a \emph{replacement} is a relation $v=v'$
between nonempty words of equal length, usable in either direction
inside any word. A \emph{zero declaration} is a relation $v=0$ for a
nonempty word $v$. The \emph{length} of a replacement is the common
length of its two sides. For a word $w$, write $|w|$ for its length
and $[w]$ for its image in $R$.

\begin{proposition}[The companion hierarchy]\label{prop:hierarchy-input}
The algebra $R$ in Proposition~\ref{prop:ring-input} is the first
algebra $A_0$ of a sequence $(A_j)_{j\geq0}$. There are positive
integers $c,k,h_0,D$ with $D>10(c+1)(k+1)$ and
\[
 h_j=h_0D^j,\qquad L_j=2\cdot2^{h_j}+1.
\]
For each $j\geq0$:
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=*]
\item The algebra $A_j$ is presented by a finite alphabet $\Sigma_j$,
finitely many replacements, and finitely many zero declarations, all
of length at most $k$. It is graded by word length.
\item There is a substitution $E_j$ sending each letter of
$\Sigma_{j+1}$ to a word of length $L_j$ over $\Sigma_j$. It extends
by concatenation to a unital algebra homomorphism $A_{j+1}\to A_j$.
\item A specified subset of $\Sigma_j$ gives \emph{cuts},
immediately before the corresponding letters. There are integers
$B_j,T_j\geq0$ and a finite set of words $\mathcal T_j$ such that
every nonzero word $w$ of length at least $T_j$ has a unique
decomposition
\begin{equation}\label{eq:token-decomposition}
 w=p_w\,t_1\cdots t_r\,s_w,
 \qquad |p_w|,|s_w|\leq B_j,\qquad t_i\in\mathcal T_j.
\end{equation}
Here $p_w$ precedes the first cut, $s_w$ begins at the last cut, and
the $t_i$ are the intervals between consecutive cuts. A cut counts
only if its following letter occurs in the word. For each
$t\in\mathcal T_j$ there is a nonempty word $u_t$ over
$\Sigma_{j+1}$ with
\begin{equation}\label{eq:token-upper-word}
 1\leq |u_t|\leq k,\qquad
 |t|=L_j|u_t|,\qquad [t]=[E_j(u_t)]\text{ in }A_j.
\end{equation}
\end{enumerate}
\end{proposition}

These are the applicable parts of \cite[Proposition~2.2 and the
parameter choices in Section~3]{Companion}. Because cuts are
recognized from individual letters, restricting to a subword cannot
create or move a cut.

We recall the elementary word-basis lemma
\cite[Lemma~2.1]{Companion}. It also identifies which equalities of
long words must preserve their phases. Form a graph whose vertices are
all words, including the empty word, and whose edges are replacements
in arbitrary contexts. Quotienting the free algebra by the
replacement relations leaves one basis vector per connected
component: the defining ideal is exactly spanned by differences of
words in the same component. Imposing the zero declarations then
kills precisely the components containing a word with a
zero-declared subword. Hence, inside every $A_j$, the nonzero
replacement classes form a basis, and
\begin{equation}\label{eq:word-zero-criterion}
 [w]=0\quad\Longleftrightarrow\quad
 \text{$w$ can be replaced by a word containing a zero-declared subword.}
\end{equation}
In particular, a subword of a nonzero word is nonzero. The product
of two basis classes is either zero or the class of their
concatenation. We shall write the same symbol for a nonzero word and
its basis class when it occurs as an algebra entry.

The remaining task of the section is to extract the modular phases
of long words. Fix $d\geq0$ and put $L=L_0L_1\cdots L_d$.

\begin{lemma}[Phases of long words]\label{lem:phases}
There is an integer $N\geq1$ such that every nonzero word $w$ in $R$
of length at least $N$ has a \emph{start phase}
$\sigma(w)\in\Z/L\Z$ and an \emph{end phase}
\[
 \tau(w)=\sigma(w)+|w|\in\Z/L\Z
\]
with the following properties.
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=*]
\item Both phases depend only on the basis class of $w$.
\item If $v$ is a contiguous subword of $w$, has length at least $N$,
and starts at offset $b$ from the beginning of $w$, then
\begin{equation}\label{eq:phase-subword}
 \sigma(v)=\sigma(w)+b.
\end{equation}
\end{enumerate}
The integers $L$ so obtained are odd and arbitrarily large.
\end{lemma}

\begin{proof}
We first construct phases for literal nonzero words, without
invariance under replacements. Work backward through the levels
$j=d,d-1,\ldots,0$. At level $d+1$, use the trivial phase modulo the
empty product $1$, with threshold $1$. Suppose phases with the
subword property have been constructed at level $j+1$ modulo
$L'=L_{j+1}\cdots L_d$, above a threshold $n_{j+1}\geq1$.

For every token $t\in\mathcal T_j$, choose one word $u_t$ satisfying
\eqref{eq:token-upper-word}, and use this same choice at every
occurrence of $t$. For the decomposition
\eqref{eq:token-decomposition}, set $U(w)=u_{t_1}\cdots u_{t_r}$ and
choose
\begin{equation}\label{eq:literal-threshold}
 n_j\geq\max\{1,T_j,\,2B_j+L_j n_{j+1}\}.
\end{equation}
Every nonzero $w$ of length at least $n_j$ then has such a
decomposition, and
\[
 |U(w)|=\frac{|w|-|p_w|-|s_w|}{L_j}\geq n_{j+1}.
\]
Moreover, $U(w)$ is nonzero in $A_{j+1}$. The substitution
homomorphism and the token equalities give
$[w]=[p_w]\,[E_j(U(w))]\,[s_w]$, so $[U(w)]=0$ would force $[w]=0$.
Define the literal start phase by
\begin{equation}\label{eq:phase-induction}
 \sigma_j(w)=L_j\sigma_{j+1}(U(w))-|p_w|
 \pmod{L_jL'}.
\end{equation}
Multiplication by $L_j$ makes this independent of the representative
chosen for the residue $\sigma_{j+1}(U(w))$ modulo $L'$.

We verify the subword property at level $j$. Let $v$ be a subword of
$w$ of length at least $n_j$, starting at offset $b$. Its cuts are
exactly the cuts of $w$ whose following letters lie in $v$.
Therefore its complete tokens form a consecutive subsequence of the
complete tokens of $w$. Our fixed token choices make $U(v)$ a literal
subword of $U(w)$. If $a$ is its upper-level offset, the lower-level
distance to its first complete token can be computed in two ways:
\begin{equation}\label{eq:token-offset}
 b+|p_v|=|p_w|+L_j a.
\end{equation}
Both upper words meet the upper-level threshold. Applying the
inductive subword property and then \eqref{eq:token-offset} gives
\begin{align*}
 \sigma_j(v)
 &=L_j\bigl(\sigma_{j+1}(U(w))+a\bigr)-|p_v|\\
 &=L_j\sigma_{j+1}(U(w))-|p_w|+b
 =\sigma_j(w)+b \pmod{L_jL'}.
\end{align*}
This completes the backward induction.

At level zero we have literal phases modulo $L$, with the subword
property for lengths at least $n_0$. Invariance under replacements
remains. Choose $N=2n_0+k$. A single replacement inside a nonzero
word of length at least $N$ affects at most $k$ letters, so an
unchanged prefix or unchanged suffix has length at least $n_0$. That
common subword has the same offset before and after the replacement,
because the replacement preserves length. Applying the literal
subword property to both words shows that their phases agree. Every
intermediate word in a nonzero replacement class is nonzero.
Chaining these equalities shows that the start phase is constant on
each class. Length is constant on each class too, so the end phase is
also invariant, and the subword property, already established for
literal words, still holds after raising the threshold from $n_0$ to
$N$. Finally, every $L_j$ is odd and greater than one, so the
products $L_0\cdots L_d$ are odd and unbounded.
\end{proof}

\begin{corollary}\label{cor:phase-products}
Fix $L,N,\sigma,\tau$ as in Lemma~\ref{lem:phases}. If $u,v$ are
nonzero basis words of lengths at least $N$, then
\[
 \tau(u)\ne\sigma(v)\quad\Longrightarrow\quad uv=0.
\]
If $uv\ne0$, then $\sigma(uv)=\sigma(u)$ and $\tau(uv)=\tau(v)$.
\end{corollary}

\begin{proof}
If the concatenation is nonzero, apply \eqref{eq:phase-subword} to its
prefix $u$ and its suffix $v$. This gives
$\sigma(uv)=\sigma(u)$ and
$\sigma(v)=\sigma(uv)+|u|=\tau(u)$; adding $|v|$ gives the end-phase
identity.
\end{proof}
==== END An-infinite-finitely-presented-residually-finite-2-group//ring-phases.tex ====

==== BEGIN An-infinite-finitely-presented-residually-finite-2-group//steinberg.tex ====
\section{Steinberg groups and faithful strips}\label{sec:steinberg}

The Steinberg group records elementary matrix operations as formal
symbols before they are collapsed to matrices. Its kernel over the
matrix group is the piece that finite quotients must detect. This
section sets out the definitions and the three elementary lemmas used
throughout the paper: faithful strips, centrality of the kernel, and
coordinate permutations. All algebras here are unital associative
$\F_2$-algebras.

