text:
\section{Introduction}
This article introduces the concept of hyperreflection groups, which are a fruitful generalization of
Coxeter groups (also called reflection groups). A hyperreflection is a kind of multiple reflection.
In the case of reflection symmetry on a connected space, the fixed points of the reflection separate
the space into two components which are interchanged by the reflection. A hyperreflection is a group
action on a connected space whose fixed points separate the space into many components, and for any two
components there is a unique group element that maps one to the other. A hyperreflection group is
a group that is generated by hyperreflections.

Since hyperreflections need not have order two, they are much more general than reflections.
We will prove that graph products of groups are hyperreflection groups. The graph product is
a very general construction that includes the weak direct product and the free product as special cases.
We will also show that Coxeter groups are hyperreflection groups.

On the other hand, hyperreflection groups are not hopelessly general. They retain many of the
properties of Coxeter groups, and many results in the theory of Coxeter groups can be translated
to this more general setting.

\section{Hypergraphs and Cayley Hypergraphs}

In this section, we state the basic definitions concerning hypergraphs, and we define the Cayley hypergraph.
The  reader who wishes to learn more about hypergraphs is referred to~\cite{Berge89}.
It should be noted that the term ``Cayley hypergraph'' is not standard, but it was named by analogy to Cayley graphs,
which will be discussed in section \ref{Coxeter}.

A \emph{hypergraph} is a pair $(V,E)$ where $E$ is a set of nonempty subsets of $V$.
$V$ is called the vertex set and $E$ is called the edge set. A hypergraph differs from a graph insofar as
an edge of a hypergraph can contain an arbitrary number of vertices, but an edge of a graph always
contains exactly two vertices.
A subgraph of a hypergraph $(V,E)$ is a hypergraph $(V', E')$ such that
$V' \subseteq V$ and $E' \subseteq E$.
A hypergraph $(V,E)$ is \emph{disconnected} if there exists a partition
of $V$ into two nonempty disjoint subsets such that
no edge contains elements from both subsets.
A hypergraph is \emph{connected} if it is not disconnected.
A \emph{component} of a hypergraph is a maximal connected subgraph that contains at least one vertex.
The vertex set of a component will also be called a component.
A disconnected hypergraph is the disjoint union of its components.

A \emph{walk} in $(V,E)$ is an alternating sequence of vertices and edges
$$p = (v_0, e_1, v_1, e_2, \ldots, v_n)$$
such that $\{v_{i-1}, v_i\} \subseteq  e_i$ for all $1 \le i \le n$.
A walk may have repeated vertices or edges.
We say that the $p$ is a walk of length $n$ from $v_0$ to $v_n$. Two vertices $u$
and $v$ belong to the same component if and only if there exists a walk from $u$ to $v$.

If $G$ is a group and $\Sigma$ is a collection of nontrivial subgroups of $G$,
then the \emph{Cayley hypergraph} of $(G,\Sigma)$, denoted $\Cay(G,\Sigma)$, is the hypergraph whose vertex set is $G$
and whose edge set is
$$\{ gS \colon (g,S) \in G\times {\S}\}.$$
The Cayley hypergraph is connected if and only $\bigcup\S$ generates $G$.
$G$ acts on $\Cay(G,\S)$ by left multiplication, and the action is simply transitive on the vertex set.

\section{Hyperreflections}

In this section, we define hyperreflections and introduce their most elementary properties. We will use hyperreflections in
the next section to define hyperreflection groups, which are groups that are generated by hyperreflections.

Let $X$ be a connected space. The term ``space'' is meant to include
graphs, hypergraphs, and topological spaces, but it may refer to any sort of
geometric object for which a notion of connectivity can be meaningfully defined (see~\cite{Muscat}).
If $G$ is a group of automorphisms of $X$ then let
$\Fix(G)$ denote the fixed set of $G$, i.e.\ $$\Fix(G) = \{ x \in X : \forall g \in G,
g x = x\}.$$

A nontrivial subgroup $R$ of $\Aut(X)$ is a called a \emph{hyperreflection} if $R$ acts simply
transitively on the components of $X \setminus \Fix(R)$. This means that $\Fix(R) \ne X$,
and if $C_1$ and $C_2$ are two components of $X \setminus \Fix(R)$ then there is a unique $r
\in R$ such that $r C_1 = C_2$. The fixed sets of hyperreflections are called \emph{walls}.
A hyperreflection of order two is called a \emph{reflection}.

\begin{theorem}
Let $R$ be a hyperreflection on $X$ and let $1 \ne r_1 \in R$. Then $\Fix(R) = \Fix(r_1) \ceq \{x \in
X : r_1 x = x\}$. \label{L1}
\end{theorem}

\begin{proof}
If $x \in \Fix(R)$ then $r x = x$ for all $r \in R$, so in particular $r_1 x = x$.
Now suppose that $x \notin \Fix(R)$, and let $C$ be the component of $X \setminus \Fix(R)$
that contains $x$. Then $r_1C$ is also a component of $X \setminus \Fix(R)$. Since $R$ acts
simply transitively on the set of components, $r_1C \neq C$. Therefore $r_1C \cap C = \emptyset$,
hence $r_1 x \ne x$.
\end{proof}

\begin{theorem}
If $R$ and $S$ are hyperreflections on $X$, and there exists $t \in R \cap S$ with $t\ne1$, then $\Fix(R) = \Fix(S)$.
\label{L2}
\end{theorem}
\begin{proof}
By Theorem~\ref{L1}, $\Fix(R) = \Fix(t) = \Fix(S)$.
\end{proof}

\begin{theorem}
If $R$ is a hyperreflection on $X$ and if $\sigma \in \Aut(X)$ then $R^\sigma \ceq
\sigma R \sigma^{-1}$ is a hyperreflection on $X$. \label{L3}
\end{theorem}
\begin{proof}
Note that $\Fix(R^\sigma) = \sigma \Fix(R)$, since $x \in \Fix(R) \iff R x = x
\iff \sigma R \sigma^{-1} \cdot \sigma x = \sigma x$. Also, $X\setminus\Fix(R^\sigma)$ is disconnected
since $X\setminus\Fix(R^\sigma) = \sigma(X\setminus\Fix(R))$. Let $C_1$ and $C_2$ be components of
$X\setminus\Fix(R^\sigma)$. Then $\sigma^{-1} C_1$ and $\sigma^{-1} C_2$ are components of
$X\setminus\Fix(R^\sigma)$. So, there is a unique $r \in R$ such that $r \cdot \sigma^{-1} C_1 = \sigma^{-1} C_2$.
Let $s = \sigma r \sigma^{-1} \in R^\sigma$. Then $s C_1= C_2$, and $s$ is unique, because $r$ is unique.
Therefore $R^\sigma$ is a hyperreflection on $X$.
\end{proof}

