\documentclass[11pt,letterpaper]{article}
\usepackage[T1]{fontenc}
\usepackage[margin=1.05in]{geometry}
\usepackage{amsmath,amsthm,amssymb,mathtools}
\usepackage{microtype}
\usepackage{booktabs,array}
\usepackage{enumitem}
\usepackage{aliascnt}
\usepackage{xcolor}
\definecolor{paperlink}{RGB}{0,0,114}
\usepackage[colorlinks=true,linkcolor=paperlink,citecolor=paperlink,
  urlcolor=paperlink,pdfstartview=Fit]{hyperref}
\usepackage[nameinlink,noabbrev]{cleveref}
\usepackage{url}
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}
\fancyhead[L]{\small The third unit in the adjoint tower}
\fancyhead[R]{\small\thepage}
\setlength{\headheight}{14pt}
\setlength{\parindent}{1.25em}
\setlength{\parskip}{0pt}

\newtheorem{theorem}{Theorem}[section]
\newaliascnt{proposition}{theorem}
\newtheorem{proposition}[proposition]{Proposition}
\aliascntresetthe{proposition}
\newaliascnt{lemma}{theorem}
\newtheorem{lemma}[lemma]{Lemma}
\aliascntresetthe{lemma}
\newaliascnt{corollary}{theorem}
\newtheorem{corollary}[corollary]{Corollary}
\aliascntresetthe{corollary}
\theoremstyle{definition}
\newaliascnt{definition}{theorem}
\newtheorem{definition}[definition]{Definition}
\aliascntresetthe{definition}
\theoremstyle{remark}
\newaliascnt{remark}{theorem}
\newtheorem{remark}[remark]{Remark}
\aliascntresetthe{remark}
\numberwithin{equation}{section}

\newcommand{\Grp}{\mathbf{Grp}}
\newcommand{\Set}{\mathbf{Set}}
\newcommand{\UKei}{\mathbf{UKei}}
\newcommand{\FF}{\mathbb F}
\newcommand{\ZZ}{\mathbb Z}
\newcommand{\Om}{\Omega}
\newcommand{\Ione}{\mathcal I_1}
\newcommand{\Itwo}{\mathcal I_2}
\newcommand{\id}{\operatorname{id}}
\newcommand{\Aut}{\operatorname{Aut}}
\newcommand{\Hom}{\operatorname{Hom}}
\newcommand{\kei}{\mathbin{\triangleright}}
\newcommand{\gen}[1]{\langle #1\rangle}
\newcommand{\lab}[1]{\widehat{#1}}
\newcommand{\E}{\mathcal E}
\crefname{proposition}{Proposition}{Propositions}
\crefname{lemma}{Lemma}{Lemmas}
\crefname{theorem}{Theorem}{Theorems}
\crefname{corollary}{Corollary}{Corollaries}
\crefname{definition}{Definition}{Definitions}
\crefname{remark}{Remark}{Remarks}
\crefname{section}{Section}{Sections}
\crefname{appendix}{Appendix}{Appendices}

\setlist[enumerate]{label=\textup{(\roman*)},leftmargin=2.2em}
\setlist[itemize]{leftmargin=1.7em}
\emergencystretch=1.5em

\title{The third unit in the adjoint tower of group involutions}
\author{Zhenyu Sun\\[.3em]
  \small School of Mathematics and Statistics\\
  \small Northeastern University at Qinhuangdao, China\\
  \small\href{mailto:sunzhenyu@neuq.edu.cn}{\texttt{sunzhenyu@neuq.edu.cn}}}
\hypersetup{
  pdftitle={The third unit in the adjoint tower of group involutions},
  pdfauthor={Zhenyu Sun},
  pdfsubject={A finite counterexample to stabilization at the third adjunction},
  pdfkeywords={adjoint tower, involutory quandle, central extension}
}
\date{}

\begin{document}
\maketitle
\begin{abstract}
The functor $G\mapsto\Hom(C_2,G)$ determines an adjoint tower
through successive Eilenberg--Moore comparison functors. We prove
that the unit of the third adjunction need not be an isomorphism,
answering a question of Bergman. For the generalized dihedral
group $H=(C_4\times C_4)\rtimes C_2$, with inversion action, we
compute $F_3V_3H\cong H\times C_2$, with counit the first
projection. The unit at $V_3H$ is injective, with underlying
domain and codomain of cardinalities $20$ and $40$.
For an arbitrary group $G$, let $E(G)=F_2V_2G$ and let $K_G$
be the kernel of the second counit. We prove that $K_G$ is central
and that $F_3V_3G\cong E(G)/K_G[2]$, where $K_G[2]$ consists of
the elements of $K_G$ whose square is one. The third unit at
$V_3G$ is an isomorphism precisely when $K_G$ has no element of
order four. A central extension of $H$ of order $128$ supplies
such an element and yields the stated description of $F_3V_3H$.
\end{abstract}
\noindent\textit{Keywords.} Adjoint tower, monadic decomposition,
involutory quandle, central extension, generalized dihedral group.

\section{Introduction}

For a group $G$, put
\begin{equation*}
  \Om(G)=\{g\in G:g^2=1\}.
\end{equation*}
We use \emph{involution} for an element of order two, so $\Om(G)$
consists of the identity and the involutions of $G$.
Evaluation at the nonidentity element of $C_2$
identifies $\Om(G)$ with $\Hom_{\Grp}(C_2,G)$. Consider the
functor
\begin{equation*}
  V_1\colon\Grp\longrightarrow\Set,\qquad V_1G=\Om(G),
\end{equation*}
and its left adjoint $F_1$. Form the monad and its
Eilenberg--Moore category
\begin{equation*}
  M_1=V_1F_1,\qquad \Ione=\Set^{M_1},
\end{equation*}
and let $F_2\dashv V_2$ be the adjunction associated with the
comparison functor $V_2\colon\Grp\to\Ione$. Repeating the
construction gives
\begin{equation*}
  M_2=V_2F_2,\qquad \Itwo=\Ione^{M_2},\qquad
  F_3\dashv V_3,\qquad V_3\colon\Grp\to\Itwo.
\end{equation*}
Write $\eta_i$ and $\varepsilon_i$ for the units and
counits. The forgetful functors are
$U_1\colon\Ione\to\Set$ and $U_2\colon\Itwo\to\Ione$.
The left adjoints $F_2$ and $F_3$ exist because $\Grp$ has coequalizers;
the construction is recalled in~\cref{lem:coequalizer}.