For such an algebra $B$ and an integer $n\ge3$, the group $\St_n(B)$
has generators $x_{ij}(a)$ with $a\in B$ and $1\le i,j\le n$,
$i\ne j$, subject to the relations
\begin{align}
 x_{ij}(a)x_{ij}(b)&=x_{ij}(a+b),\label{eq:root-addition}\\
 [x_{ij}(a),x_{lt}(b)]&=1
       &&(j\ne l,\ i\ne t),\label{eq:root-commuting}\\
 [x_{ij}(a),x_{jt}(b)]&=x_{it}(ab)
       &&(i,j,t\text{ distinct}).\label{eq:root-product}
\end{align}
We use $[X,Y]=XYX^{-1}Y^{-1}$. In particular, $x_{ij}(0)=1$ and every
root element has square $1$. The elementary matrix $e_{ij}(a)$ is
the identity with $a$ added in position $(i,j)$. Elementary matrices
obey \eqref{eq:root-addition}--\eqref{eq:root-product}, so there is
an epimorphism
\[
 \pi_n:\St_n(B)\longrightarrow\E_n(B)
       :=\langle e_{ij}(a):a\in B,\ i\ne j\rangle.
\]
Write $J_n(B)=\ker\pi_n$. The same presentation with a countable
index set defines $\St(B)$, and we set
\[
 \E(B)=\varinjlim_n\E_n(B),\qquad
 J(B)=\ker\bigl(\St(B)\longrightarrow\E(B)\bigr).
\]
Thus $\St(B)=\varinjlim_n\St_n(B)$, because every word and every
finite derivation uses only finitely many indices. These are direct
limits of discrete groups; no completion enters. The fixed-rank
injectivity needed for the companion's ring is proved separately in
Section~\ref{sec:stability}.

The presentation, and its relation to algebraic $K$-theory, are
classical; see \cite{Milnor,WeibelK}. We use three elementary
consequences repeatedly. They also explain when an identity of
matrices implies the corresponding identity in the Steinberg group.

\begin{lemma}[Faithful strips]\label{lem:strip}
Let $S,T$ be disjoint finite sets of coordinates. The subgroup
generated by $x_{ij}(a)$ with $i\in S$, $j\in T$ maps
isomorphically onto the additive group of matrices supported in
$S\times T$ under $u\mapsto\pi(u)-\Id$. A word supported on $S$
normalizes this subgroup and acts on its entries by left
multiplication by its matrix on $S$. On the reverse strip $T\times S$
it acts by right multiplication by the inverse matrix on $S$.
\end{lemma}

\begin{proof}
All roots in $S\times T$ commute by \eqref{eq:root-commuting}. Root
addition collects a word into one factor per coordinate pair. The
corresponding entry matrices have pairwise zero products, so the
matrix image reads each collected coefficient directly. This proves
injectivity and identifies the group law with addition.

For one root supported on $S$, conjugation on an outgoing strip
follows from \eqref{eq:root-product}: it performs the corresponding
elementary row addition. The same relations on the reverse strip
perform the inverse elementary column operation. Iterating over the
factors of a word gives the stated left and right actions, in the
order prescribed by the possibly noncommutative algebra $B$.
\end{proof}

\begin{lemma}[Stable centrality]\label{lem:stable-center}
For every $B$, the kernel $J(B)$ is central in $\St(B)$.
\end{lemma}

\begin{proof}
Let $z\in J(B)$ and choose a finite set $S$ containing the support of
a representative of $z$. To show that $z$ commutes with a specified
root, enlarge $S$ to contain both indices of that root. Choose
$h\notin S$. By Lemma~\ref{lem:strip}, $z$ acts trivially on the
strips $S\times\{h\}$ and $\{h\}\times S$, since its matrix image is
the identity. For $i,j\in S$ with $i\ne j$, the identity
$x_{ij}(a)=[x_{ih}(a),x_{hj}(1)]$ therefore implies that $z$
commutes with $x_{ij}(a)$. These roots generate the group.
\end{proof}

\begin{lemma}[Coordinate permutations]\label{lem:permutations}
In characteristic two, every permutation of a finite set of
coordinates is induced on roots by conjugation in the Steinberg
group after adjoining an outside coordinate if necessary. In
particular, relabeling the finitely many coordinates in a
representative of $z\in J(B)$ leaves $z$ unchanged in $\St(B)$.
\end{lemma}

\begin{proof}
For a transposition of coordinates $a,b$, use
$x_{ab}(1)x_{ba}(1)x_{ab}(1)$. On a column strip from $\{a,b\}$ to an
outside coordinate, its three elementary row additions exchange the
two entries over $\F_2$; Lemma~\ref{lem:strip} gives this as an exact
group action. The reverse strip is exchanged the same way. Roots
with both indices outside $\{a,b\}$ commute with this element. The
remaining two root positions can be written, with $c\notin\{a,b\}$, as
\[
 x_{ab}(u)=[x_{ac}(u),x_{cb}(1)],\qquad
 x_{ba}(u)=[x_{bc}(u),x_{ca}(1)].
\]
The strip actions exchange these as well. Products of transpositions
give arbitrary finite permutations. Every injective relabeling of a
finite set extends to a permutation of a larger finite set, so the
last assertion follows from Lemma~\ref{lem:stable-center}.
\end{proof}

The proof of Lemma~\ref{lem:permutations} is the argument of
\cite[Lemma~6.3]{Companion}. The exact action on roots, not only on
their matrix images, matters below.

\begin{lemma}[A commutator identity]\label{lem:commutator}
In any group, let $A=[X,Y]$ and $B'=[Y,Z]$. If
\[
 [X,Z]=[A,X]=[B',Z]=[A,B']=1,
\]
then $[A,Z]=[X,B']$. This common commutator commutes with $A,Z,X,B'$.
\end{lemma}

\begin{proof}
The subgroups $H=\langle A,Z\rangle$ and $K=\langle X,B'\rangle$
commute elementwise by the four hypotheses. Conjugating $[X,Z]=1$ by
$Y$ gives $[A^{-1}X,B'Z]=1$. Separating the commuting $H$ and $K$
factors yields $[A^{-1},Z][X,B']=1$. Thus $[X,B']=[A^{-1},Z]^{-1}$
lies in $H\cap K$, which is central in both $H$ and $K$. Centrality in
$H$ then gives $[A,Z]=[A^{-1},Z]^{-1}=[X,B']$.
\end{proof}

Finally, $\St_n(B)$ and $\E_n(B)$ are perfect for $n\ge3$: each
generating root is a commutator in \eqref{eq:root-product}, with one
coefficient equal to $1$. If $B$ is infinite, both groups are
infinite, since distinct coefficients at a fixed root position give
distinct elementary matrices.
==== END An-infinite-finitely-presented-residually-finite-2-group//steinberg.tex ====

==== BEGIN An-infinite-finitely-presented-residually-finite-2-group//shifts.tex ====
\section{Cyclic shifts and the relative central kernel}\label{sec:shifts}

Consider an element $z\in J(R)$ whose reduction to $\St(\F_2)$ is
trivial. The grading of $R$ lets us evaluate such an element at a
finite numerical matrix. We will prove two facts. First, evaluation
at a sufficiently large cyclic shift vanishes. Second, evaluation at
a cyclic shift of $2$-power size gives a power of the original
element. Together these imply $2$-primary torsion in the relative
central kernel. Later we will compare the vanishing of the first
fact with the odd power from the second, using the phases of
Section~\ref{sec:ring-phases}.

The matrix substitutions we use are elementary versions of
Weibel's graded transfer operations \cite[Section~1]{Weibel1987}.
At the level of stable algebraic $K$-theory, the relative torsion
conclusion also follows from his continuous Witt-vector action
\cite[Theorem~0.1]{Weibel1987}. We reproduce the concrete
Steinberg-group arguments of the companion
\cite[Lemmas~6.4--6.7]{Companion}, since their exact identities, not
only identities of matrix images, are needed here.

For the next three lemmas let $B$ be any unital $\F_2$-algebra. Fix
$n\ge3$ and $b\ge1$, and label the $nb$ coordinates by $(i,s)$ with
$1\le i\le n$ and $0\le s<b$. We call $i$ the \emph{outer} index and
$s$ the \emph{inner} index. For $A=(a_{st})\in\Mat_b(B)$ and $i\ne j$
define the \emph{block root}
\[
 X_{ij}(A)=\prod_{s,t=0}^{b-1}
           x_{(i,s),(j,t)}(a_{st})\in\St_{nb}(B).
\]
All factors commute, because their source outer index is $i$ and
their target outer index is the different index $j$.

\begin{lemma}[Block roots and coordinate changes]\label{lem:block-laws}
Block roots satisfy the Steinberg relations over $\Mat_b(B)$:
\begin{align*}
 X_{ij}(A+A')&=X_{ij}(A)X_{ij}(A'),\\
 [X_{ij}(A),X_{kl}(A')]&=1
       &&\text{if }j\ne k\text{ and }i\ne l,\\
 [X_{ij}(A),X_{jk}(A')]&=X_{ik}(AA')
       &&\text{if }i,j,k\text{ are distinct}.
\end{align*}
Consequently they define a homomorphism
$\St_n(\Mat_b(B))\to\St_{nb}(B)$. Moreover, for every
$P\in\GL_b(\F_2)$ there is an element $q\in\St_{nb}(B)$ such that
\[
 q^{-1}X_{ij}(A)q=X_{ij}(P^{-1}AP)
 \qquad(i\ne j).
\]
Its matrix image is the block diagonal matrix with a copy of $P$ in
every outer block.
\end{lemma}

\begin{proof}
Additivity and the commuting relation follow factor by factor. For
the consecutive commutator, the only nontrivial elementary
commutators are
\[
 [x_{(i,s),(j,t)}(a_{st}),
   x_{(j,t),(k,v)}(a'_{tv})]
   =x_{(i,s),(k,v)}(a_{st}a'_{tv}).
\]
Each resulting root commutes with both input blocks. Expanding the
commutator and using additivity at each position therefore gives the
entries $\sum_t a_{st}a'_{tv}$ of $AA'$.