\begin{theorem}
If $R$ and $S$ are hyperreflections on $X$ and if $R \subseteq S$ then $R = S$. \label{L4}
\end{theorem}

\begin{proof}
$\Fix(R) = \Fix(S)$ by Theorem~\ref{L2}. Let $1 \ne s \in S$ and let $C$ be a component
of $X\setminus\Fix(S)$. Then $C$ and $sC$ are distinct components of $X\setminus\Fix(S)$,
so they are also distinct components of $X\setminus\Fix(R)$. Therefore there is a unique element
$r\in R$ such that $rC = sC$. Since $S$ acts freely on the components of $X\setminus\Fix(S)$, it
follows that $r=s$. Therefore $S\subseteq R$, hence $R=S$.
\end{proof}

It is not true that distinct hyperreflections are always disjoint. For example, let $X$ be the union of the coordinate axes
in the plane, viewed as a topological space. Let $R$ be the subgroup of $\Aut(X)$ that is generated by the $90^\circ$
rotation $r(x,y) = (-y,x)$, and let $S$ be the subgroup generated by $s(x,y) = (-2y,x/2)$. Then $R$ and $S$ are both
hyperreflections on $X$, but $r^2 = s^2 \ne 1$.
\section{Hyperreflection Systems}

Let $G$ be a group, let $\Sigma$ be a set of nontrivial subgroups of $G$ whose union generates $G$, and let
$X = \Cay(G, \Sigma)$. We say that $(G, \Sigma)$ is a \emph{hyperreflection system}  if the action of each
element of $\Sigma$ by left multiplication on $X$ is a hyperreflection. If $(G, \Sigma)$ is a hyperreflection
system, then the elements of $\Sigma$ are called \emph{fundamental hyperreflections}.

Note that $(G, \{G\})$ is a hyperreflection system for any nontrivial group $G$. Such a hyperreflection system
is called \emph{trivial}. We say that $G$ is a \emph{hyperreflection group} if there exists a set $\Sigma$
of \emph{proper} subgroups of $G$ such that $(G, \Sigma)$ is a hyperreflection system.

We assume for the remainder of the section that $(G, \Sigma)$ is a hyperreflection system.

\begin{theorem}
A subgroup $T$ of $G$ fixes the edge $gS$ if and only if $T \subseteq S^g$. \label{L5}
\end{theorem}

\begin{proof}
$T\ \text{fixes}\ gS \iff TgS = gS \iff TgSg^{-1} = gSg^{-1} \iff T \subseteq S^g$.
\end{proof}
We will assume for the remainder of this section that $(G,\Sigma)$ is a hyperreflection system.
\begin{theorem}
Let $A, B \in \Sigma$ and let $h,k \in G$. Then either $A^h =
B^k$ or $A^h \cap B^k = \{1\}$. In particular, if $A \ne B$ then $A\cap B\ne\{1\}$.\label{L6}
\end{theorem}

\begin{proof}
Suppose that $1 \ne g \in A^h \cap B^k$.
Then $g$ fixes the edges $hA$ and $kB$ by Theorem~\ref{L5}.
By Theorem~\ref{L3}, $A^h$ and $B^k$ are hyperreflections.
By Theorem~\ref{L2}, $A^h$ fixes the edge $kB$, and $B^k$ fixes the edge $hA$.
Therefore, $A^h \subseteq B^k$ and $B^k \subseteq A^h$ by Theorem~\ref{L5},
hence $A^h = B^k$.
\end{proof}

\begin{theorem}
If $T$ is a subgroup of $G$, and $T$ is a hyperreflection that fixes the edge $gA$, then $T = A^g$.
In particular, $T$ is a hyperreflection if and only if it is a conjugate of a fundamental hyperreflection.
\label{L7}
\end{theorem}

\begin{proof}
$T \subseteq A^g$ by Theorem~\ref{L5}, therefore $T = A^g$ by Theorems~\ref{L3} and \ref{L4}.
Since every hyperreflection fixes an edge, and conjugates of hyperreflections are hyperreflections,
it follows that $T$ is a hyperreflection if and only if it is conjugate to some $A \in \Sigma$.
\end{proof}

\begin{theorem}
If $R, S \in \Sigma$, \ $g,h \in G$, and $\Fix(R^g) = \Fix(S^h)$, then $R^g = S^h$.
\label{L8}
\end{theorem}

\begin{proof}
If $e \in \Fix(R^g) \cap \Fix(S^h)$ then $e = kA$ for some $A \in \Sigma$ and $k \in G$. Therefore $R^g = A^k$
and $S^h = A^k$ by Theorem~\ref{L7}, hence $R^g = S^h$.
\end{proof}

\section{Words and Reduced Words}
\label{hyperwords}

Let $G$ be a group and let $\Sigma$ be a set of nontrivial subgroups of $G$.
A \emph{word} in $(G,\Sigma)$ of length $n$ is a pair of sequences
\begin{eqnarray*}
\bm{s}  & = & ((s_1, \ldots, s_n),\ (S_1, \ldots, S_n))
\end{eqnarray*}
such that $1 \ne s_i \in S_i \in \Sigma$ for all $i$.
The elements $s_i$ are called letters.
If the subgroups in $\Sigma$ are pairwise disjoint then the $S_i$ are determined uniquely
by the $s_i$. In that case, we will call $(s_1, \ldots, s_n)$ a word, since there is no ambiguity.
Recall that if $(G,\Sigma)$ is a hyperreflection system then the elements of $\Sigma$ are disjoint by Theorem~\ref{L6}.

A word determines certain other important sequences, which we will describe.
The word $\bm{s}$ determines a sequence of partial products $(g_0, \ldots, g_n)$ which may be defined recursively as follows:
\begin{eqnarray*}
g_0 & = & 1_G, \\
g_i & = & g_{i-1} s_i\quad \text{for $1\le i\le n$.}
\end{eqnarray*}
The word $\bm{s}$ is said to \emph{represent} $g_n$. Two words are \emph{equivalent} if they represent the same group element. A word is \emph{reduced} if there is no shorter word that represents the same element.
The length of a group element $g$ is denoted $\ell(g)$, and is defined as the length of a reduced word that represents $g$.