The adjoint tower originates in the work of Applegate and
Tierney~\cite{ApplegateTierney1970}. We follow Bergman's indexing~\cite[pp.~290--292]{Bergman1975}; a recent treatment of iterated
comparison constructions is given by Ardizzoni and Menini~\cite{ArdizzoniMenini2022}. In his 1974 ICM lecture, published in
1975, Bergman asked whether
\begin{equation*}
  \eta_{3,A}\colon A\longrightarrow V_3F_3A
\end{equation*}
is an isomorphism for every $A\in\Itwo$~\cite[p.~292]{Bergman1975}. Equivalently, he asked whether
$F_3$ is fully faithful, so that it identifies $\Itwo$ with
a full coreflective subcategory of $\Grp$.

\begin{theorem}\label{thm:main}
Let
\begin{equation*}
  H=(C_4\times C_4)\rtimes C_2,
\end{equation*}
where the generator of $C_2$ acts by inversion. There is an
isomorphism
\begin{equation*}
  F_3V_3H\cong H\times C_2
\end{equation*}
under which the counit $\varepsilon_{3,H}$ is the first
projection. The underlying set map of $\eta_{3,V_3H}$ is
injective, with domain of cardinality $20$ and codomain of
cardinality $40$. Hence $\eta_3$ is not a natural
isomorphism, and $F_3$ is not fully faithful.
\end{theorem}

The proof uses a description of the third unit on comparison
objects. For any group $G$, set
\begin{equation*}
  E(G)=F_2V_2G,\qquad
  \epsilon_G=\varepsilon_{2,G}\colon E(G)\to G,\qquad
  K_G=\ker\epsilon_G.
\end{equation*}
The group $K_G$ is central. For an abelian group $K$, write
$K[m]=\{k\in K:k^m=1\}$. We prove natural isomorphisms
\begin{equation}\label{eq:introquotient}
  F_3V_3G\cong E(G)/K_G[2]
\end{equation}
in $\Grp$, and
\begin{equation}\label{eq:introproduct}
  \Om\bigl(E(G)/K_G[2]\bigr)
  \cong \Om(G)\times K_G[4]/K_G[2]
\end{equation}
in the category of unital keis. The second factor has the
trivial operation $b\kei c=c$ and distinguished element
$0$. In these coordinates $U_2\eta_{3,V_3G}$ is
$t\mapsto(t,0)$. Thus an element of order four in $K_G$ is
exactly the obstruction to the third unit being an isomorphism
at $V_3G$.

To obtain the counterexample, we construct a group $P$ of order
$128$ and a central quotient $P\to H$ with cyclic kernel of order
four. The map on the sets of elements whose square is one has a
section that preserves the identity and conjugation, and whose
image generates $P$. The universal property of $E(H)$ therefore
gives a surjection $E(H)\to P$. Since $H$ has nilpotency class
two, $E(H)$ has exponent dividing four; the chosen generator
of the kernel of $P\to H$ then lifts to an element of order four
in $K_H$. We then prove
$E(H)\cong P\times C_2^3$ over $H$ and compute the quotient in~\eqref{eq:introquotient}.

The first comparison category is the variety of involutory
quandles with a distinguished unit, also called unital keis.
This description occurs in Bergman's discussion of algebraic
structure~\cite[\S10.11]{Bergman2015} and in~\cite[Remark~2.14]{Szymik2026}. For a nonempty set of generators, the free kei is the union
of their conjugacy classes in the corresponding free product
of groups of order two; see~\cite[Corollary~10.3]{Joyce1982} and~\cite[Propositions~2.11 and~2.13]{Szymik2026}. We give the
free-object argument, identify the monad, and derive~\eqref{eq:introquotient} from the universal property of the
comparison adjunction.

The group $E(G)$ also has established descriptions in group
and quandle theory. Its presentation is the active sum of the
cyclic subgroups of order two in $G$~\cite{DiazBarrigaEtAl2004,Romero2015}. On the quandle side, it
is obtained by imposing generator-square relations in an
associated group and then setting the identity label equal to
one. Hasegawa~\cite{Hasegawa2024} studies the former quotient
for involutive quandles, and Akita and Shikata~\cite{AkitaShikata2025} treat it within the theory of symmetric
quandles. Related central-extension descriptions for structure
groups of conjugation quandles appear in~\cite{Lebed2026}.

For the group $H$ in~\cref{thm:main}, the noncentral
involutions form the Takasaki quandle on $(\ZZ/4)^2$, with
operation $x\kei y=2x-y$. It has four orbits. Associated
groups of Alexander quandles were computed by Clauwens~\cite{Clauwens2010} under an invertibility hypothesis and by
Ishikawa~\cite{Ishikawa2025} in general. Together with the
results of Hasegawa and Akita--Shikata, Ishikawa's formula gives
the order $128$ of the involutory associated group in this
example. \Cref{rem:associated} explains this specialization.
The construction below determines the group explicitly,
identifies the kernel of its map to $H$, and calculates the
third unit.

\Cref{sec:kei} identifies $\Ione$ and constructs the enveloping
group. \Cref{sec:third} proves~\eqref{eq:introquotient}--\eqref{eq:introproduct}.
\Cref{sec:extension} constructs a central extension of $H$
and obtains an element of order four in $K_H$.
\Cref{sec:exact} computes $E(H)$, the third counit, and the
third unit. The appendices give coordinates and describe the
finite computation. Throughout, commutators are
$[x,y]=x^{-1}y^{-1}xy$, and kei operations use left translations.

\section{Unital keis and enveloping groups}\label{sec:kei}

Conjugation and the identity element give $\Om(G)$ the structure needed
for the first comparison category. We identify this category with
$\UKei$ and describe the left adjoint $F_2$ to
$V_2\colon\Grp\to\UKei$ by a group presentation.

\subsection{The first comparison category}

\begin{definition}
A \emph{unital kei} is a set $X$ with an element $1_X$ and a
binary operation $\kei$ satisfying
\begin{align}
  x\kei x&=x,&
  x\kei(x\kei y)&=y,\label{eq:kei1}\\
  x\kei(y\kei z)&=(x\kei y)\kei(x\kei z),\label{eq:kei2}\\
  1_X\kei x&=x,&x\kei1_X&=1_X.\label{eq:kei3}
\end{align}
A morphism preserves the operation and the distinguished element.
We denote this category by $\UKei$.
\end{definition}

The operation on $\Om(G)$ is
\begin{equation*}
  s\kei t=sts,
\end{equation*}
with distinguished element $1_G$. The left translation
$L_x(y)=x\kei y$ in a unital kei is an automorphism satisfying
$L_x^2=\id$. It fixes $1_X$, and every automorphism $\alpha$ satisfies
$\alpha L_x\alpha^{-1}=L_{\alpha(x)}$.

\begin{lemma}\label{lem:freekei}
For a set $S$, let
\begin{equation*}
  D(S)=\mathop{*}_{s\in S}C_2
       =\gen{s\ (s\in S)\mid s^2=1}.
\end{equation*}
Then $\Om(D(S))$, with its canonical map from $S$, is the
free unital kei on $S$.
\end{lemma}

\begin{proof}
If $S$ is empty, then $D(S)$ is trivial and $\Om(D(S))$ is the
one-element unital kei, which is initial. Suppose that $S$ is nonempty.
Reduced words in the free product give two facts. Every
element of order two is conjugate to a distinguished generator,
and distinct distinguished generators are not conjugate.
Indeed, a cyclically reduced word of length at least two has
infinite order. Also
\begin{equation*}
  C_{D(S)}(s)=\gen{s}.
\end{equation*}
To see the latter assertion, write a word $w$ as $vs^\epsilon$,
where $v$ is empty or its last letter differs from $s$. Then
$wsw^{-1}=vsv^{-1}$ is reduced of length $2|v|+1$, unless
$v$ is empty. The same argument distinguishes the conjugacy
classes of different generators.