An elementary operation supported inside outer block $i$ acts on the
outgoing strip to block $j$ by the corresponding row operation; an
operation inside block $j$ acts by the inverse column operation.
These are exact strip actions by Lemma~\ref{lem:strip}. An operation
inside any other outer block centralizes the block root. Gaussian
elimination over $\F_2$ writes $P$ as a product of elementary
matrices: every nonzero pivot is $1$, and a row transposition is a
product of three row additions. Lift those factors inside each outer
block and apply the strip actions successively. The case $b=1$ has
$P=1$ and is immediate.
\end{proof}

If $v(t)$ is a root word over the polynomial algebra $B[t]$ with $t$
central, and $T\in\Mat_b(\F_2)$, let $W_v(T)$ be its evaluation at
$t=T$, followed by the block-root homomorphism. The substitution
$B[t]\to\Mat_b(B)$ is a unital algebra homomorphism, since the
numerical matrix $T$ commutes with every scalar matrix from $B$. In
particular, if $v(t)$ has identity matrix image over $B[t]$, then
$W_v(T)$ has identity matrix image over $B$. For each inner index
$s$, also write
\[
 \iota_s:\St_n(B)\longrightarrow\St_{nb}(B),\qquad
 x_{ij}(a)\longmapsto x_{(i,s),(j,s)}(a).
\]
These coordinate homomorphisms have commuting images for distinct
$s$; we do not assume them injective.

\begin{lemma}[Triangular evaluation]\label{lem:triangular}
Let $U$ be generated by the roots $x_{(i,s),(j,t)}(a)$ with $s<t$,
allowing $i=j$. Then $U$ injects into its matrix image and is
normalized by every root whose inner indices agree. If a root word
$v(t)$ over $B[t]$ has identity matrix image, and
$T\in\Mat_b(\F_2)$ is upper triangular with constant diagonal entry
$\lambda$, then
\begin{equation}\label{eq:triangular-evaluation}
 W_v(T)=\prod_{s=0}^{b-1}\iota_s\bigl(v(\lambda)\bigr).
\end{equation}
\end{lemma}

\begin{proof}
The \emph{gap} of a generator of $U$ is $t-s>0$. Two such roots
either commute or have a commutator root whose gap is the sum of
their gaps, by the consecutive-root relations in either order; they
cannot be opposite. For $g\ge1$, let $U_g$ be the subgroup generated
by roots of gap at least $g$, so $U_1=U$ and $U_b=1$. Conjugating a
generator of $U_g$ by a generator of $U$ adds only roots of larger
gap, so $U_g$ is normal in $U$. Modulo $U_{g+1}$, the group $U_g$ is
generated by the commuting roots of gap $g$.

Consequently, successively for $g=1,\ldots,b-1$, every element of
$U$ is an ordered product with one factor at each coordinate pair,
in increasing gap; repeated coefficients at a given pair are added.
This collection is finite, since there are finitely many coordinate
pairs and $U_b=1$. If the matrix image is the identity, the entries
of gap one are precisely the collected coefficients, so they vanish.
After coefficients of smaller gap have vanished, the same assertion
holds at gap $g$: any product of two remaining entry matrices has
gap greater than $g$. Induction makes every coefficient zero and
proves faithfulness.

A root with equal inner indices cannot be opposite to a positive-gap
root. Its commutator with a positive-gap root is either trivial or
has the same positive gap. This proves it normalizes $U$.

For a polynomial entry $a(t)$, the matrix $a(T)$ is upper triangular
with diagonal $a(\lambda)$. Split its block root into its commuting
diagonal and strictly upper factors. The diagonal factors normalize
$U$, so the strictly upper factors in the complete word can be
collected on the right. Diagonal factors at different inner indices
commute, giving
\[
 W_v(T)=\left(\prod_s\iota_s(v(\lambda))\right)u,
 \qquad u\in U.
\]
The left side and all words $v(\lambda)$ have identity matrix image.
Hence $u$ has identity matrix image, and faithfulness of $U$ gives
$u=1$.
\end{proof}

\subsection*{Removing the wrap of a cyclic shift}

Let $C_b$ be the numerical cyclic shift from row $s$ to column $s+1$
modulo $b$, and let $S_b$ be the shift with the same entries except
for the entry from row $b-1$ to column $0$, which is removed. Thus
$S_b$ is strictly upper triangular. The next lemma replaces $C_b$ by
$S_b$ without changing the evaluated Steinberg element. Its proof
confines the difference to a faithful strip between two disjoint
bands of inner indices.

\begin{lemma}[Removing the wrap]\label{lem:wrap}
Let $v(t)$ be a product of $m\ge1$ root factors over $B[t]$ whose
polynomial entries have degrees at most $d\ge0$. Suppose its matrix
image is the identity over $B[t]$. Put $\delta=\max\{1,d\}$. If
$b>2(m+1)\delta$, then $W_v(C_b)=W_v(S_b)$ in $\St_{nb}(B)$.
\end{lemma}

\begin{proof}
For each factor, split its cyclic evaluation as $D_hF_h$, where $D_h$
is the evaluation at $S_b$ and $F_h$ consists of the entries wrapping
past inner index $b-1$. The splitting is exact because all roots in
one outer rectangle commute. Since $d<b$, every term of degree $a\le
d$ goes from $s$ to $s+a$ and can wrap only once. Every root in $F_h$
therefore goes from some source in $\{b-\delta,\ldots,b-1\}$ to some
target in $\{0,\ldots,\delta-1\}$.

For $r\ge1$, set $H_r=\{b-r\delta,\ldots,b-1\}$ and
$L_r=\{0,\ldots,r\delta-1\}$. When $b>2r\delta$, the roots from inner
indices in $H_r$ to inner indices in $L_r$, with arbitrary outer
indices, form an abelian strip $V_r$ that injects into matrices by
Lemma~\ref{lem:strip}. We show that conjugation by one $D_h$
enlarges both endpoints by at most $\delta$.

More precisely, for $r\le m$ we claim
\begin{equation}\label{eq:strip-enlargement}
 D_h^{-1}V_rD_h\subseteq V_{r+1}.
\end{equation}
Fix a root of $V_r$. The constituents of $D_h$ all have the same
outer source $i$ and outer target $j\ne i$, and inner displacements
between $0$ and $d$; they commute with one another. A constituent
commuting with the initial root also commutes with every conjugate of
it by the other constituents, so it can be omitted.

An interacting constituent changes the source endpoint only when its
target is the old source; the new source is at most $d$ positions
further left. It changes the target endpoint only when its source is
the old target; the new target is at most $d$ positions further
right. These statements follow directly from the consecutive-root
relations, including their reverse order. An opposite-root
interaction cannot occur. Initially the source inner index minus the
target inner index is greater than $b-2r\delta$. After one endpoint
has moved, the difference is still greater than
$b-(2r+1)\delta>\delta\ge d$. So no constituent, whose forward
displacement is at most $d$, can be opposite to either the initial
root or a root obtained after one endpoint move.

Nor can displacements accumulate repeatedly at one endpoint. A
changed source has outer index $i$, whereas every constituent has
target outer index $j\ne i$; the source cannot change again. A
changed target has outer index $j$, whereas every constituent has
source outer index $i$; the target cannot change again. Both
endpoints may change, but each changes at most once. A root with
both endpoints changed has outer indices $i,j$ and commutes with
every constituent. All produced roots lie in $V_{r+1}$, proving
\eqref{eq:strip-enlargement}. This reasoning applies to a product of
strip roots as well.