The word $\bm{s}$ also determines a \emph{dual word}
\begin{eqnarray*}
\bm{t} = ((t_1, \ldots, t_n),\ (T_1, \ldots, T_n))
\end{eqnarray*}
defined as follows:
\begin{eqnarray*}
t_i & = & g_i g^{-1}_{i-1} = g_{i-1} s_i g^{-1}_{i-1},\\
T_i &=& g_{i-1} S_i g^{-1}_{i-1}.
\end{eqnarray*}
The reader may verify that
$$g_i = s_1 \cdots s_i = t_i \cdots t_1$$
for all $1 \le i \le n$, and that the dual word of $\bm{t}$ is $\bm{s}$.

These definitions are best understood in the context of the Cayley hypergraph $\Cay(G, \Sigma)$. The word $\bm{s}$ corresponds to a walk $(g_0, e_1, g_1, e_2, \ldots, g_n)$
from $1$ to $g_n$, where $e_i = g_{i-1} S_i = g_i S_i$.
Since $g_i = g_{i-1} s_i = t_i g_{i-1}$,
we have two different ways to describe how we move from one vertex $g_{i-1}$ to the next vertex $g_i$.
We can either multiply by $s_i$ on the right, or multiply
by $t_i$ on the left. The subgroup $T_i$ is the stabilizer of the $i$th edge along the walk.

We will maintain these notations for the remainder of the article, so if a word $\bm{s}$ is defined, then we consider the sequences $(S_i)$, $(g_i)$, $(t_i)$, and $(T_i)$
to be defined as well.

\section{The Deletion and Exchange Conditions}
In this section we describe two conditions, called the Deletion and Exchange Conditions,
that are satisfied by any hyperreflection system.
These conditions illustrate the important role of dual words in the reduction of words in hyperreflection systems.
We will assume throughout this section that $(G, \Sigma)$ is a hyperreflection system.

The first theorem shows that any word in a hyperreflection system can be reduced by successive deletion and
replacement of letters. We call this theorem the Deletion Condition, because it generalizes the Deletion Condition
for Coxeter groups~\cite{Humphreys90,Davis08}.

\begin{theorem}
Let $\bm{s} = (s_1, \cdots, s_n)$ be a word representing $g \in G$. Then the following statements hold.
\begin{enumerate}
\item If $t_i = t_j^{-1}$ and $i<j$, then $g = s_1 \cdots \widehat{s_i} \cdots \widehat{s_j} \cdots s_n$,
where the hats indicate that the letters $s_i$ and $s_j$ are to be deleted.
\item If $T_i = T_j$, $t_i \ne t_j^{-1}$, and $i < j$, then there exists $1 \ne \widetilde{s_i}\in S_i$ such that
$g = s_1 \cdots \widetilde{s_i} \cdots \widehat{s_j} \cdots s_n$. In other words, we replace $s_i$ with another
non-identity element of $S_i$, and we delete $s_j$.
\item If $T_i = T_j$, $t_i \ne t_j^{-1}$, and $i < j$, then there exists $1 \ne \widetilde{s_j}\in S_j$ such that
$g = s_1 \cdots \widehat{s_i} \cdots \widetilde{s_j} \cdots s_n$.
\item $\bm{s}$ is reduced if and only if $T_i \ne T_j$ for all $i\ne j$.
\end{enumerate}
\label{L9}
\end{theorem}

\begin{proof}
We begin by observing that $s_1 \cdots s_n = t_j t_i s_1 \cdots \widehat{s_i} \cdots \widehat{s_j} \cdots s_n$.
\begin{eqnarray*}
s_1 \cdots s_n
& = & (s_1 \cdots s_{j-1}) s_j (s_{j+1} \cdots s_n) \\
&= & t_j (s_1 \cdots s_{j-1}) (s_{j+1} \cdots s_n) \\
&=& t_j (s_1 \cdots s_{i-1}) s_i (s_{i+1} \cdots s_{j-1}) (s_{j+1} \cdots s_n) \\
&=& t_j t_i (s_1 \cdots s_{i-1}) (s_{i+1} \cdots s_{j-1}) (s_{j+1} \cdots s_n) \\
&=& t_j t_i s_1 \cdots \widehat{s_i} \cdots \widehat{s_j} \cdots s_n. \label{del_eq}
\end{eqnarray*}

If $t_i = t_j^{-1}$ then the last expression is equal to $s_1 \cdots \widehat{s_i} \cdots \widehat{s_j} \cdots s_n$,
which proves (1).

Suppose that $T_i = T_j$ and $t_i \ne t_j^{-1}$. Then
$t_j t_i g_{i-1} S_i \in T_i g_{i-1} S_i = g_{i-1} S_i$,
so there exists $1\ne\widetilde{s_i} \in S_i$ such that $t_j t_i g_{i-1} = g_{i-1} \widetilde{s_i}$.
Consequently, $s_1 \cdots s_n = s_1 \cdots \widetilde{s_i} \cdots \widehat{s_j} \cdots s_n$, which proves (2).

One can prove part (3) by applying part (2) to the inverse word $\bm{s}^{-1} = (s_n^{-1}, \ldots, s_1^{-1})$.
If $\bm{s}$ is reduced then $T_i \ne T_j$ for all $i\ne s$, since otherwise we could reduce the word length
by applying (2) or (3). The converse is proved in the next theorem.
\end{proof}

\begin{theorem}
Let $\bm{s} = (s_1, \cdots, s_m)$ and $\bm{s'} = (s'_1, \cdots, s_n)$ be two words representing $g$.
\begin{enumerate}
\item If $T_i \ne T_j$ for all $i \ne j$ then $\{T_1, \cdots, T_m\}\subseteq\{T'_1,\cdots,T'_n\}$
\item If $T_i \ne T_j$ for all $i \ne j$ then $\bm{s}$ is reduced.
\item If $\bm{s}$ and $\bm{s'}$ are both reduced, then $m = n$ and $\{T_1, \cdots, T_m\}=\{T'_1,\cdots,T'_n\}$
\end{enumerate}
\label{L10}
\end{theorem}

\begin{proof}
Suppose that $T_i \ne T_j$ for all $i < j$. The word $\bm{s}$ corresponds
to a walk $(g_0, e_1, g_1, \ldots, g_n)$ in $\Cay(G, \Sigma)$, and this walk crosses each wall
$\Fix(T_i)$ exactly once. Since these walls separate $1$ and $g$, they must be crossed by every walk
from $1$ to $g$. Therefore $T_i \in \{T'_1,\cdots,T'_n\}$ for all $i$, which implies that
$\{T_1, \cdots, T_m\}\subseteq\{T'_1,\cdots,T'_n\}$. This also implies that $m \le n$, which proves (2).