Let $X$ be a unital kei and $f\colon S\to X$ a function.
Since $L_{f(s)}^2=\id$, there is a homomorphism
\begin{equation*}
  \rho\colon D(S)\longrightarrow\Aut(X),\qquad
  s\longmapsto L_{f(s)}.
\end{equation*}
Define
\begin{equation*}
  \widetilde f(1)=1_X,\qquad
  \widetilde f(wsw^{-1})=\rho(w)f(s).
\end{equation*}
If $wsw^{-1}=vtv^{-1}$, then $s=t$ and
$v^{-1}w\in\gen{s}$. As $L_{f(s)}$ fixes $f(s)$, the
definition is independent of the chosen expression.

For $g\in D(S)$ and $q\in\Om(D(S))$, it gives
\begin{equation*}
  \widetilde f(gqg^{-1})=\rho(g)\widetilde f(q).
\end{equation*}
If $p=wsw^{-1}$, then
\begin{equation*}
  \rho(p)=\rho(w)L_{f(s)}\rho(w)^{-1}
         =L_{\widetilde f(p)}.
\end{equation*}
It follows that
$\widetilde f(pqp)=\widetilde f(p)\kei\widetilde f(q)$.
This identity also holds for $p=1$. Thus $\widetilde f$ is a
unital-kei morphism. Every involution of $D(S)$ is
an iterated conjugate of generators, so this extension is unique.
\end{proof}

The free construction also identifies the algebra structure on
$V_2G$. This requires compatibility with the monad multiplication,
as well as the universal property of the free object.

\begin{proposition}\label{prop:Ione}
The monad $M_1=V_1F_1$ is the monad of free unital keis.
Consequently, $\Ione$ is equivalent to $\UKei$, and $V_2G$
is identified with the unital kei $\Om(G)$.
\end{proposition}

\begin{proof}
The first left adjoint is $F_1S=D(S)$, so~\cref{lem:freekei} identifies the underlying free-object functors
and their units. The multiplication of $M_1$ is induced by the
homomorphism
\begin{equation*}
  D(\Om(D(S)))\longrightarrow D(S)
\end{equation*}
which evaluates the generator indexed by $q$ at $q$. Restricting this
homomorphism to elements whose square is one gives free-kei
evaluation.
The identifications therefore preserve monad multiplication.
Algebras for this term monad are exactly the sets with operations
satisfying~\eqref{eq:kei1}--\eqref{eq:kei3}. On the comparison
algebra $V_2G$, evaluation is induced by the group counit.
Its distinguished element is therefore $1_G$, and its
operation is $s\kei t=sts$.
\end{proof}

\subsection{The enveloping group}

For a unital kei $X$, define
\begin{equation}\label{eq:envelope}
  \E(X)=
  \left\langle
     \lab{x}\ (x\in X)\ \middle|\
     \lab{1_X}=1,\quad \lab{x}^{\,2}=1,\quad
     \lab{x\kei y}=\lab{x}\lab{y}\lab{x}
  \right\rangle.
\end{equation}
The hat distinguishes a labelled generator from the element
labelling it.

\begin{proposition}\label{prop:Ftwo}
The functor $\E\colon\UKei\to\Grp$ is left adjoint to
$G\mapsto\Om(G)$. Thus $F_2X=\E(X)$.
\end{proposition}

\begin{proof}
A homomorphism $\E(X)\to G$ is specified by the images of its
labelled generators. The relations in~\eqref{eq:envelope}
say exactly that these images define a unital-kei morphism
$X\to\Om(G)$. This correspondence is natural.
\end{proof}

For a group $G$, we henceforth write
\begin{equation}\label{eq:EG}
  E(G)=
  \left\langle
     \lab{t}\ (t\in\Om(G))\ \middle|\
     \lab{1}=1,\quad\lab{t}^{\,2}=1,\quad
     \lab{sts}=\lab{s}\lab{t}\lab{s}
  \right\rangle.
\end{equation}
The counit $\epsilon_G\colon E(G)\to G$ sends
$\lab{t}$ to $t$. Its image is the subgroup generated by
$\Om(G)$.

\begin{lemma}\label{lem:conjugation}
For every $g\in E(G)$ and $t\in\Om(G)$,
\begin{equation}\label{eq:conjugation}
  g\lab{t}g^{-1}
  =\lab{\epsilon_G(g)t\epsilon_G(g)^{-1}}.
\end{equation}
In particular,
\begin{equation*}
  \epsilon_G^{-1}(Z(G))\subseteq Z(E(G)),
  \qquad K_G\subseteq Z(E(G)).
\end{equation*}
\end{lemma}

\begin{proof}
Equation~\eqref{eq:conjugation} is a defining relation when $g$
is a labelled generator. These generators square to one and
generate $E(G)$, so induction on word length proves the
equation for every $g$. If $\epsilon_G(g)$ is central, the
equation says that $g$ commutes with every generator.
\end{proof}

\section{The third comparison unit}\label{sec:third}

The next comparison adjunction imposes relations between elements of
$E(G)$ whose squares are one and the labelled generators with the same
image in $G$. Centrality of $K_G$ makes these relations explicit.

\subsection{The comparison coequalizer}

Let $L\colon\mathcal C\rightleftarrows\mathcal D\colon R$ be an
adjunction, with $L\dashv R$, and assume that $\mathcal D$
has coequalizers. Denote the unit and counit of the adjunction by
$\eta$ and $\varepsilon$. Its monad is $T=RL$, with multiplication
$\mu=R\varepsilon L$. A $T$-algebra is a pair $(X,a)$ with
$a\colon TX\to X$ satisfying $a\eta_X=\id_X$ and
$a\,Ta=a\mu_X$. A morphism $g\colon(X,a)\to(Y,b)$ satisfies
$ga=b\,Tg$. Let
\begin{equation*}
  J\colon\mathcal D\longrightarrow\mathcal C^T,\qquad
  JD=(RD,R\varepsilon_D)
\end{equation*}
be the comparison functor.