To collect the corrections without conjugating them by one another,
put $D_{>h}=D_{h+1}\cdots D_m$ and $D_{>m}=1$. The exact group
identity
\[
 \prod_{h=1}^m(D_hF_h)
  =(D_1\cdots D_m)
       \prod_{h=1}^m(D_{>h}^{-1}F_hD_{>h}),
\]
with the final product taken in increasing order of $h$, does this.
Every $F_h$ starts in $V_1$, and every suffix has at most $m$
factors. By \eqref{eq:strip-enlargement}, all final corrections
belong to $V_{m+1}$. Therefore $W_v(C_b)=W_v(S_b)u$ with
$u\in V_{m+1}$. Both evaluated words have identity matrix image. The
same is true of $u$, and the disjointness of $H_{m+1}$ and $L_{m+1}$
makes $V_{m+1}$ faithful. Hence $u=1$.
\end{proof}

\subsection*{Graded evaluation and powers of central elements}

Return to the graded algebra $R=\bigoplus_{a\ge0}R_a$ with
$R_0=\F_2$. The map
\[
 \gamma:R\longrightarrow R[t],\qquad
       \sum_a r_a\longmapsto\sum_a r_a t^a
       \quad(r_a\in R_a)
\]
is a unital algebra homomorphism. Applying it to a root word
representing $z\in J(R)$ gives a word $z(t)$ with identity matrix
image over $R[t]$. Its evaluations at $1$ and $0$ are $z$ and the
degree-zero reduction of $z$, respectively.

For $T\in\Mat_b(\F_2)$, fix an injection from the pairs $(i,s)$,
where $i$ ranges over the stable index set and $0\le s<b$, into the
stable index set. Define $W_z(T)$ by applying the graded
substitution and evaluating by block roots in those coordinates.
This is a homomorphism on $\St(R)$ by Lemma~\ref{lem:block-laws}, so
$W_z(T)$ depends only on $z$. The element $W_z(T)$ belongs to
$J(R)$. All words and their block evaluations use finitely many
coordinates, and choosing a different injection gives the same
element by centrality and Lemma~\ref{lem:permutations}.

\begin{lemma}[Vanishing at long cyclic shifts]\label{lem:cyclic-vanishing}
If $z\in J(R)$ maps to $1$ in $\St(\F_2)$, then
\[
 W_z(C_b)=1
 \qquad\text{for every sufficiently large integer }b.
\]
More precisely, if the graded word $z(t)$ has $m\ge1$ factors of
degrees at most $d$, it suffices that $b>2(m+1)\max\{1,d\}$.
\end{lemma}

\begin{proof}
Choose one finite root word for $z$, and apply Lemma~\ref{lem:wrap}
to its graded word. Since $S_b$ is strictly upper triangular,
Lemma~\ref{lem:triangular} gives, in the stable group,
\[
 W_z(C_b)=W_z(S_b)
       =\prod_{s=0}^{b-1}\iota_s(z(0))=1.
\]
The fixed injection extends each $\iota_s$ to a homomorphism on the
stable group. Since $z(0)$ is trivial there by hypothesis, every
displayed coordinate image is trivial. The empty word satisfies the
conclusion for all $b$.
\end{proof}

\begin{proposition}[Relative $2$-primary torsion]\label{prop:relative-torsion}
Every element of
\[
 J(R)\cap\ker\bigl(\St(R)\longrightarrow\St(\F_2)\bigr)
\]
has finite $2$-power order.
\end{proposition}

\begin{proof}
Let $z$ be such an element. If $b$ is a power of two, then $C_b^b=\Id$
and characteristic two gives $(C_b-\Id)^b=0$. A basis adapted to the
successive kernels of the nilpotent map $C_b-\Id$ makes $C_b$ upper
triangular with all diagonal entries $1$. Thus some
$P\in\GL_b(\F_2)$ has this effect. By Lemma~\ref{lem:block-laws},
conjugation by a Steinberg lift of the repeated basis change takes
$W_z(C_b)$ to $W_z(P^{-1}C_bP)$. Centrality of $J(R)$ makes these
elements equal. Lemma~\ref{lem:triangular} then gives
\begin{equation}\label{eq:cyclic-two-power}
 W_z(C_b)=\prod_{s=0}^{b-1}\iota_s(z)=z^b.
\end{equation}
The last equality uses Lemma~\ref{lem:permutations}: every
coordinate image of the central element $z$ equals $z$ in the stable
group. Choose a power of two $b$ large enough for
Lemma~\ref{lem:cyclic-vanishing}. Then $z^b=1$.
\end{proof}

Only the positive grading and the degree-zero field were used in this
section. Detecting central elements by finite degree truncations needs
the additional phase structure of the specific algebra $R$.
Section~\ref{sec:long-entries} supplies the commutator control the
phases require; Section~\ref{sec:detection} assembles both into an
odd-power counterpart of \eqref{eq:cyclic-two-power}.
==== END An-infinite-finitely-presented-residually-finite-2-group//shifts.tex ====

==== BEGIN An-infinite-finitely-presented-residually-finite-2-group//long-entries.tex ====
\section{Commutators of long entries}\label{sec:long-entries}

The phase property will let us distribute a long word among several
coordinate blocks. Doing this inside the Steinberg group requires
control over the extra factors produced by moving its roots. The main
point of this section is that these factors lie in an abelian
subgroup on which the matrix map is injective. Opposite roots require
particular care: their commutation cannot be inferred from their
matrix images alone.

Fix an odd integer $L$ and a threshold $N\ge1$ from
Lemma~\ref{lem:phases}. Let $k$ bound the lengths of the defining word
rules, and fix
\begin{equation}\label{eq:long-threshold}
 M\ge 2N+k+1.
\end{equation}
All groups in this section are stable groups over $R$. We identify a
nonzero word class with its basis element in $R$, and write $|w|$ for
its length. Write $E_{ij}$ for the matrix unit in position $(i,j)$.
Phases always take values in $\Z/L$.

\subsection*{Opposite roots and movement of a cut}

\begin{lemma}\label{lem:opposite}
Suppose $a,c\in R$ satisfy $ca=0$. For distinct indices $i,h,j$ and
every $b\in R$,
\begin{equation}\label{eq:cut-movement}
 [x_{ij}(ab),x_{ji}(c)]
   =[x_{ih}(a),x_{hi}(bc)].
\end{equation}
In particular, $[x_{ij}(a),x_{ji}(c)]$ is independent of $j\ne i$.
It commutes with both defining roots, has square $1$, and has matrix
image $\Id+acE_{ii}$.
\end{lemma}

\begin{proof}
Apply Lemma~\ref{lem:commutator} to $X=x_{ih}(a)$, $Y=x_{hj}(b)$,
$Z=x_{ji}(c)$. Then $A=[X,Y]=x_{ij}(ab)$ and $B'=[Y,Z]=x_{hi}(bc)$.
The Steinberg relations give $[A,X]=[B',Z]=1$. They also give
$[X,Z]=1$ because $ca=0$, and $[A,B']=1$ because
$(bc)(ab)=b(ca)b=0$. This proves \eqref{eq:cut-movement}.

Taking $b=1$ proves independence of the auxiliary index. Express the
commutator using an index $h$ different from $i,j$. Each of its two
factors then commutes with $x_{ij}(a)$ and with $x_{ji}(c)$: the only
potential nonzero consecutive-root coefficient is $ca$. Thus the
commutator commutes with both defining roots. Since each root has
square $1$, expanding $[x_{ij}(a)^2,x_{ji}(c)]$ shows that its square
is $1$. Finally, multiplication of the four elementary matrices,
using $ca=0$, gives $\Id+acE_{ii}$.
\end{proof}

\begin{lemma}\label{lem:long-commutator}
For each coordinate $i$ and every nonzero word class $w$ satisfying
$|w|\ge M$ and $\sigma(w)\ne\tau(w)$, there is an element
$H_i(w)\in\St(R)$ with the following property. If $p,q$ are nonzero
words with
\[
 |p|,|q|\ge N,\qquad |p|+|q|\ge M,
 \qquad \sigma(p)\ne\tau(q),
\]
then, for every $j\ne i$,
\begin{equation}\label{eq:long-opposite}
 [x_{ij}(p),x_{ji}(q)]
 =\begin{cases}
    H_i(w),& pq\text{ represents the nonzero class }w,\\
    1,& pq=0\text{ in }R.
   \end{cases}
\end{equation}
The element $H_i(w)$ has square $1$ and matrix image $\Id+wE_{ii}$.
\end{lemma}

\begin{proof}
The phase mismatch gives $qp=0$, so Lemma~\ref{lem:opposite}
applies. We first show that the commutator does not change when the
cut between $p$ and $q$ moves, provided both new parts still have
length at least $N$.

For a move to the right, write the original word as $abc$ with $a=p$
and $bc=q$, and with $|c|\ge N$. The word $c$ is nonzero and has
$\tau(c)=\tau(q)$. Hence $ca=0$, and \eqref{eq:cut-movement},
together with auxiliary-index independence, identifies the
commutators for the cuts $a\mid bc$ and $ab\mid c$. This equality
remains valid if $ab=0$, in which case the commutator is $1$.
Otherwise $\sigma(ab)=\sigma(a)$, so the new cut retains the
mismatch. For a move to the left, write $p=ab$ and $q=c$ with
$|a|\ge N$. Now $a$ is nonzero with $\sigma(a)=\sigma(p)$, so $ca=0$
again, and the same identity compares $ab\mid c$ with $a\mid bc$. If
the new suffix is zero we again obtain $1$; otherwise its end phase
is unchanged. This proves invariance under every allowed movement of
the cut. If a part becomes zero, the equalities already obtained show
that the original commutator was $1$, and the argument stops.