If $\bm{s}$ and $\bm{s'}$ are both reduced, then $T_i \ne T_j$ for all $i<j$, and $T'_i \ne T'_j$ for all $i<j$.
Part (1) implies that $\{T_1, \cdots, T_m\}\subseteq\{T'_1,\cdots,T'_n\}$ and
$\{T'_1, \cdots, T'_n\}\subseteq\{T_1,\cdots,T_m\}$, hence the two sets are equal.
\end{proof}

\begin{theorem}
If $(s_1, \ldots, s_n)$ and $(s'_1, \ldots, s'_n)$ are two reduced words representing $g$
then $\{t_1, \ldots, t_n\} = \{t'_1, \ldots, t'_n\}$.
\label{L11}
\end{theorem}
\begin{proof}
Let $i \in \{1,2,\ldots,n\}$.
The walk $(g_0, e_0, g_1, \ldots, g_n)$ crosses each wall $\Fix(T_i)$ exactly once, since $T_i \ne T_j$ for $i \ne j$
by Theorem~\ref{L9}.
The walk $(g'_0, g'_1, \ldots, g'_n)$ must also cross each wall $\Fix(T_i)$ exactly once, since $T'_i \ne T'_j$ for $i \ne j$.
Both walks travel from $C_i$ to $t_i C_i$, so there exists $j$ such that $g'_{j-1} \in C_i$ and $g'_j \in t_i C_i$,
which implies that $t'_j = t_i$.

Since $i$ was arbitrary, it follows that $\{t_1, \ldots, t_n\} \subseteq \{t'_1, \ldots, t'_n\}$, therefore
$\{t_1, \ldots, t_n\} = \{t'_1, \ldots, t'_n\}$ by symmetry.
\end{proof}

\begin{theorem}
If $g \in G$ and $S\in\Sigma$ then the right coset $Sg$ has a unique element of minimal length,
and the left coset $gS$ also has a unique element of minimal length. (cf.\ Theorem~\ref{L19}.)
\label{L12}
\end{theorem}

\begin{proof}
Let $w$ and $w'$ be two distinct elements of minimal length in $Sg$, and choose a reduced word
$(s_1, \ldots, s_n)$ representing $w$. Then there exists $1\ne s\in S$ such that
$w' = sw = ss_1\ldots s_n$.

The word $(s,s_1,\ldots, s_n)$ is not reduced, so we can reduce the length via deletion. The deletion must involve
the first letter $s$, since $(s_1, \ldots, s_n)$ is reduced. Therefore, there exists $1 \le i \le n$ and $s'\in S$
such that $w' = s' s_1 \ldots \widehat{s_i} \ldots s_n$. But $(s')^{-1} w' \in Sg$ and $\ell((s')^{-1} w')  < n$, which
is a contradiction. This proves that $Sg$ has a unique element of minimal length.

Note that if $w$ is the unique element of minimal length in $Sg$ then $w^{-1}$ is the unique element
of minimal length in $g^{-1}S$, and so every left coset of a fundamental hyperreflection has a unique
element of minimal length.
\end{proof}

We conclude the section with a theorem that we call the Exchange Condition, which is a generalization of the
Exchange Condition for Coxeter groups~\cite{Davis08,Humphreys90}.
\begin{theorem}
Let $\bm{s} = (s_1, \ldots, s_n)$ be a reduced word for $g$, and let $1 \ne s_0 \in S_o \in \Sigma$.
Then $|\ell(s_0^{-1}g) - \ell(g)| \le 1$, and the following statements hold.
\begin{enumerate}
\item If $\ell(s_0^{-1} g) = \ell(g) - 1$ then there exists $1 \le i \le n$ such that
$g = s_0 s_1 \ldots \widehat{s_i} \ldots s_n$.
\item If $\ell(s_0^{-1} g) = \ell(g)$ then there exists $1 \le i \le n$ and $\widetilde{s_i} \in S_i\setminus\{1,s_i\}$
such that $g = s_0 s_1 \ldots \widetilde{s_i} \ldots s_n$.
\item If $\ell(s_0^{-1} g) = \ell(g)+1$ then no reduced word for $g$ begins with an element of $S_0$.
\label{L13}
\end{enumerate}
\end{theorem}

\begin{proof}
$|\ell(s^{-1}g) - \ell(g)| \le 1$ is a consequence of the triangle inequality for word length.

Suppose that $\ell(s_0^{-1} g) = \ell(g) - 1$.  Note that $s_0^{-1} g$ is the unique element of minimal length
in $S_0g$. The word $(s_0^{-1}, s_1, \ldots, s_n)$ for $s_0^{-1} g$
is not reduced, so it can be reduced using the Deletion Condition. Since $(s_1, \ldots, s_n)$ is reduced, the
deletion must involve the first letter. Therefore, there exists $1 \le i \le n$ and $s \in S_0$ such that
$s_0^{-1} g = ss_1\cdots\widehat{s_i}\cdots s_n$, hence $g = (s_0s) s_1 \cdots \widehat{s_i} \ldots s_n$.

Let $s' = (s_0 s)^{-1}$. Then $s'g \in S_0g$ and $\ell(s'g) \le n-1$, so $s'g$ has minimal length in $S_0g$.
Therefore $s' = s_0^{-1}$, which implies $s = 1$.
Thus, $g = s_1 \cdots \widehat{s_i} \cdots s_n$, which proves part (1).