\begin{lemma}\label{lem:coequalizer}
The functor $J$ has a left adjoint $L'$, given on a
$T$-algebra $(X,a)$ by
\begin{equation}\label{eq:comparisoncoeq}
  LTX
  \mathrel{\substack{\xrightarrow{\ La\ }\\[-.5ex]
                     \xrightarrow[\ \varepsilon_{LX}\ ]{}}}
  LX\xrightarrow{q}L'(X,a).
\end{equation}
The underlying $\mathcal C$-map of the unit is
$Rq\circ\eta_X$.
\end{lemma}

\begin{proof}
A morphism $f\colon LX\to D$ factors through the displayed
coequalizer precisely when $fLa=f\varepsilon_{LX}$. Under
adjunction, write $g=Rf\circ\eta_X\colon X\to RD$. The
transposes of these two composites are respectively
\begin{equation*}
  ga,\qquad R\varepsilon_D\circ Tg.
\end{equation*}
For the second equality, the triangle identity gives the transpose
$Rf$, while naturality of $\varepsilon$ gives
\begin{equation*}
  R\varepsilon_D\circ Tg
  =Rf\circ R(\varepsilon_{LX}\circ L\eta_X)=Rf.
\end{equation*}
Equality is exactly the condition that $g$ be a $T$-algebra
morphism $(X,a)\to JD$. This gives the adjunction, and the
transpose of $q$ gives the stated unit.
\end{proof}

\begin{proposition}\label{prop:thirdquotient}
For every group $G$, there is an isomorphism, natural in $G$,
\begin{equation}\label{eq:thirdquotient}
  F_3V_3G\cong E(G)/K_G[2].
\end{equation}
The counit $\varepsilon_{3,G}$ is induced by $\epsilon_G$.
The morphism $U_2\eta_{3,V_3G}$ is given by
\begin{equation}\label{eq:thirdunit}
  \Om(G)\longrightarrow\Om(E(G)/K_G[2]),
  \qquad t\longmapsto\lab{t}K_G[2].
\end{equation}
\end{proposition}

\begin{proof}
Apply~\cref{lem:coequalizer} to $L=F_2$, $R=V_2$, and the
comparison algebra $V_3G=(V_2G,V_2\epsilon_G)$.
The middle group is $E=E(G)$, and the left group is $E(E)$.
On the generator indexed by $u\in\Om(E)$, the two maps take
the respective values
\begin{equation*}
  \lab{\epsilon_G(u)},\qquad u.
\end{equation*}
Thus $F_3V_3G$ is the quotient of $E$ by the normal subgroup
generated by
\begin{equation*}
  u\lab{\epsilon_G(u)}^{-1}\qquad(u\in\Om(E)).
\end{equation*}
Each such element $d$ lies in $K=K_G$, which is central by~\cref{lem:conjugation}. Since
$u=d\lab{\epsilon_G(u)}$ and both $u$ and the labelled
generator square to one, $d^2=1$. All the relations therefore
belong to $K[2]$.

Conversely, every $k\in K[2]$ is in $\Om(E)$, and its
corresponding relation is
\begin{equation*}
  k\lab{\epsilon_G(k)}^{-1}=k\lab{1}^{-1}=k.
\end{equation*}
The normal subgroup is exactly $K[2]$, proving~\eqref{eq:thirdquotient}. The map $\epsilon_G$ coequalizes
the two arrows and is the $F_2\dashv V_2$ transpose of
$\id_{V_2G}$. Its induced factor therefore corresponds,
under the adjunction in~\cref{lem:coequalizer}, to
$\id_{V_3G}$, and is $\varepsilon_{3,G}$. The unit
formula follows from the last assertion of~\cref{lem:coequalizer} and the unit $t\mapsto\lab t$ of
$F_2\dashv V_2$. For a homomorphism $f\colon G\to G'$,
the map $E(f)$ sends $\lab t$ to $\lab{f(t)}$ and maps
$K_G[2]$ into $K_{G'}[2]$. The quotient isomorphism
therefore commutes with the maps induced by $f$.
\end{proof}

\subsection{The obstruction from elements of order four}

Quotienting by $K_G[2]$ can create new elements whose square is one.
To describe them, write a lift of an element of $\Om(G)$ as its
labelled generator times an element of the central kernel.

For an abelian group $K$, the quotient $K[4]/K[2]$ is an
elementary abelian two-group. We regard it as a unital kei with
distinguished element $0=K[2]$ and operation $b\kei c=c$.

\begin{theorem}\label{thm:obstruction}
For every group $G$, the map
\begin{equation}\label{eq:product}
  \Theta_G\colon\Om(G)\times K_G[4]/K_G[2]
   \longrightarrow\Om(E(G)/K_G[2]),
  \qquad
  (t,kK_G[2])\longmapsto\lab t\,kK_G[2]
\end{equation}
is an isomorphism of unital keis, natural in $G$. Under
this isomorphism, $U_2\eta_{3,V_3G}$ is $t\mapsto(t,0)$.
The following conditions are equivalent.
\begin{enumerate}
\item $\eta_{3,V_3G}$ is an isomorphism in $\Itwo$;
\item $K_G[4]=K_G[2]$;
\item $K_G$ has no element of order four.
\end{enumerate}
\end{theorem}

\begin{proof}
Abbreviate $K=K_G$, $E=E(G)$, and $\epsilon=\epsilon_G$.
Centrality gives
\begin{equation*}
  (\lab t\,k)^2=k^2\in K[2]\qquad(k\in K[4]),
\end{equation*}
so the displayed map is well defined. Conversely, if
$uK[2]\in\Om(E/K[2])$, then $u^2\in K[2]$.
Put $t=\epsilon(u)$, which lies in $\Om(G)$, and write
$u=\lab t\,k$ with $k\in K$. Then
\begin{equation*}
  k^2=u^2\in K[2],\qquad k^4=1.
\end{equation*}
This proves surjectivity.

If $\lab s\,kK[2]=\lab t\,\ell K[2]$, applying the induced map
to $G$ gives $s=t$, after which cancellation gives
$kK[2]=\ell K[2]$. Thus the map is bijective.
In the quotient,
\begin{equation*}
  (\lab s\,k)(\lab t\,\ell)(\lab s\,k)
   =\lab s\,\lab t\,\lab s\,k^2\ell
   =\lab{sts}\,\ell.
\end{equation*}
This proves compatibility with the product-kei operation.
The distinguished elements also correspond.

For a homomorphism $f\colon G\to G'$, naturality of the
counit gives $\epsilon_{G'}E(f)=f\epsilon_G$. Thus $E(f)$
induces maps on both $E(G)/K_G[2]$ and $K_G[4]/K_G[2]$.
Naturality of $\Theta_G$ follows from
\begin{equation*}
  E(f)(\lab t\,k)=\lab{f(t)}\,E(f)(k).
\end{equation*}
The formula for the unit follows from~\eqref{eq:thirdunit}. It is bijective precisely when
$K[4]/K[2]$ is trivial. The forgetful functors
$U_2$ and $U_1$ reflect isomorphisms, so bijectivity
of this unit is equivalent to its being an isomorphism in
$\Itwo$. Finally, $K[4]\setminus K[2]$ consists exactly of
the elements of order four.
\end{proof}

\section{A finite central extension}\label{sec:extension}

We construct a central extension of the group in~\cref{thm:main}
whose kernel contains an element of order four. A section of
$\Om(P)\to\Om(H)$ preserving conjugation then yields an element
of order four in $K_H$.

