==== 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 ====