We prove part (2) by applying the Deletion Condition to $s_0^{-1} g = s_0^{-1} s_1 \cdots s_n$.
As before, the deletion must involve the first letter, so $s_0^{-1} g = s_1 \cdots \widetilde{s_i} \cdots s_n$
for some $\widetilde{s_i} \in S_i$. Therefore,  $g = s_0 s_1 \cdots \widetilde{s_i} \cdots s_n$. But
$\widetilde{s_i} \ne 1$ because $\ell(g) = n$, and $\widetilde{s_i} \ne s_i$ because $s_0 \ne 1$.

If $g$ has a reduced word of length $n$ that begins with $s \in S_0$, then
$s_0^{-1} g$ has a word of length $n$ that begins with $s_0^{-1} s$, which proves part (3).

\end{proof}

\section{Special Subgroups and Sectors}
Let $(G, \Sigma)$ be a hyperreflection system. If $\A \subseteq \S$ then let $G_{\A}$ denote the
subgroup of $G$ that is generated by $\bigcup \A$.  If $\A = \{A\}$ then we will sometimes
write $G_A$ instead of $G_{\{A\}}$. We define $G_\emptyset$ to
be the identity subgroup. A subgroup of the form $G_{\A}$ is called a \emph{special subgroup}.

\begin{theorem}
If $\A\subseteq\S$,  $R\in\S$, and $G_{\A}\cap R \ne \{1\}$ then $R \in \A$.
\label{L14}
\end{theorem}

\begin{proof}
Let $1 \ne g \in R \subseteq G_{\A}$, and let $(s_1, \ldots, s_n)$ be a word of minimal length such that
$1 \ne s_i \in S_i \in \A$ and $g = s_1 \cdots s_n$.

We claim that $(s_1, \ldots, s_n)$ is reduced. Suppose not; then by Theorem~\ref{L9} we can reduce the word
to a shorter word by successive deletions. But each letter in the shorter word also belongs to
$\bigcup\A$, and this contradicts minimality.
The shortest possible word for $g$ has length 1, since $g \in R \in \Sigma$.
Therefore, $g \in R \cap S_1$, so $R = S_1 \in \A$ by Theorem~\ref{L6}.
\end{proof}

\begin{theorem}
If $(s_1, \ldots, s_n)$ and $(s'_1, \ldots, s'_n)$ are two reduced words for $g$ then
$\{S_1, \ldots, S_n\} = \{S'_1, \ldots, S'_n\}$.
\label{L15}
\end{theorem}

\begin{proof}
Let $\A = \{S_1, \ldots, S_n\}$ and $\B = \{S'_1, \ldots, S'_n\}$. Then $t_i \in G_{\A}$ for all $i$, so
$t'_i \in G_{\A}$ for all $i$ by Theorem~\ref{L11}. Since $w'_i = t'_i \cdots t'_1$ for all $i$, it follows
that $w'_i \in G_{\A}$ for all $i$. But $w'_i = w'_{i-1} s'_i$, thus $s'_i \in G_{\A}$ for all $i$.
Therefore, $S'_i \cap G_{\A} \ne \{1\}$ for all $i$, hence $S'_i \in \A$ for all $i$ by Theorem~\ref{L14}.
This shows that $\B \subseteq \A$. By symmetry, $\A \subseteq \B$, therefore $\A = \B$.
\end{proof}

\begin{theorem}
If $(s_1, \cdots, s_m)$ is a reduced word for $g$, and $(s'_1, \cdots, s'_n)$ is any other word representing $g$,
not necessarily reduced, then $\{S_1, \cdots, S_m\}\subseteq\{S'_1,\cdots,S'_n\}$.
\label{L16}

\end{theorem}

\begin{proof}
If $(s'_1, \cdots, s'_n)$ is not reduced then it may be transformed to an equivalent reduced word
$(s''_1, \ldots, s''_p)$ by successive deletions (Theorem~\ref{L9}).
It is clear that $\{S''_1, \cdots, S''_p\} \subseteq \{S'_1, \cdots, S'_n\}$. By Theorem~\ref{L15},
$\{S''_1, \cdots, S''_p\} = \{S_1, \cdots, S_m\}$. Therefore, $\{S_1, \cdots, S_m\}\subseteq\{S'_1,\cdots,S'_n\}$.
\end{proof}

\begin{theorem}
If $\A\subseteq\S$ then $(G_{\A},\A)$ is a hyperreflection system.
\label{L17}
\end{theorem}
\begin{proof}
Let $X_0 = \Cay(G_{\A}, \A)$ and $X = \Cay(G, \Sigma)$. Note that $X_0$ is a connected subgraph of $X$.
Let $A\in\A$ be given. Let $C$ be the identity component of $X\setminus\Fix(A)$ and let $C_0 = C \cap X_0$.

Now $X = \Fix(A) \cup \bigcup_{a\in A} aC$, so $X_0 = (\Fix(A)\cap X_0) \cup \bigcup_{a\in A} (aC \cap X_0)
= (\Fix(A)\cap X_0) \cup \bigcup_{a\in A} aC_0$. Moreover, $\Fix(A) \cap X_0$ is identical to the fixed set of the
action of $A$ on $X_0$, since $X_0$ is invariant under $A$.
Since the unions in the expression of $X$ are disjoint, the unions in the
expression of $X_1$ are also disjoint. So it only remains to show that $C_0$ is connected.

Let $g$ be any vertex of $C_0$. This implies that $g \in G_{\A}$, and there exists a word
$(s_1, \ldots, s_m)$ representing $g$ whose path does not cross $\Fix(A)$, i.e.\ $T_i \ne A$ for all $1\le i \le n$.
Let $(s'_1, \ldots, s'_n)$ be a reduced word representing $g$.
Then $S'_i \in \A$ for all $i$ by Theorem~\ref{L16}, and $T'_i \ne A$ for all $i$ by Theorem~\ref{L10}.
Therefore, the walk $(g'_0, e'_1, g'_1, \ldots, g'_n)$ lies in $C_0$, hence $C_0$ is connected.
\end{proof}

\begin{theorem}
If $\A, \B \subseteq \S$ then $G_{\A\cap\B} = G_{\A}\cap G_{\B}$.
\label{L18}
\end{theorem}
\begin{proof}
Since $\A\cap\B \subseteq \A$ and $\A\cap\B \subseteq \B$ it follows that $G_{\A\cap\B} \subseteq G_{\A}$ and
$G_{\A\cap\B} \subseteq G_{\B}$. Therefore, $G_{\A\cap\B} \subseteq G_{\A} \cap G_{\B}$.