\subsection{The extension and its quotient}

Define $P$ by
\begin{equation}\label{eq:Ppresentation}
\begin{split}
P=\langle x_1,x_2,x_3,x_4,a,b,c\mid {}&
 x_i^2=a^2=b^2=c^2=1,\quad a,b,c\text{ central},\\
&[x_1,x_2]=a,\quad[x_1,x_3]=b,\quad[x_2,x_3]=c,\\
&[x_1,x_4]=ab,\quad[x_2,x_4]=ac,\quad[x_3,x_4]=bc\rangle.
\end{split}
\end{equation}
Since $a,b,c$ are commutators of the $x_i$, the four
involutions $x_1,x_2,x_3,x_4$ generate $P$.

\begin{lemma}\label{lem:P}
The group $P$ has order $128$, and
$W=\gen{a,b,c}\cong C_2^3$. The element
\begin{equation*}
  r=x_1x_2x_3x_4
\end{equation*}
is central of order four, with
\begin{equation*}
  r^2=n:=abc,\qquad \gen r\cap W=\gen n.
\end{equation*}
\end{lemma}

\begin{proof}
The relations put every element into the form
\begin{equation}\label{eq:Pnormal}
  x_1^{v_1}x_2^{v_2}x_3^{v_3}x_4^{v_4}
       a^\alpha b^\beta c^\gamma,
  \qquad v_i,\alpha,\beta,\gamma\in\FF_2.
\end{equation}
To prove that these $128$ forms are distinct, set
$V=\FF_2^4$ and use additive notation for
$W_0=\FF_2^3$, with basis $a,b,c$. Define
\begin{align*}
 b_{12}&=a,& b_{13}&=b,& b_{14}&=a+b,\\
 b_{23}&=c,& b_{24}&=a+c,& b_{34}&=b+c,
\end{align*}
and
\begin{equation*}
  f_P(v,u)=\sum_{i<j}v_ju_i b_{ij}.
\end{equation*}
On $V\times W_0$, put
\begin{equation}\label{eq:Pmodel}
  (v,w)(u,z)=(v+u,w+z+f_P(v,u)).
\end{equation}
Bilinearity of $f_P$ gives
\begin{equation*}
  f_P(v,u)+f_P(v+u,t)=f_P(u,t)+f_P(v,u+t),
\end{equation*}
which proves associativity. The identity is $(0,0)$, and
$(v,w)^{-1}=(v,w+f_P(v,v))$.
The elements $x_i=(e_i,0)$ and the central basis vectors
satisfy~\eqref{eq:Ppresentation}. They generate all of
$V\times W_0$, so the normal forms are distinct.

The product of the commutators of any $x_i$ with the other
three generators is one. Hence $r$ is central. The square
formula in a group of class at most two gives
\begin{equation*}
  r^2=\prod_{i<j}[x_i,x_j]=abc=n\ne1.
\end{equation*}
In the coordinate model, $r=((1,1,1,1),0)$.
Its odd powers are outside $W$, and its even powers are
$1,n$, proving the intersection assertion.
\end{proof}

We now realize the group $H$ of~\cref{thm:main} as a quotient of $P$.
Set
\begin{equation}\label{eq:Hquotient}
  H=P/\gen r,\qquad
  \pi\colon P\longrightarrow H,\qquad h_i=\pi(x_i).
\end{equation}
Bars will denote the images of central elements of $P$, so
$\overline W=W/\gen n$.

\begin{lemma}\label{lem:H}
The group $H$ is isomorphic to
$(C_4\times C_4)\rtimes C_2$, with inversion action. More
precisely, set
\begin{equation*}
  s=h_1,\qquad u=h_2h_1,\qquad v=h_3h_1.
\end{equation*}
Then $\gen{u,v}\cong C_4^2$, and
\begin{equation*}
  sus=u^{-1},\quad svs=v^{-1},\quad
  Z(H)=[H,H]=\overline W=\gen{u^2,v^2}.
\end{equation*}
The set $\Om(H)$ is the disjoint union
\begin{equation}\label{eq:Hinvolutions}
  \Om(H)=\overline W\sqcup
          h_1\overline W\sqcup h_2\overline W
          \sqcup h_3\overline W\sqcup h_4\overline W.
\end{equation}
Each of the last four sets is a conjugacy class of size four.
\end{lemma}

\begin{proof}
By~\cref{lem:P}, $|H|=32$. The relations imply
\begin{equation*}
  u^2=\bar a,\qquad v^2=\bar b,\qquad
  [u,v]=\overline{abc}=1,
\end{equation*}
as well as the two inversion identities. Since
$h_4=(h_1h_2h_3)^{-1}$, the group $H$ is generated by
$s,u,v$. The subgroup $A=\gen{u,v}$ is an abelian quotient
of $C_4^2$ and is normalized by $s$. Hence
\begin{equation*}
  32=|H|\leq 2|A|\leq32.
\end{equation*}
Both inequalities are equalities. Thus $A\cong C_4^2$,
$s\notin A$, and $H=A\rtimes\gen s$.

In this description,
\begin{equation*}
  h_1=s,\qquad h_2=us,\qquad h_3=vs,\qquad h_4=u^{-1}vs.
\end{equation*}
The central subgroup is the two-torsion subgroup
$\gen{u,v}[2]=\gen{u^2,v^2}$, which is also the commutator
subgroup. All sixteen elements of $\gen{u,v}s$ square to one,
whereas exactly four elements of $\gen{u,v}$ do so.
Conjugating a reflection changes its translation coordinate
by an element of $2(\ZZ/4)^2$. The four parity classes give
the four conjugacy classes in~\eqref{eq:Hinvolutions}.
\end{proof}

\subsection{A section preserving conjugation}

Each noncentral conjugacy class in $\Om(H)$ lifts bijectively to a
conjugacy class of involutions in $P$. Choosing central lifts for
the remaining elements gives a section compatible with conjugation.

\begin{proposition}\label{prop:section}
There is a unital-kei morphism
\begin{equation*}
  \sigma\colon\Om(H)\longrightarrow\Om(P)
\end{equation*}
such that $\pi\sigma=\id$ and $\sigma(h_i)=x_i$ for
$1\leq i\leq4$. More generally,
\begin{equation}\label{eq:sectionequivariant}
  \sigma(\pi(p)t\pi(p)^{-1})=p\sigma(t)p^{-1}
  \qquad(p\in P,\ t\in\Om(H)).
\end{equation}
\end{proposition}

\begin{proof}
The four subgroups $[x_i,P]$ are
\begin{equation}\label{eq:planes}
\begin{aligned}
  W_1=[x_1,P]&=\gen{a,b},&
  W_2=[x_2,P]&=\gen{a,c},\\
  W_3=[x_3,P]&=\gen{b,c},&
  W_4=[x_4,P]&=\gen{ab,ac}.
\end{aligned}
\end{equation}
Each has order four and does not contain $n=abc$. Since
$\ker(W\to\overline W)=\gen n$, restriction gives
\begin{equation*}
  W_i\xrightarrow{\sim}\overline W.
\end{equation*}
The conjugacy class $x_iW_i$ consequently maps bijectively
to the conjugacy class $h_i\overline W$.

Define $\sigma$ on $h_i\overline W$ to be the inverse of
this bijection. On $\overline W$, take the unique lift in
$\gen{a,b}$, which is a complement to $\gen n$ in $W$.
All selected elements square to one, $\sigma(1)=1$, and
$\sigma(h_i)=x_i$.