Next consider a single defining replacement inside a word of length
$n\ge M$. The allowed cut positions are the integers in $[N,n-N]$. A
rule of length at most $k$ can straddle at most $k-1$ of these
positions, whereas \eqref{eq:long-threshold} supplies more than $k$
allowed positions. We can therefore move the cut so that the entire
replacement lies in one part. That part has the same value in $R$
before and after the replacement. If it is nonzero, its phases are
unchanged because phases depend only on the word class. The other
part is unchanged. The commutator consequently survives the
replacement with the same value and mismatch, unless a zero part has
already proved the original commutator to be $1$.

If the concatenation is nonzero, every part of every equivalent word
is nonzero. Cut movement and the preceding replacement argument show
that the commutator depends only on the class of the concatenation.
Conversely, a nonzero class $w$ as in the statement can be cut into
two parts of lengths at least $N$. The subword assertion of
Lemma~\ref{lem:phases} makes their start and end phases $\sigma(w)$
and $\tau(w)$, respectively. Define $H_i(w)$ by any such cut. The
independence just proved makes this definition unambiguous.

If the concatenation represents zero, the criterion
\eqref{eq:word-zero-criterion} gives a sequence of replacements
ending at a word that contains a zero-declared subword. Follow this
sequence with the commutator. Either a part has already become zero,
or the cut can be moved once more to put the final zero-declared
subword wholly in one part. That part is zero in $R$, and the
commutator is $1$. This proves the zero case as well. The order and
matrix-image assertions follow from Lemma~\ref{lem:opposite}.
\end{proof}

\subsection*{A subgroup detected by its matrix entries}

Let $S$ be a finite set of coordinates, with a phase $P(i)\in\Z/L$
assigned to each $i\in S$. Call an entry $w$ from $i$ to $j$
\emph{consistent} if it is a nonzero word class of length at least $M$
with $\sigma(w)=P(i)$ and $\tau(w)=P(j)$. Call it a \emph{wrong-row
entry} if instead
\begin{equation}\label{eq:wrong-row}
 |w|\ge M,\qquad \sigma(w)\ne P(i),\qquad \tau(w)=P(j).
\end{equation}
For $i\ne j$, a wrong-row entry defines the root $x_{ij}(w)$. For
$i=j$, it defines $H_i(w)$ by Lemma~\ref{lem:long-commutator}. Let
$U(S,P)$ be the subgroup generated by all these elements.

\begin{lemma}\label{lem:wrong-row}
The subgroup $U(S,P)$ is abelian of exponent at most $2$, and the
matrix map is injective on it. Every root with a consistent entry
normalizes $U(S,P)$.
\end{lemma}

\begin{proof}
Write $U=U(S,P)$.

First consider an off-diagonal generator $x_{ij}(a)$ of $U$ and an
off-diagonal root $x_{ef}(b)$ whose entry is either wrong-row or
consistent.

If $j=e$ and $i\ne f$, the commutator is $x_{if}(ab)$. The product is
zero when $b$ is wrong-row, since $\tau(a)=P(j)\ne\sigma(b)$. When
$b$ is consistent and $ab\ne0$, the product inherits $\sigma(a)$ and
$\tau(b)$, so it is a wrong-row entry from $i$ to $f$.

If $i=f$ and $j\ne e$, the reverse product $ba$ is zero, because
$\tau(b)=P(i)\ne\sigma(a)$, so the roots commute.

If the two positions are opposite, so that $e=j$ and $f=i$, the same
mismatch $\sigma(a)\ne\tau(b)$ allows us to use
\eqref{eq:long-opposite}. The commutator is $H_i(ab)$ when $ab\ne0$,
and $1$ when $ab=0$. In the first case $ab$ is a wrong-row entry from
$i$ to itself. In particular the commutator is $1$ whenever $b$ is
wrong-row.

All remaining pairs of positions commute by the defining Steinberg
relations.

Now take a diagonal generator $H_i(w)$. Represent it as a commutator
via an auxiliary coordinate $h\notin S$. To compare it with a root
$x_{ef}(b)$ on $S$, choose $h$ outside its indices as well. If
neither index is $i$, the supports are disjoint and the elements
commute. If one index is $i$, the root lies in a strip between
$\{i,h\}$ and its complement. By Lemma~\ref{lem:strip}, conjugation
by $H_i(w)$ acts on that strip by its matrix $\Id+wE_{ii}$. This
matrix is its own inverse, since $w^2=0$ by the phase mismatch.
Consequently, for $f\ne i$ and $e\ne i$,
\begin{align*}
 H_i(w)x_{if}(b)H_i(w)^{-1}&=x_{if}(b+wb),\\
 H_i(w)x_{ei}(b)H_i(w)^{-1}&=x_{ei}(b+bw).
\end{align*}
The incoming product $bw$ is always zero, because
$\tau(b)=P(i)\ne\sigma(w)$. The outgoing product $wb$ is zero when
$b$ is wrong-row; when $b$ is consistent and $wb\ne0$, it is a
wrong-row entry from $i$ to $f$. These formulas show that diagonal
generators commute with off-diagonal generators of $U$, and that
their commutators with consistent roots belong to $U$.

Two diagonal generators at different coordinates can be represented
using disjoint pairs of coordinates, and hence commute. For two at
the same coordinate $i$, write the second as
\[
 H_i(w')=[x_{it}(a),x_{ti}(c)],\qquad ac=w',\quad |a|,|c|\ge N,
\]
using $t$ distinct from the auxiliary coordinate for the first. Here
$\sigma(a)=\sigma(w')\ne P(i)$ and $\tau(c)=\tau(w')=P(i)$. The phase
rule therefore gives $wa=0$ and $cw=0$. Applying the same strip
action to the two displayed factors shows that $H_i(w)$ fixes both,
and thus commutes with $H_i(w')$. Parts of length at least $N$
suffice for these two applications of the phase rule.

We proved that all generators of $U$ commute. They have square $1$ by
root addition and Lemma~\ref{lem:long-commutator}. Their matrix
images are $\Id+wE_{ij}$, indexed by positions and nonzero word
classes satisfying \eqref{eq:wrong-row}. Any two added matrices
multiply to zero: a potentially nonzero product requires matching
positions, and then the end phase of the first entry is the
designated row phase of the second, which differs from its start
phase. So a product of distinct generators has matrix image
\[
 \Id+\sum_{(i,j,w)}wE_{ij}.
\]
The word classes form a basis of $R$, so this is the identity only
when the sum is empty. This proves injectivity on $U$.

Finally, the commutator calculations show that conjugation by any
consistent root sends every generator of $U$ into $U$. Such a root is
its own inverse, so it normalizes $U$.
\end{proof}
==== END An-infinite-finitely-presented-residually-finite-2-group//long-entries.tex ====

==== BEGIN An-infinite-finitely-presented-residually-finite-2-group//detection.tex ====
\section{Detection by degree truncations}\label{sec:detection}

We now show that degree truncations detect every element of $\St(R)$.
The central kernel is the essential case. For a representative whose
entries are sufficiently long, the phases turn cyclic-shift
evaluation into an odd power of the original element. For an element
invisible in all truncations, Lemma~\ref{lem:cyclic-vanishing} makes
the same evaluation trivial, and the already established 2-power
order forces the element itself to be trivial.

Keep $L,N,M$ as in Section~\ref{sec:long-entries}. For a root word
with entries of length at least $M$, the following lemma places the
start and end phases directly in the coordinate labels.

\begin{lemma}[Relocation by phases]\label{lem:relocation}
Suppose $z\in J(R)$ is represented by a finite word
\begin{equation}\label{eq:high-word}
 z=\prod_{\nu=1}^m x_{i_\nu j_\nu}(w_\nu),
 \qquad |w_\nu|\ge M,
\end{equation}
where every $w_\nu$ is a nonzero word class. Label new coordinates
by $(i,s)$ with $s\in\Z/L$. In the stable group,
\begin{equation}\label{eq:relocation}
 z=\prod_{\nu=1}^m
 x_{(i_\nu,\sigma(w_\nu)),(j_\nu,\tau(w_\nu))}(w_\nu).
\end{equation}
\end{lemma}

\begin{proof}
The empty word presents no issue, so assume $m\ge1$, and let $S$
contain the finitely many original indices and at least three
coordinates. Put $A=\F_2\oplus R_{\ge M}\subseteq R$. We compare
several evaluations of the word \eqref{eq:high-word}.

Choose numerical columns $u_s\in\F_2^L$ and rows $v_s\in(\F_2^L)^*$
such that $v_su_s=1$ for every $s\in\Z/L$. On the word basis of the
tail, set
\begin{equation}\label{eq:phase-rank-one}
 \rho(w)=u_{\sigma(w)}v_{\tau(w)}\,w,
 \qquad \rho(1)=\Id_L,
\end{equation}
and extend linearly to $A$. This is a unital algebra homomorphism
$A\to\Mat_L(R)$. Indeed, a nonzero product $ww'$ has matching middle
phases and inherits the outside phases, so
\[
 \rho(w)\rho(w')
 =u_{\sigma(w)}
   \bigl(v_{\tau(w)}u_{\sigma(w')}\bigr)
   v_{\tau(w')}ww'
 =\rho(ww').
\]
If $ww'=0$, the product of the two entry matrices is zero too,
regardless of the middle numerical pairing. Scalar products are
immediate.