To prove the reverse inclusion, let $g \in G_{\A} \cap G_{\B}$, and let $(s_1, \ldots, s_n)$ be a reduced word for $g$.
Then $S_i \in \A$ for all $i$ by Theorem~\ref{L16}, and $S_i \in \B$ for the same reason. Therefore, $S_i \in \A\cap\B$ for all $i$,
hence $g\in G_{\A\cap\B}$.
\end{proof}

\begin{theorem}
\emph{(cf.\ \cite[p. 47]{Davis08}).}
Suppose that $\A, \B \subseteq \S$ and that $w$ is an element of minimum length in the double coset $G_{\A} w G_{\B}$.
Then any element $w'$ in this double coset can be written in the form $w' = awb$ where $a \in G_{\A}$, $b \in G_{\B}$,
and $\ell(w') = \ell(a) + \ell(w) + \ell(b)$. In particular, the double coset has a unique element of minimum length.
\label{L19}
\end{theorem}

\begin{proof}
Choose $a\in G_{\A}$ and $b \in G_{\B}$ such that $w' = awb$ and $\ell(a)+\ell(b)$ is as small as possible. Choose reduced words
$\bm{r}$, $\bm{s}$, and $\bm{t}$ for $a$, $w$, and $b$, respectively. If the concatenation $\bm{rst}$ is not reduced, then we can produce an equivalent word of shorter length by deleting two letters, or by deleting one letter and replacing another letter.
The two letters cannot occur in the same subword ($\bm{r}$, $\bm{s}$, or $\bm{t}$), because these subwords are reduced. There are three cases to consider:
\begin{enumerate}
\item Delete a letter from $\bm{s}$, and delete or replace a letter from $\bm{r}$.
\item Delete a letter from $\bm{s}$, and delete or replace a letter from $\bm{t}$.
\item Delete a letter from $\bm{r}$, and delete or replace a letter from $\bm{t}$.
\end{enumerate}
Note that deleting or replacing a letter from $\bm{r}$ or $\bm{t}$ yields another element of the same special subgroup. Thus, the first two cases are impossible as they would yield an element of the double coset that is shorter than $w$, and the third case is impossible because it contradicts the minimality of $\ell(a)+\ell(b)$. Therefore, $\bm{rst}$ is reduced, hence
$$\ell(w') = \ell(awb) = \ell(a) + \ell(w) + \ell(b).$$
If $w$ and $w'$ both have minimal length in the double coset, then $\ell(a) = \ell(b) = 0$, thus $w' = w$.
\end{proof}

Given a subset $\A$ of $\S$, let $G^{\A}$ be the set of all $g \in G$ such that $g$ is the unique
element of minimal length in the coset $gG_{\A}$.  $G^{\A}$ is called the \emph{fundamental $\A$-sector}.

\begin{theorem}
If $g \in G$ and $\A \subseteq \Sigma$ then there exist unique elements $h \in G^{\A}$ and $k \in G_{\A}$
such that $g = hk$.
\label{L20}
\end{theorem}

\begin{proof}
Let $h$ be the unique minimal element $gG_{\A}$. Then $h \in G^{\A}$ because $gG_{\A} = hG_{\A}$.
Since $g \in hG_{\A}$, there exists a unique $k \in G_{\A}$ such that $h = hk$.
\end{proof}

\section{Coxeter Groups as Hyperreflection Systems}
\label{Coxeter}

A Coxeter group is a group having a presentation of the form
    $$W = \langle S\ |\ (st)^{m(s,t)} = 1\ (s,t\in S) \rangle$$
where $S$ is a finite set of generators of $W$, $m(s,s)=1$ for all
$s\in S$, and $m(s,t)=m(t,s)\in \{2,3,4,...,\infty \}$ for all
$s,t\in S$ with $s\neq t$. If $m(s,t)=\infty$ then the corresponding
relation is omitted. It can be shown that $m(s,t)$
is the order of $st$ in $W$~\cite[p.\ 110]{Humphreys90}.
The pair $(W,S)$ is called a Coxeter system, and $S$ is a set of Coxeter
generators for $W$.

Coxeter groups appear in nature as the symmetry groups of regular polytopes, and they are important in
the theory of Lie algebras, where they arise as subgroups of the isometry groups of root systems~\cite{Humphreys78}.
The reader who wishes to learn more about Coxeter groups is referred to~\cite{Davis08} or~\cite{Humphreys90}.

There is an alternative characterization of Coxeter groups, due to Michael Davis. He proves in~\cite[Thm 3.3.4]{Davis08}
that $(W,S)$ is a Coxeter system if and only if the Cayley graph of $(W,S)$ is a reflection system.
The Cayley graph $\Cay(W,S)$ is defined as the graph whose vertex set is $W$ and whose edge set is $\{\{w, ws\} :
w \in W, s \in S\}$. Observe that $\{w,ws\} = w\langle s \rangle$, so we may identify $\Cay(W,S)$ with
$\Cay(W,\Sigma)$ where $\Sigma = \{\langle s \rangle : s \in S\}$. The definition of a reflection system is rather involved,
but in the case of a Cayley graph it reduces to the assertion that each element of $S$ acts by reflection on the Cayley graph.
Therefore, if $(W,S)$ is a Coxeter system then $(W, \Sigma)$ is a hyperreflection system.

\section{Graph Products of Groups}

\emph{The next two sections incorporate material from a preprint by the present author~\cite{Radcliffe02}.}

Given a graph with nontrivial groups as vertices, a group is formed by taking the free
product of the vertex groups, with added relations implying that elements of adjacent groups
commute. This group is said to be the \emph{graph product} of the vertex groups. If the graph is
discrete then the graph product is the free product of the vertex groups; while if the graph is
complete then the graph product is the weak direct product of the vertex groups. See~\cite{hungerford}
for the definitions of the free product and weak direct product of groups. Graph products
were first defined in Elizabeth Green's Ph.D. thesis~\cite{green}, and have been studied by many other
authors~\cite{hermiller-meier, LohSen, chiswell}. In this section, we will characterize graph products of
groups by a universal mapping property, and we will present two constructions of the graph product.