Each class bijection is equivariant for the action of $P$
by conjugation on the source and through $\pi$ on the
target. Its inverse is therefore equivariant. The selected
lifts of central elements are central, so the same statement
holds there. This proves~\eqref{eq:sectionequivariant}.
Taking $p=\sigma(s)$ gives
\begin{equation*}
  \sigma(sts)=\sigma(s)\sigma(t)\sigma(s)
  \qquad(s,t\in\Om(H)),
\end{equation*}
as required.
\end{proof}

\subsection{An element of order four in the kernel}

The section induces a surjection $E(H)\to P$. We first bound the
exponent of $E(H)$ so that a preimage of $r$ has order exactly four.

\begin{lemma}\label{lem:class2}
If $G$ has nilpotency class at most two, then $E(G)$ has
nilpotency class at most two and exponent dividing four.
\end{lemma}

\begin{proof}
For $a,b\in E(G)$,
\begin{equation*}
  \epsilon_G([a,b])\in[G,G]\subseteq Z(G).
\end{equation*}
\Cref{lem:conjugation} gives $[E(G),E(G)]\subseteq Z(E(G))$.
In a group of class at most two, commutators of involutory
generators satisfy
\begin{equation*}
  [a,b]^2=[a^2,b]=1.
\end{equation*}
Hence the commutator subgroup has exponent dividing two.
The square of a word in the involutory generators belongs to
that subgroup, so every fourth power is one.
\end{proof}

Let $E=E(H)$, $K=K_H$, and write
\begin{equation*}
  y_i=\lab{h_i},\qquad k=y_1y_2y_3y_4.
\end{equation*}

\begin{proposition}\label{prop:order4}
The element $k$ belongs to $K$ and has order four.
Consequently, $\eta_{3,V_3H}$ is not an isomorphism.
\end{proposition}

\begin{proof}
The section in~\cref{prop:section} induces a homomorphism
\begin{equation*}
  \Phi\colon E\longrightarrow P,\qquad
  \lab t\longmapsto\sigma(t).
\end{equation*}
Its image contains all four generators $x_i$, so it is
surjective. The identity $\pi\Phi=\epsilon_H$ holds on every
labelled generator. In particular,
\begin{equation*}
  \epsilon_H(k)=h_1h_2h_3h_4=1,\qquad \Phi(k)=r.
\end{equation*}
Hence $k\in K$. The group $H$ has nilpotency class at most
two, so~\cref{lem:class2} gives $k^4=1$. On the other hand,
\begin{equation*}
  \Phi(k^2)=r^2=n\ne1.
\end{equation*}
Thus $k^2\ne1$, and $k$ has order four. \Cref{thm:obstruction} now applies.
\end{proof}

\section{The enveloping group and the third unit}\label{sec:exact}

The homomorphism $\Phi\colon E(H)\to P$ in~\cref{prop:order4} determines a direct-product decomposition
of $E(H)$. Its kernel is generated by three central
involutions arising from the nonidentity elements of $Z(H)$.
This decomposition determines both the third counit and its unit.

\subsection{The enveloping group}

\begin{proposition}\label{prop:exactenvelope}
There is an isomorphism over $H$
\begin{equation}\label{eq:exactenvelope}
  E(H)\cong P\times C_2^3,
\end{equation}
under which $\epsilon_H(p,b)=\pi(p)$. In particular,
\begin{equation*}
  K_H\cong\gen r\times C_2^3,\qquad
  K_H[2]\cong\gen n\times C_2^3.
\end{equation*}
\end{proposition}

\begin{proof}
We first split $\Phi$ and identify a central complement, then prove
that the complement has rank three.
By~\cref{lem:class2}, the group $E=E(H)$ has nilpotency class
at most two. Put
\begin{equation*}
  a_E=[y_1,y_2],\qquad
  b_E=[y_1,y_3],\qquad
  c_E=[y_2,y_3].
\end{equation*}
These elements are central and have square one. The element
$k=y_1y_2y_3y_4$ lies in the central kernel $K_H$.
Expanding $[y_i,k]=1$ gives
\begin{equation*}
  [y_1,y_4]=a_Eb_E,\qquad
  [y_2,y_4]=a_Ec_E,\qquad
  [y_3,y_4]=b_Ec_E.
\end{equation*}
The relations of $P$ therefore define a homomorphism
\begin{equation*}
  f\colon P\to E,\qquad f(x_i)=y_i.
\end{equation*}
The composite $\Phi f$ fixes each $x_i$, so
$\Phi f=\id_P$.

Consider the elements
\begin{equation}\label{eq:extracentral}
  z_a=\lab{\bar a}\,a_E,\qquad
  z_b=\lab{\bar b}\,b_E,\qquad
  z_{ab}=\lab{\overline{ab}}\,a_Eb_E.
\end{equation}
They are central and have square one by~\cref{lem:conjugation}. The choice of $\sigma$ on
$\overline W$ also gives $\Phi(z_a)=\Phi(z_b)=\Phi(z_{ab})=1$.
Thus
\begin{equation*}
  B_0=\gen{z_a,z_b,z_{ab}}\subseteq Z(E)\cap\ker\Phi
\end{equation*}
is an elementary abelian two-group of rank at most three.

Every noncentral element of $\Om(H)$ is conjugate to one of
the $h_i$. Since $h_1,h_2,h_3$ generate $H$,~\eqref{eq:conjugation} places its labelled generator in
$f(P)$. The three remaining nonidentity labels lie in
$f(P)B_0$ by~\eqref{eq:extracentral}. Consequently,
\begin{equation*}
  E=f(P)B_0.
\end{equation*}
The identity $\Phi f=\id_P$ implies
$f(P)\cap B_0=1$. Since $B_0$ is central, this proves
$E\cong P\times B_0$, with $\Phi$ as the first projection.

To determine the rank of $B_0$, let $B=C_2^3$ have basis
$e_a,e_b,e_{ab}$. Define
$\tau\colon\Om(H)\to\Om(P\times B)$ by
$\tau(t)=(\sigma(t),0)$ on the noncentral elements and on
the identity, and by
\begin{equation*}
  \tau(\bar a)=(a,e_a),\qquad
  \tau(\bar b)=(b,e_b),\qquad
  \tau(\overline{ab})=(ab,e_{ab})
\end{equation*}
on the three remaining elements. The added $B$-coordinate
is constant on conjugacy classes. Since $B$ is central of
exponent two, $\tau$ is a unital-kei morphism and induces
\begin{equation*}
  \Psi\colon E\to P\times B.
\end{equation*}
For example,
\begin{equation*}
  \Psi(z_a)=(a,e_a)(a,0)=(1,e_a).
\end{equation*}
The other two generators similarly map to $(1,e_b)$ and
$(1,e_{ab})$. They are therefore independent, and
$B_0\cong C_2^3$.