Apply $\rho$ to the entries in \eqref{eq:high-word} and then the
block roots of Lemma~\ref{lem:block-laws}. The result lies in $J(R)$:
the original word is already the identity matrix over the subalgebra
$A$, because its entries belong to $A$ and $A\hookrightarrow R$ is
injective. This uses only the matrix identity; no injectivity
assertion about $\St(A)\to\St(R)$ is needed. The word we have
produced therefore has identity matrix image over $R$, so it lies in
the kernel of the matrix map.

Write $\mathbf e_s$ for the standard numerical columns and
$\mathbf e_s^*$ for their rows. Start with $u_s=\mathbf e_0$ and
$v_s=\mathbf e_0^*$ for every $s$. The evaluated word is $z$ supported
on the coordinates $(i,0)$, which is the same stable element by
Lemma~\ref{lem:permutations} and centrality. We change this
evaluation in three steps.

\emph{Add entries in unused columns.} Keep $u_s=\mathbf e_0$, and
change the rows to
\[
 v_0=\mathbf e_0^*,\qquad
 v_t=\mathbf e_0^*+\mathbf e_t^*\quad(t\ne0).
\]
Every pairing $v_tu_s$ is still $1$. Each block root acquires at most
one additional factor, going from a coordinate $(i,0)$ to an unused
coordinate $(j,t)$ with $t\ne0$. These factors lie in the strip from
$S\times\{0\}$ to $S\times(\Z/L\setminus\{0\})$. The original
factors, supported on $S\times\{0\}$, normalize this strip by
Lemma~\ref{lem:strip}, so we can collect the additional factors into
a single strip element. Both old and new evaluated words have
identity matrix image, so that strip element does too. Strip
injectivity makes it trivial, and the new evaluation is still $z$.

\emph{Change the numerical basis.} The rows $v_t$ just defined form
a basis. Let $Q$ be the numerical matrix with these rows, so that
$v_tQ^{-1}=\mathbf e_t^*$. Also $Q\mathbf e_0=\sum_s\mathbf e_s$,
because $v_t\mathbf e_0=1$ for every $t$. Conjugating all entry
matrices by $Q$ therefore changes the columns and rows to
\[
 u_s=\sum_{r\in\Z/L}\mathbf e_r\quad\text{for every }s,
 \qquad v_t=\mathbf e_t^*.
\]
Apply the exact block-conjugation assertion of
Lemma~\ref{lem:block-laws} with $P=Q^{-1}$. This simultaneous basis
change is induced by conjugation in $\St(R)$. The evaluated word
belongs to the central kernel, so it remains $z$.

\emph{Remove the factors in the wrong rows.} Give coordinate $(i,s)$
the designated phase $P((i,s))=s$. A block root with entry $w$ now
factors as
\[
 \prod_{r\in\Z/L}x_{(i,r),(j,\tau(w))}(w).
\]
Its factors commute. The factor with $r=\sigma(w)$ is the root in
\eqref{eq:relocation}; the others are wrong-row generators of
$U(S\times\Z/L,P)$. Lemma~\ref{lem:wrong-row} says that the
retained, consistent roots normalize this subgroup. Collecting the
other factors expresses the evaluated word as the right side of
\eqref{eq:relocation} times an element $u\in U(S\times\Z/L,P)$. The
retained word itself has identity matrix image: apply
\eqref{eq:phase-rank-one} with $u_s=\mathbf e_s$ and
$v_t=\mathbf e_t^*$. Hence $u$ has identity matrix image. Injectivity
on the wrong-row subgroup gives $u=1$, proving \eqref{eq:relocation}.
\end{proof}

\begin{proposition}\label{prop:odd-power}
Under the hypotheses of Lemma~\ref{lem:relocation},
\begin{equation}\label{eq:odd-power}
 W_z(C_L)=z^L.
\end{equation}
\end{proposition}

\begin{proof}
For a homogeneous word $w$ of degree $d=|w|$, the cyclic-shift
evaluation of $x_{ij}(w)$ is the block root with entry $wC_L^d$.
Since $\tau(w)-\sigma(w)=d$ in $\Z/L$, its factors are
\begin{equation}\label{eq:offset-factors}
 x_{(i,\sigma(w)+a),(j,\tau(w)+a)}(w),
 \qquad a\in\Z/L.
\end{equation}
We claim that any two such factors arising from different offsets
$a\ne a'$ commute, even when they come from different roots of the
original word.

Write their entries as $w,w'$. A match between the column
coordinate of the first factor and the row coordinate of the second
requires $\tau(w)+a=\sigma(w')+a'$. Since $a\ne a'$, this implies
$\tau(w)\ne\sigma(w')$, and hence $ww'=0$. A match in the reverse
direction similarly implies $w'w=0$. Thus the Steinberg relations
prove the claim unless the two roots are opposite. In that case both
products vanish, and the reverse match also gives
$\sigma(w)\ne\tau(w')$. Lemma~\ref{lem:long-commutator}, whose length
hypotheses hold because $|w|,|w'|\ge M\ge N$, then gives
$[x_{uv}(w),x_{vu}(w')]=1$. This proves the claim in every case.

We may consequently collect the product by offsets, preserving
within each offset the order of the original word. The offset-zero
word is precisely the relocated word of Lemma~\ref{lem:relocation},
and every other offset word is its coordinate permutation. They all
give $z$, since $z\in J(R)$. There are $L$ offsets, giving
\eqref{eq:odd-power}.
\end{proof}

It remains to obtain a long-entry representative from triviality in a
degree truncation. Stable coordinates make this possible: a spare
coordinate separates the two factors of a long word while it is being
conjugated.

\begin{lemma}\label{lem:high-entries}
For every integer $M\ge1$, each element of
\[
 \ker\bigl(\St(R)\longrightarrow\St(R/R_{\ge2M})\bigr)
\]
can be represented by a finite product of roots whose entries are
nonzero word classes of lengths at least $M$.
\end{lemma}

\begin{proof}
For any two-sided ideal $K$ of a unital ring, the Steinberg
presentation shows that the kernel of $\St(R)\to\St(R/K)$ is the
normal closure of the roots $x_{ij}(a)$ with $a\in K$. Indeed, after
these roots are killed, root addition makes $x_{ij}(a)$ depend only
on the class of $a$ modulo $K$, and the remaining relations are
exactly the Steinberg relations over $R/K$. For $K=R_{\ge2M}$, root
addition expands every such entry in the homogeneous word basis. It
is therefore enough to handle a conjugate $g x_{ij}(w)g^{-1}$ with
$|w|\ge2M$.

Choose a representative word $w=ab$ with $|a|,|b|\ge M$. Both parts
are nonzero because $w$ is nonzero. Let $S$ contain $i,j$ and every
coordinate used by a fixed root word for $g$, and choose $h\notin S$.
The consecutive-root relation gives
\[
 g x_{ij}(w)g^{-1}
   =[g x_{ih}(a)g^{-1},\,g x_{hj}(b)g^{-1}].
\]
The two conjugated factors lie in the strips from $S$ to $\{h\}$ and
from $\{h\}$ to $S$. Write $A_g=\pi(g)|_S$. By Lemma~\ref{lem:strip},
\begin{align*}
 g x_{ih}(a)g^{-1}
   &=\prod_{r\in S}x_{rh}\bigl((A_g)_{ri}a\bigr),\\
 g x_{hj}(b)g^{-1}
   &=\prod_{s\in S}x_{hs}\bigl(b(A_g^{-1})_{js}\bigr).
\end{align*}
Since $R_{\ge M}$ is a two-sided ideal, every displayed entry
belongs to $R_{\ge M}$. Expand the two strip elements into roots,
expand each entry in the word basis, and then expand their
commutator. Every resulting root entry is either zero or a
concatenation of two entries of length at least $M$, hence a word of
length at least $M$. Doing this for the finitely many conjugates in
a normal-closure expression proves the lemma.
\end{proof}

\begin{theorem}\label{thm:detection}
Degree truncations detect the stable Steinberg group:
\begin{equation}\label{eq:stable-detection}
 \bigcap_{d\ge1}
 \ker\bigl(\St(R)\longrightarrow\St(R/R_{\ge d})\bigr)=1.
\end{equation}
\end{theorem}

\begin{proof}
Let $z$ lie in the displayed intersection. Its matrix image becomes
the identity modulo every $R_{\ge d}$. The grading has
$\bigcap_dR_{\ge d}=0$, so its matrix image is the identity over $R$,
and $z\in J(R)$. Taking $d=1$ also shows that $z$ maps to $1$ in
$\St(\F_2)$.

By Lemma~\ref{lem:cyclic-vanishing}, $W_z(C_b)=1$ for every
sufficiently large integer $b$. Choose an odd phase modulus $L$ from
Lemma~\ref{lem:phases} beyond this threshold. This choice uses a
fixed representative of $z$ and precedes every choice of a long-entry
representative. Now choose its phase threshold $N$ and an integer
$M\ge2N+k+1$.