Let $\Gamma = (V,E)$ be a graph, and let $\{G_v\}_{v\in V}$ be a collection of groups which is
indexed by the vertex set of $\Gamma$. We say that $(\Gamma, \{G_v\}_{v\in V})$ is a \emph{graph of
groups.} (This differs from the usual definition, which has vertex groups and edge groups,
together with monomorphisms from the edge groups to the vertex groups. See~\cite{baumslag}.)

A \emph{graph product} of a graph of groups consists of a group $G$ and a collection of
homomorphisms $e_v \colon G_v \rightarrow G$ such that the following conditions hold.
\begin{enumerate}
 \item If $\{u,v\} \in E$ then $[e_u(x),e_v(y)] = 1$ for all $x\in G_u$, $y\in G_v$.
 \item
If $h_v\colon G_v \rightarrow H$ is a collection of homomorphisms such that $[h_u(x),h_v(y)] = 1$
whenever $\{u,v\}\in E$, then there is a unique homomorphism $\phi\colon G \rightarrow H$ such that
$\phi\circ e_v = h_v$ for all $v\in V$.
\end{enumerate}

\begin{theorem} The homomorphisms $e_v$ in the definition of graph product are injective, and $G$ is
generated by the union of the images of the $e_v$.
\end{theorem}

\begin{proof}
Let $v \in V$, and define a homomorphism $h_u \colon G_u \rightarrow G_v$ for each $u \in V$ as
follows. If $u=v$ then $h_u(g) = g$ for all $g \in G_u$, otherwise $h_u(g) = 1$ for all $g \in
G_u$. Since $[h_u(x), h_w(y)] = 1$ for all $u\ne w$, there is a
unique homomorphism $\phi\colon G \rightarrow G_v$ such that $\phi \circ e_u = h_u$ for all $u$. But
$h_v$ is injective, hence $e_v$ is also injective. (Note that it also follows that $\phi$ is
surjective.)

As for the other assertion, let $G_0$ denote the subgroup of $G$ which is generated by the union
of the images of the $e_v$. It is required to prove that $G_0 = G$. Let $j \colon G_0 \rightarrow
G$ be the inclusion homomorphism, and let $h_v$ denote the co-restriction of $e_v$ to $G_0$. That
is, $j \circ h_v = e_v$ for all $v\in V$. By the universal mapping property of graph products,
there is a unique homomorphism $\phi \colon G \rightarrow G_0$ such that $\phi \circ e_v = h_v$ for
all $v \in V$. Therefore $(j \circ \phi) \circ e_v = j \circ h_v = e_v$ for all $v \in V$. But
$\mathit{id}_G \circ e_v = e_v$ so it follows from the universal mapping property that $j \circ
\phi = id_G$. Therefore $G_0 = G$ as claimed.
\end{proof}

\begin{theorem} If $(G,e_v)$ and $(H,f_v)$ are two graph products of $(\Gamma, \{G_v\}_{v\in V})$ then
there exists an isomorphism $\phi\colon G \rightarrow H$ such that $\phi\circ e_v = f_v$. In other
words, the graph product of a graph of groups is unique up to isomorphism.
\end{theorem}

\begin{proof}
By the definition of graph product there is a unique homomorphism $\phi \colon G \rightarrow H$ so
that $f_v = \phi \circ e_v$, and there is a unique homomorphism $\psi \colon H \rightarrow G$ so
that $e_v = \psi \circ f_v$. Therefore $(\psi \circ \phi) \circ e_v = e_v$ and $(\phi \circ \psi)
\circ f_v = f_v$. On the other hand, $\mathit{id}_G \circ e_v = e_v$ and $\mathit{id}_H \circ f_v
= f_v$, so by the uniqueness property we have $\psi \circ \phi = \mathit{id}_G$ and $\phi \circ
\psi = \mathit{id}_H$. Therefore $\phi$ is an isomorphism from $G$ to $H$.
\end{proof}

If each $G_v$ is a subgroup of $G$, and each $e_v$ is an inclusion homomorphism, then we say that
$G$ is the \emph{internal graph product} of the $G_v$. In this case we suppress mention of the
$e_v$, and say that $G$ is the graph product of the subgroups $G_v$. In general, if $(G, e_v)$ is
the graph product of $(\Gamma, \{G_v\}_{v\in V})$, then $G$ is the internal graph product of the subgroups
$e_v(G_v)$.

It remains to prove that graph products exist. We will give two different constructions.

Let $F$ denote the free product of the $G_v$. By definition, there exist monomorphisms
$\iota_v\colon G_v \rightarrow F$ such that the following condition is satisfied: for any family of
homomorphisms $h_v\colon G_v \rightarrow H$ there is a unique homomorphism $\psi \colon F
\rightarrow H$ such that $\psi \circ \iota_v = h_v$ for all $v\in V$.

Let $N$ denote the normal closure in $F$ of the set of all commutators $[\iota_u(x),\iota_v(y)]$
where $u$ and $v$ are adjacent vertices, $x\in G_u$ and $y\in G_v$. Let $\pi \colon F \rightarrow
F/N$ be the quotient homomorphism and let $e_v = \pi \circ \iota_v$ for all $v\in V$.

\begin{theorem} With the above definitions, $(F/N,e_v)$ is the graph product of $(\Gamma, \{G_v\}_{v\in V})$.
\end{theorem}

\begin{proof}
Let $H$ be a group, and let $h_v \colon G_v \rightarrow H$ be a collection of homomorphisms so
that $[h_u(x),h_v(y)]=1$ whenever $\{u,v\} \in E$, $u \in G_u$ and $v \in G_v$. By the definition
of free product there is a unique homomorphism $\psi\colon F \rightarrow H$ such that $\psi \circ
\iota_v = h_v$ for all $v\in V$.

If $\{u,v\}\in E$, $x\in G_u$ and $y\in G_v$ then $\psi([\iota_u(x),\iota_v(y)]) = [h_u(x),h_v(y)]
= 1$, so $N \subseteq \ker(\psi)$. Therefore there is an induced homomorphism $\phi \colon F/N
\rightarrow H$ such that $\phi = \psi \circ \pi$. It follows that $\phi \circ e_v = h_v$ for all $v
\in V$, as $\phi \circ e_v = \phi \circ \pi \circ \iota_v = \psi \circ \iota_v = h_v$.