The identity $\epsilon_H=\pi\Phi$ gives
$\epsilon_H(p,b)=\pi(p)$. Since $\ker\pi=\gen r$ and
$r^2=n$, the two formulas for the kernel follow.
\end{proof}

\subsection{The unit in direct-product coordinates}

\begin{proposition}\label{prop:directproduct}
There is an isomorphism over $H$
\begin{equation*}
  F_3V_3H\cong P/\gen n\cong H\times C_2.
\end{equation*}
Write
\begin{equation*}
  H=\gen{u,v}\rtimes\gen s
\end{equation*}
as in~\cref{lem:H}, and identify $C_2$ with $\FF_2$.
Under this isomorphism, the underlying set map of the third
unit is
\begin{equation}\label{eq:graph}
  t\longmapsto(t,\delta(t)),
\end{equation}
where
\begin{equation}\label{eq:delta}
  \delta(t)=
  \begin{cases}
    1,&t\in h_4\overline W,\\
    0,&t\in\overline W\cup h_1\overline W
                      \cup h_2\overline W\cup h_3\overline W.
  \end{cases}
\end{equation}
Equivalently, for $u^iv^js^\epsilon\in\Om(H)$,
\begin{equation*}
  \delta(u^iv^js^\epsilon)=(i\bmod2)(j\bmod2).
\end{equation*}
\end{proposition}

\begin{proof}
\Cref{prop:thirdquotient,prop:exactenvelope} give
\begin{equation*}
  F_3V_3H
  \cong
  (P\times C_2^3)/(\gen n\times C_2^3)
  \cong P/\gen n.
\end{equation*}
Define $\ell\colon P\to\FF_2$ by
\begin{equation*}
  \ell(x_4)=1,\qquad
  \ell(x_1)=\ell(x_2)=\ell(x_3)=\ell(a)=\ell(b)=\ell(c)=0.
\end{equation*}
The relations show that this is a homomorphism; in~\eqref{eq:Pmodel} it is the fourth coordinate of $v$.
Consider
\begin{equation*}
  (\pi,\ell)\colon P\longrightarrow H\times\FF_2.
\end{equation*}
Its kernel is
\begin{equation*}
  \gen r\cap\ker\ell=\gen{r^2}=\gen n,
\end{equation*}
since $\ell(r)=1$. The elements $x_1,x_2,x_3$ map to
generators of $H\times\{0\}$, while $r$ maps to $(1,1)$.
Thus $(\pi,\ell)$ is surjective, and
\begin{equation*}
  p\gen n\longmapsto(\pi(p),\ell(p))
\end{equation*}
is the required isomorphism over $H$.

Under the isomorphism of~\cref{prop:exactenvelope}, a labelled
generator $\lab t$ has first coordinate $\sigma(t)$.
Its second coordinate in $C_2^3$ is killed on passing to
$F_3V_3H$. Thus~\eqref{eq:thirdunit} becomes
\begin{equation*}
  t\longmapsto(t,\ell(\sigma(t))).
\end{equation*}
The chosen representatives in~\cref{prop:section} have fourth
coordinate one exactly on $h_4\overline W$. This proves~\eqref{eq:graph} and~\eqref{eq:delta}. Since
$h_4=u^{-1}vs$, this class consists of the reflections
with both translation coordinates odd.
\end{proof}

\begin{proof}[Proof of~\cref{thm:main}]
The isomorphism over $H$ and the description of the counit
follow from~\cref{prop:directproduct}. By~\cref{lem:H}, $|\Om(H)|=20$, whereas
\begin{equation*}
  \Om(H\times C_2)=\Om(H)\times C_2
\end{equation*}
has cardinality $40$. Formula~\eqref{eq:graph} is injective,
and the element $(1,1)$ is absent from its image.
The unit is therefore not an isomorphism. A left adjoint is
fully faithful if and only if its unit is an isomorphism,
so $F_3$ is not fully faithful.
\end{proof}

\begin{remark}\label{rem:coordinates}
The group coordinates in~\cref{prop:directproduct} express the
unit as the graph of $\delta$. Identify
$K_H[4]/K_H[2]$ with $\FF_2$ by sending $kK_H[2]$ to
$1$. The natural kei coordinates of~\cref{thm:obstruction} then express the unit as
$t\mapsto(t,0)$. The change from kei coordinates to group
coordinates is
\begin{equation*}
  (t,b)\longmapsto(t,b+\delta(t)).
\end{equation*}
\end{remark}

\subsection{Associated groups and cellularity}

\begin{remark}\label{rem:associated}
The order of $P$ also follows from general results on
associated groups. Let $M=(\ZZ/4)^2$, and let
$Q$ be the kei on $M$ with $x\kei y=2x-y$. Its ordinary
associated group is
\begin{equation*}
  A=\operatorname{As}(Q)
   =\gen{e_x\ (x\in Q)\mid e_{x\kei y}=e_xe_ye_x^{-1}}.
\end{equation*}
Ishikawa uses the right operation $x*y=y\kei x$.
His associated group is identified with $A$ by sending each
labelled generator to its inverse.
Put $R=\operatorname{As}(Q,\id)$, the quotient obtained by
imposing $e_x^2=1$. Identifying $M$ with $\gen{u,v}$, the map
$x\mapsto xs$ identifies $Q$ with the reflections of $H$.
It induces a homomorphism $i\colon R\to E(H)$, given by
$i(e_x)=\lab{xs}$. Define $\rho\colon E(H)\to R$ on labels by
$\rho(\lab{xs})=e_x$ and $\rho(\lab z)=1$ for
$z\in\overline W$. The reflection relations become the
defining relations of $R$, and relations involving a central
label become identities. Thus $\rho$ is well defined and
$\rho i=\id_R$, so $i$ is injective.
The preceding proof identifies $i(R)$ with
$f(P)$, and $\Phi f=\id_P$ identifies this subgroup with
$P$. Hence $R\cong P$.

The kei $Q$ has four orbits, and its inner automorphism
group is generated by the maps $z\mapsto-z+2x$. Thus
$\operatorname{Inn}(Q)\cong C_2^3$. If $\alpha$ is the
abelianization of $A$ and $\psi$ its action on $Q$,
then $\ker(\alpha,\psi)=A'$, because
$\operatorname{Inn}(Q)$ is abelian. Put $T=-\id_M$ and
$N=(1-T)M=2M$. Ishikawa's formula~\cite[arXiv version, Theorem~4.1]{Ishikawa2025} specializes to
\begin{equation*}
  A'\cong(\Lambda^2N)_T\oplus N^T.
\end{equation*}
Here the subscript denotes coinvariants and the superscript denotes
fixed points for the action of $T$. Since $N\cong C_2^2$ and $T$
acts trivially on $N$, this gives
\begin{equation*}
  A'\cong\Lambda^2(C_2^2)\oplus C_2^2\cong C_2^3.
\end{equation*}