The element $z$ is trivial in the truncation at degree $2M$.
Lemma~\ref{lem:high-entries} supplies a representative with nonzero
word entries of lengths at least $M$. Applying
Proposition~\ref{prop:odd-power} to this representative yields
\[
 z^L=W_z(C_L)=1.
\]
On the other hand, Proposition~\ref{prop:relative-torsion} gives
$z^{2^a}=1$ for some $a\ge0$, since $z\in J(R)$ has trivial
reduction to $\St(\F_2)$. The integers $L$ and $2^a$ are coprime.
Thus $z=1$.
\end{proof}
==== END An-infinite-finitely-presented-residually-finite-2-group//detection.tex ====

==== BEGIN An-infinite-finitely-presented-residually-finite-2-group//stability.tex ====
\section{Integral stability in a fixed number of coordinates}
\label{sec:stability}

The detection argument used additional coordinates in the stable
Steinberg group. To apply Theorem~\ref{thm:detection} to a finitely
presented group in fixed rank $12$, we must show that adjoining these
coordinates does not kill an element already present. This section
proves that rank $12$ suffices. All homology here has integral
coefficients.

Throughout this section, let $B=\F_2\oplus I_B$ be a unital
associative $\F_2$-algebra whose ideal $I_B$ is matrix-nil. No
grading or finite-generation assumption is needed. We often suppress
$B$ in the notation for its matrix and Steinberg groups.

\begin{proposition}[Finite-rank inputs]\label{prop:finite-rank-input}
For such an algebra $B$:
\begin{enumerate}[label=\textup{(\roman*)}]
\item Every tuple of columns of $B^m$ with linearly independent
reductions in $\F_2^m$ extends to an invertible matrix.
\item The kernel $J_r(B)$ is central in $\St_r(B)$ for $r\ge6$.
\item For $r\ge5$, the group $\E_r(B)$ is normal in $\GL_r(B)$, and
the natural maps
\[
 \GL_r(B)/\E_r(B)\longrightarrow
 \GL_{r+1}(B)/\E_{r+1}(B)
\]
are isomorphisms.
\end{enumerate}
\end{proposition}

These are, respectively, Lemma~5.1, Proposition~5.4, and Lemma~5.5
of \cite{Companion}. They have exactly the hypotheses stated above.
The integral conclusion we need is the following.

\begin{theorem}\label{thm:stability}
Let $B=\F_2\oplus I_B$ satisfy the hypotheses of
Proposition~\ref{prop:finite-rank-input}. For every $n\ge12$, the
standard stabilization map $J_n(B)\to J_{n+1}(B)$ is an isomorphism
and the natural homomorphism $\St_n(B)\to\St(B)$ is injective.
\end{theorem}

The ring $B$ is local: its elements outside $I_B$ are units, whereas
its elements in $I_B$ are nilpotent. The injectivity therefore also
follows, with a better rank bound, from the classical stability
theorem of Suslin and Tulenbaev
\cite[Theorem~4.1]{SuslinTulenbaev1976}. We give a direct proof to
make the integral homology comparison and its natural maps explicit.
It compares successive ranks by their action on ordered frames, in the
classical method of homological stability for linear groups; see van
der Kallen~\cite{vanderKallen1980}. Two features matter here: ordered
(not unordered) frames, so that the chain complex is integral and the
boundary maps have fixed signs; and the identification of the second
homology of stabilizers with universal central extensions, which
supplies a natural way to compare $J_r(B)$ across ranks.

\subsection*{Central extensions of frame stabilizers}

A group is \emph{perfect} if it equals its commutator subgroup. A
\emph{universal central extension} of a perfect group $H$ is a
central extension $\widetilde H\to H$ admitting a unique
homomorphism over $H$ to every central extension of $H$. Its kernel
is naturally $H_2(H;\Z)$; the identification is natural for
homomorphisms lifted between universal central extensions. We use
this standard connection between Steinberg groups and second
homology in its integral form; see
\cite[Theorems~III.5.4 and III.5.5]{WeibelK}.

For $r\ge6$ and $l\ge0$ put
\begin{equation}\label{eq:frame-stabilizer}
 S_{r,l}=
 \left\{\begin{pmatrix}A&0\\ C&\Id_l\end{pmatrix}:
 A\in\E_r(B),\ C\in\Mat_{l\times r}(B)\right\}.
\end{equation}
This is the semidirect product of its additive lower-left strip with
$\E_r(B)$. Replace $\E_r(B)$ in this semidirect product by $\St_r(B)$,
acting through its matrix image, and denote the result
$\widetilde S_{r,l}$. There is a natural surjection
$\widetilde S_{r,l}\to S_{r,l}$ with kernel $J_r(B)$.

\begin{lemma}\label{lem:stabilizer-uce}
The map $\widetilde S_{r,l}\to S_{r,l}$ is a universal central
extension. Consequently, the inclusion of the complementary block
induces a natural isomorphism
\begin{equation}\label{eq:stabilizer-homology}
 H_2(\E_r(B);\Z)\xrightarrow{\ \sim\ }
 H_2(S_{r,l};\Z),
\end{equation}
and both groups identify naturally with $J_r(B)$.
\end{lemma}

\begin{proof}
The extension is central by
Proposition~\ref{prop:finite-rank-input}\textup{(ii)}, because its
kernel acts trivially on the strip. Both source and target are
perfect: an internal root is a consecutive-root commutator, and a
strip root has the same expression using an intermediate index among
the first $r$ coordinates.

Consider an arbitrary central extension $T\to S_{r,l}$. The roots
available in $S_{r,l}$ are $e_{ij}(a)$ with $j\le r$ and $i\ne j$,
where $i$ may also be one of the last $l$ coordinates. Choose
arbitrary lifts $y_{ij}(a)\in T$. We first show that lifts of roots
appearing in a Steinberg commuting relation commute:
\begin{equation}\label{eq:lifts-commute}
 [y_{ij}(a),y_{kt}(b)]=1
 \qquad\text{if }j\ne k\text{ and }i\ne t.
\end{equation}
Choose $h\le r$ distinct from $i,j,k,t$. In matrices
$e_{ij}(a)=[e_{ih}(a),e_{hj}(1)]$, and both factors on the right
commute with $e_{kt}(b)$. In $T$, their commutators with $y_{kt}(b)$
are therefore central. Conjugating each factor by $y_{kt}(b)$ changes
it only by a central factor, so it fixes their commutator. Since that
commutator differs from $y_{ij}(a)$ by a central element,
\eqref{eq:lifts-commute} follows. The bound $r\ge6$ supplies every
spare index used below as well.

Define corrected lifts by
\begin{equation}\label{eq:canonical-root-lift}
 \widehat y_{ij}(a)=[y_{ih}(a),y_{hj}(1)],
 \qquad h\le r,\quad h\ne i,j.
\end{equation}
They are independent of $h$ and satisfy
\begin{equation}\label{eq:lifts-product}
 [y_{ih}(a),y_{hj}(b)]=\widehat y_{ij}(ab)
 \qquad(h\le r;\ i,h,j\text{ distinct}).
\end{equation}
To prove this, choose a fourth distinct index $f\le r$ and apply
Lemma~\ref{lem:commutator} to $X=y_{ih}(a)$, $Y=y_{hf}(b)$,
$Z=y_{fj}(1)$. All its commuting hypotheses follow from
\eqref{eq:lifts-commute}, since $[X,Y]$ and $[Y,Z]$ lift,
respectively, $e_{if}(ab)$ and $e_{hj}(b)$. The resulting identity
equates the left side of \eqref{eq:lifts-product} with
$[y_{if}(ab),y_{fj}(1)]$; replacing a lift by a central multiple does
not change a commutator. Setting $b=1$ proves independence in
\eqref{eq:canonical-root-lift}; the same identity then proves
\eqref{eq:lifts-product} for arbitrary $b$.

The corrected lifts are additive in their parameters. In
\eqref{eq:canonical-root-lift}, use $y_{ih}(a)y_{ih}(a')$ as a lift
of $e_{ih}(a+a')$ and expand its commutator with $y_{hj}(1)$.
Equation~\eqref{eq:lifts-commute} allows the commutator factors to
pass the $ih$ factors, giving
\[
 \widehat y_{ij}(a+a')=\widehat y_{ij}(a)\widehat y_{ij}(a').
\]
Here $\widehat y_{ij}(0)=1$, since any lift of an identity matrix is
central. Together with \eqref{eq:lifts-commute} and
\eqref{eq:lifts-product}, this proves all Steinberg relations for the
corrected lifts on the available roots.

On the first $r$ coordinates they therefore define a homomorphism
$\St_r(B)\to T$. On the strip they define a homomorphism from the
additive group $\Mat_{l\times r}(B)$ to $T$. These homomorphisms
respect the semidirect-product action: conjugation of a strip root by
an internal root is determined by a commuting relation, except when
their indices form a consecutive path, in which case it is determined
by \eqref{eq:lifts-product}. These are exactly the matrix conjugation
rules. So they give a homomorphism $\widetilde S_{r,l}\to T$ over
$S_{r,l}$. It is unique, since two such homomorphisms differ by a
homomorphism from the perfect group $\widetilde S_{r,l}$ to the
central, abelian kernel of $T\to S_{r,l}$. This proves universality.