The passage from $A$ to $R$ preserves the commutator subgroup and
gives $R_{\mathrm{ab}}\cong C_2^4$~\cite[Lemma~3.5 and Theorem~3.8]{Hasegawa2024};
see also~\cite{AkitaShikata2025}. In this case these facts follow
directly from the presentation. Because $Q$ is a kei, conjugation by
$e_x^2$ fixes every generator of $A$. Thus each $e_x^2$ is central,
and the relation $e_{x\kei y}^2=e_y^2$ makes these squares equal
within each orbit. Let $S$ be the subgroup generated by the four
orbit squares. Abelianizing the presentation gives
$A_{\mathrm{ab}}\cong\ZZ^4$, with one basis vector per orbit.
The images of the four generators of $S$ are twice these basis
vectors, so $S\to A_{\mathrm{ab}}$ is injective and
$S\cap A'=1$. Since $R=A/S$, it follows that $R'\cong A'$ and
$R_{\mathrm{ab}}\cong C_2^4$. Consequently, $|R|=2^3\,2^4=128$.
The presentation~\eqref{eq:Ppresentation} specifies the map to
$H$ and its cyclic kernel $\gen r$, which determine the
third unit.
\end{remark}

\begin{remark}
A group is $C_2$-cellular if it belongs to the smallest class
containing $C_2$ and closed under colimits in $\Grp$.
Romero's theorem on active sums implies that every group
$E(G)$ is $C_2$-cellular~\cite[Definition~1 and Theorem~5]{Romero2015}. Every finite
$2$-group generated by involutions is also $C_2$-cellular~\cite[Corollary~4.4]{FloresMuro2019}.
Thus $H$ is $C_2$-cellular. Its third counit is the projection
$H\times C_2\to H$ computed above.
\end{remark}

\appendix
\section{Coordinate formulas}\label{app:coordinates}

The model~\eqref{eq:Pmodel} uses $v\in\FF_2^4$ and
$w\in W_0=\FF_2^3$. Write $e_1,\ldots,e_4$ for the
standard basis of the first factor and $a,b,c$ for the basis
of $W_0$. The quotient $H$ has a model on
$\FF_2^3\times\FF_2^2$. Write $f_1,f_2,f_3$ for the
standard basis of its first factor and $\bar a,\bar b$ for
the basis of its second factor. These coordinates represent the
normal form
\begin{equation*}
  h_1^{v_1}h_2^{v_2}h_3^{v_3}\bar a^\alpha\bar b^\beta.
\end{equation*}
The commutators $[h_1,h_2]=\bar a$, $[h_1,h_3]=\bar b$ and
$[h_2,h_3]=\bar a\bar b$ give the bilinear map
\begin{equation*}
  f_H(v,u)=v_2u_1\bar a+v_3u_1\bar b
                 +v_3u_2(\bar a+\bar b)
\end{equation*}
and the multiplication
\begin{equation*}
  (v,w)(u,z)=(v+u,w+z+f_H(v,u)).
\end{equation*}
Let $\rho\colon W_0\to\FF_2^2$ be given by
\begin{equation*}
  \rho(a)=\bar a,\qquad \rho(b)=\bar b,\qquad
  \rho(c)=\bar a+\bar b.
\end{equation*}
In $H$ we have $(h_1h_2h_3)^2=\overline{abc}=1$, so
$h_4=h_1h_2h_3$. Substituting this expression for each occurrence
of $x_4$ in the image of~\eqref{eq:Pnormal} and collecting the
first three generators gives the quotient map
\begin{equation}\label{eq:coordinateprojection}
  \pi(v,w)=
  \bigl((v_1,v_2,v_3)+v_4(1,1,1),\
        \rho(w)+v_4(v_2+v_3)\bar a\bigr).
\end{equation}
It sends $x_i$ to $(f_i,0)$ for $i\leq3$, and sends
$x_4$ to $((1,1,1),0)$.

The square of $(v,w)\in P$ is $(0,q(v))$, where
\begin{equation*}
  q(v)=f_P(v,v)=\sum_{i<j}v_iv_jb_{ij}.
\end{equation*}
The value is zero when $v$ has weight zero or one. At
vectors of weight two, the six values are
\begin{equation*}
  a,\quad b,\quad a+b,\quad c,\quad a+c,\quad b+c.
\end{equation*}
At vectors of weight three or four, the value is $a+b+c$.
Consequently,
\begin{equation*}
  \Om(P)=W\sqcup x_1W\sqcup x_2W\sqcup x_3W\sqcup x_4W,
  \qquad |\Om(P)|=40.
\end{equation*}

The section of~\cref{prop:section} is given explicitly by the
following table, where $\alpha,\beta\in\FF_2$.
\begin{center}
\renewcommand{\arraystretch}{1.3}
\begin{tabular}{@{}ll@{}}
\toprule
$t\in\Om(H)$ & $\sigma(t)\in\Om(P)$\\
\midrule
$(0,\alpha\bar a+\beta\bar b)$
  & $(0,\alpha a+\beta b)$\\
$(f_1,\alpha\bar a+\beta\bar b)$
  & $(e_1,\alpha a+\beta b)$\\
$(f_2,\alpha\bar a+\beta\bar b)$
  & $(e_2,(\alpha+\beta)a+\beta c)$\\
$(f_3,\alpha\bar a+\beta\bar b)$
  & $(e_3,(\alpha+\beta)b+\alpha c)$\\
$((1,1,1),\alpha\bar a+\beta\bar b)$
  & $(e_4,\beta a+\alpha b+(\alpha+\beta)c)$\\
\bottomrule
\end{tabular}
\end{center}

\section{Finite computation}\label{app:verification}

The program \texttt{anc/verify.py} implements the coordinate
models over $\FF_2$ using integers and bitwise addition.
It also constructs $H$ independently in generalized dihedral
coordinates over $\ZZ/4$. All finite data are generated from the
displayed formulas, using only Python's standard library. From the
directory containing the manuscript, the command
\begin{verbatim}
python3 anc/verify.py
\end{verbatim}
checks the cocycle identities, the quotient map~\eqref{eq:coordinateprojection}, the central cyclic kernel
$\gen r$, and the section on all of $\Om(H)$. In
particular, it checks all $128\cdot20$ conjugation
equivariance identities and all $20^2$ kei identities.
Using the description $E(H)\cong P\times C_2^3$ proved in~\cref{prop:exactenvelope}, it checks all defining relations for the labelled
generators and verifies that these labels generate the group.
It also checks the quotient and the unit in~\cref{prop:directproduct}.

The finite groups appearing in the calculation have the
following sizes.
\begin{center}
\renewcommand{\arraystretch}{1.2}
\begin{tabular}{@{}lrr@{}}
\toprule
Group & Order & $\lvert\Om(-)\rvert$\\
\midrule
$P$ & $128$ & $40$\\
$H=P/\gen r$ & $32$ & $20$\\
$E(H)\cong P\times C_2^3$ & $1024$ & $320$\\
$K_H\cong C_4\times C_2^3$ & $32$ & $16$\\
$F_3V_3H\cong H\times C_2$ & $64$ & $40$\\
\bottomrule
\end{tabular}
\end{center}
\section*{Generative AI statement}
ChatGPT was used for proofreading and language polishing.

\bibliographystyle{amsplain}
\bibliography{references}
\end{document}