Taking $l=0$ identifies $H_2(\E_r(B);\Z)$ with $J_r(B)$. The
inclusion $\E_r(B)\to S_{r,l}$ lifts by adjoining the zero strip and
is the identity on the kernels $J_r(B)$. Naturality of the universal
central extension now proves \eqref{eq:stabilizer-homology}.
\end{proof}

\subsection*{The integral complex of ordered frames}

An \emph{ordered frame} of $B^m$ is an ordered tuple of columns whose
reductions modulo $I_B$ are linearly independent. Let $C_p(m)$ be the
free abelian group on ordered frames of length $p+1$, for $p\ge0$,
with boundary the alternating sum of deletions. Set $C_{-1}(m)=\Z$,
generated by the empty frame, and augment by sending each one-column
frame to $1$. Every ordering is a separate generator; we impose no
relation that identifies two orderings by a sign.

The corresponding finite complexes are complexes of injective words
on matroid independence complexes. Their shellability is established
in \cite[Theorem~1.2 and Example~3.3]{JonssonWelker2009}. The
following direct proof gives the precise integral vanishing range we
need.

\begin{lemma}\label{lem:ordered-frames}
The augmented chain complex $C_\bullet(m)$ is exact in degrees at most
$m-2$.
\end{lemma}

\begin{proof}
Let $V$ be a finite labeled collection of nonzero vectors spanning a
vector space of dimension $m$; different labels remain different
vertices even if their vectors agree. Write $C_\bullet(V)$ for the
augmented complex on its ordered independent tuples. We prove the
stated vanishing by induction on the number of labels, allowing
$m=0$: the empty collection has its only chain group in degree $-1$,
and the asserted range is then empty.

Delete a label $v$ and let $V'=V\setminus\{v\}$. In the relative
complex $Q_\bullet=C_\bullet(V)/C_\bullet(V')$, the generators are
the independent tuples containing $v$. Filter $Q_\bullet$ increasingly
by the number $a$ of labels before $v$. Deleting $v$ is zero in the
relative complex; deleting a label before it lowers the filtration;
deleting a label after it preserves the filtration. So the associated
graded complex for fixed $a$ is a direct sum, over independent
ordered prefixes $(u_1,\ldots,u_a,v)$, of complexes in which only the
suffix varies.

For such a prefix, project all remaining labeled vectors to the
quotient by $\langle u_1,\ldots,u_a,v\rangle$ and discard the zero
projections. The projected collection spans dimension $m-a-1$ and has
fewer labels than $V$. Its augmented ordered-tuple complex, shifted up
by $a+1$ degrees, is precisely the corresponding summand of the
associated graded complex, up to the constant boundary sign
$(-1)^{a+1}$. By induction its homology vanishes through degree
\[
 (m-a-1)-2+(a+1)=m-2.
\]
This also covers a zero-dimensional quotient: its sole homology in
degree $-1$ shifts to degree $a=m-1$, outside the asserted range. The
filtration is finite, so $H_i(Q_\bullet)=0$ for $i\le m-2$.

If $V'$ still spans dimension $m$, induction and the long exact
sequence of the pair give the required vanishing for $V$. Otherwise
$V'$ spans dimension $m-1$, and $v$ is independent of the entire span
of $V'$. Prepending $v$ to every tuple of $V'$ gives a chain homotopy
from its inclusion into $C_\bullet(V)$ to zero; in augmented degree
$-1$ it sends $1$ to $(v)$. The induced maps
$H_i(C_\bullet(V'))\to H_i(C_\bullet(V))$ are therefore zero. They are
also surjective for $i\le m-2$, by the relative vanishing just proved.
The homology of $C_\bullet(V)$ vanishes in this range.

Finally, any cycle in $C_\bullet(m)$ uses finitely many columns.
Adjoin the standard basis columns to this finite collection. Their
reductions, with column labels retained, form a finite collection
spanning $\F_2^m$. The finite result supplies a bounding chain in the
asserted range, proving the lemma.
\end{proof}

\subsection*{Second homology in successive ranks}

\begin{proof}[Proof of Theorem~\ref{thm:stability}]
Fix $m\ge13$ and let $\E_m(B)$ act on ordered frames by left
multiplication. For frame lengths $l\le5$, this action is transitive:
complete a given frame to an invertible matrix, placing it in the last
$l$ columns, with $r=m-l$; by
Proposition~\ref{prop:finite-rank-input}\textup{(iii)}, the map
$\GL_r/\E_r\to\GL_m/\E_m$ is an isomorphism, so right multiplication
by a suitable complementary $\GL_r$ block brings the completion into
$\E_m$ without changing its last $l$ columns.

The pointwise stabilizer of the last $l$ standard columns is exactly
$S_{m-l,l}$ from \eqref{eq:frame-stabilizer}. Its general form has
$A\in\GL_{m-l}$ and arbitrary lower-left strip. The strip is
elementary, and the injectivity of
$\GL_{m-l}/\E_{m-l}\to\GL_m/\E_m$ says that this matrix belongs to
$\E_m$ precisely when $A\in\E_{m-l}$. Write $S_l=S_{m-l,l}$. All
these complementary ranks are at least eight, so
Lemma~\ref{lem:stabilizer-uce} applies.

Take a free right $\Z[\E_m]$-resolution $P_\bullet\to\Z$ of the
trivial module and form the first-quadrant double complex
\begin{equation}\label{eq:frames-double-complex}
 D_{p,q}=P_q\otimes_{\Z[\E_m]}C_p(m),\qquad p,q\ge0.
\end{equation}
Its total homology in degree two is $H_2(\E_m;\Z)$ via augmentation;
see \cite[Chapter~VII]{Brown}. For completeness, adjoining the
augmentation column $p=-1$ gives the same tensor product with the
augmented frame complex. Each $P_q$ is free and hence flat, so
Lemma~\ref{lem:ordered-frames} makes the rows exact in the stated
range. The augmented total complex has zero homology in degrees one
and two. The long exact sequence separating its augmentation column
from \eqref{eq:frames-double-complex} consequently gives the asserted
isomorphism.

Take vertical homology first. For $0\le p\le4$, the permutation
module on ordered $(p+1)$-frames has one orbit and stabilizer $S_{p+1}$.
Shapiro's Lemma identifies the first page as
\begin{equation}\label{eq:frames-first-page}
 E^1_{p,q}=H_q(S_{p+1};\Z),\qquad 0\le p\le4.
\end{equation}
The first differential is the alternating sum of the face maps, each
expressed by a stabilizer inclusion and a coordinate change that
returns the remaining frame to the chosen standard frame.

The row $q=1$ vanishes in these columns, because every $S_l$ is
perfect. In row $q=0$, all entries are $\Z$ and each face map is the
identity. The first differential is therefore zero for odd $p$ and
the identity for even $p$. In particular,
\begin{equation}\label{eq:vanishing-second-page}
 E^2_{1,1}=E^2_{2,1}=E^2_{2,0}=E^2_{3,0}=0.
\end{equation}

In row $q=2$, the two face maps from column one to column zero
coincide. Indeed, the inclusion $\E_{m-2}\to S_2$ induces an
isomorphism on $H_2$ by Lemma~\ref{lem:stabilizer-uce}, so it suffices
to compare the faces on this complementary block. The first standard
frame is $(e_{m-1},e_m)$. Deleting its first column gives $e_m$.
Deleting its second column gives $e_{m-1}$, which is returned to $e_m$
by swapping the last two coordinates. This swap is elementary in
characteristic two and centralizes the first $m-2$ coordinates. The
two induced maps on that block are therefore identical. Their
alternating sum vanishes, and $E^2_{0,2}=H_2(S_1;\Z)$.

No higher differential leaves position $(0,2)$. The only possible
incoming differentials are from $(2,1)$ on the second page and $(3,0)$
on the third page; both vanish by \eqref{eq:vanishing-second-page}.
The other positions of total degree two, $(1,1)$ and $(2,0)$, also
vanish. The spectral sequence thus shows that the edge map
$H_2(S_1;\Z)\to H_2(\E_m;\Z)$ is an isomorphism. This is the map
induced by inclusion: the zero-th column corresponds to the standard
column $e_m$, and its map into the total complex followed by
augmentation is the usual homology map for its stabilizer inclusion.

Combining with \eqref{eq:stabilizer-homology}, we have proved that the
natural stabilization map
\[
 H_2(\E_{m-1}(B);\Z)\longrightarrow H_2(\E_m(B);\Z)
\]
is an isomorphism for $m\ge13$. The natural identifications by
universal central extensions turn this into the isomorphism
$J_{m-1}(B)\to J_m(B)$.

Finally, an element killed by $\St_n(B)\to\St(B)$ already has identity
matrix image, since coordinate embeddings of matrix groups are
injective. It therefore lies in $J_n(B)$. If $n\ge12$, the successive
isomorphisms on these kernels show that it cannot become trivial at
any later rank. The stable group is the direct limit, so this proves
the injectivity asserted in Theorem~\ref{thm:stability}.
\end{proof}
==== END An-infinite-finitely-presented-residually-finite-2-group//stability.tex ====