It remains to show that $\phi$ is unique. To that end, let $\phi' \colon F/N \rightarrow H$ be a
homomorphism such that $\phi' \circ e_v = h_v$ for all $v\in V$, and let $\psi' = \phi' \circ
\pi$. Then $\psi' \circ \iota_v = \phi' \circ \pi \circ \iota_v = \phi' \circ e_v = h_v$ for all
$v \in V$. Therefore $\psi' = \psi$, by uniqueness of $\psi$. Since $\pi$ is surjective and $\phi
\circ \pi = \phi' \circ \pi$, it follows that $\phi = \phi'$.

Therefore $(F/N, e_v)$ is the graph product of $(\Gamma,\{G_v\}_{v\in V})$.
\end{proof}

We describe another construction of the graph product. Let $X$ denote the set of all finite
sequences $(g_1, \ldots, g_n)$ where $1 \neq g_i \in G_{v_i}$ and $v_i \in V$ for all $i$ from 1
to $n$. We assume that the $G_v$ are pairwise disjoint except for a common identity element 1. A
sequence of this type is called a \emph{word}, and each entry is a \emph{syllable}. The
\emph{length} of a word is the number of syllables. We admit the empty word $\lambda = (\ )$,
which has length 0.

Given two words $w = (g_1, \ldots g_n)$ and $x = (h_1, \ldots, h_m)$, the \emph{product} $wx$ is
defined by concatenation: $wx = (g_1, \ldots, g_n, h_1, \ldots, h_m)$. This product is associative
and it has an identity element $\lambda$, so it gives $X$ the structure of a monoid. The
\emph{inverse} of $w$ is defined as $w^{-1} = (g^{-1}_n, \ldots, g^{-1}_1)$, although it must be
noted that $w^{-1}$ is not the multiplicative inverse of $w$ in $X$. In fact $\lambda$ is the only
element of $X$ which has a multiplicative inverse.

We define non-negative integer powers by the following recursive definition.

\begin{equation*}
w^n = \left\lbrace
\begin{array}{cl}
\lambda & \text{if } n = 0, \\
ww^{n-1} & \text{if } n \geq 1.
\end{array}
\right.
\end{equation*}

This is extended to negative integer powers by defining $w^{-n} = (w^{-1})^n$ for $n \geq 2$.

Let $\mathcal{R} = \mathcal{R}_1 \cup \mathcal{R}_2 \cup \mathcal{R}_3$, where
\begin{eqnarray*}
 \mathcal{R}_1  & = & \bigcup_{v \in V} \left\{\ \left(\,(g,g^{-1})\,,\,\lambda\,\right)
            \ \colon\ g\in G_v\ \right\}, \\
 \mathcal{R}_2  & = & \bigcup_{v \in V} \left\{\ \left(\,(g,h)\,,\,(gh)\,\right)
            \ \colon\ g,h \in G_v,\ gh \neq 1\ \right\},\ \ \text{and} \\
 \mathcal{R}_3  & = & \bigcup_{\{u,v\} \in E} \left\{\ \left(\,(g,h)\,,\,(h,g)\,\right)
            \ \colon\ g \in G_u,\ h \in G_v\ \right\}.
\end{eqnarray*}

We say that two words $r$ and $s$ are \emph{elementarily equivalent}, denoted $r \approx s$, if
there exist words $w,x,y,z$ such that $r = wxz$, $s = wyz$, and either $(x,y) \in \mathcal{R}$ or
$(y,x) \in \mathcal{R}$. Furthermore, $r$ and $s$ are said to be \emph{equivalent}, denoted $r
\sim s$, if there exists a finite sequences of words $w_0, w_1, \ldots, w_n$ such that $w_0 = r$,
$w_n = s$, and $w_{i-1} \approx w_i$ for all $i$ from 1 to $n$.

In more intuitive terms, two words are equivalent if the first word can be transformed to the
second word by means of the following moves and their inverses.
\begin{enumerate}
\item If a syllable $g$ is followed by its inverse $g^{-1}$, then delete both syllables.
\item If two successive syllables $g$ and $h$ belong to the same vertex group, and if $gh \neq
1$, then replace the two syllables with the single syllable $gh$.
\item If two successive syllables $g$ and $h$ belong to adjacent vertex groups, then swap $g$ and
$h$. \label{moves}
\end{enumerate}
Sometimes we will allow words to contain the identity element $1$ as a syllable. In that case we
add another rule stating that 1's can be deleted.

It is clear that the relation defined above is an equivalence relation. Moreover, it preserves
multiplication\footnote{In other words, $\sim$ is a congruence.} --- if $w\sim x$ and $y\sim z$
then $wy\sim xz$. Let $\Omega$ be the set of equivalence classes of $X$. Then $\Omega$ inherits
from $X$ the structure of a monoid. In fact $\Omega$ is a group, since $ww^{-1} \sim \lambda$ and
$w^{-1}w \sim \lambda$ for all $w\in X$. We will write $1$ for the identity element $[\lambda]$ of
$\Omega$. Define $\pi \colon X \rightarrow \Omega$ by $\pi(x) = [x]$. There are natural
homomorphisms $e_v \colon G_v \rightarrow \Omega$ defined by $e_v(g) = [(g)]$ for $g \neq 1$ and
$e_v(1) = 1$.

\begin{theorem} With the above definitions, $(\Omega, e_v)$ is the graph product of $(\Gamma,
\{G_v\}_{v\in V})$.\label{words}
\end{theorem}

\begin{proof}
Let $\{u,v\} \in E$, $1\neq x \in G_u$, and $1\neq y\in G_v$. Then $e_v(xy) = e_v(yx)$, since
$(x,y) \approx (y,x)$. Therefore $e_v([x,y]) = 1$, and the first condition in the definition of
graph product is verified.

Let $h_v \colon G_v \rightarrow H$ be any collection of homomorphisms with the property that
$[h_u(x),h_v(y)]=1$ for all $x\in G_u$, $y \in G_v$ when $\{u,v\}\in E$.

Define $\psi\colon X \rightarrow H$ as follows. If $w = (g_1, \ldots, g_n)$ and $g_i \in G_{v_i}$
for all $i$, then let $\psi(w) = h_{v_1}(g_1) \cdots h_{v_n}(g_n)$. If $w = \lambda$ then let
$\psi(w) = 1$.

It is clear that $\psi$ is a monoid homomorphism and that it respects equivalence. Therefore there
is a group homomorphism $\phi \colon \Omega \rightarrow H$ such that $\psi = \phi \circ \pi$.

Now if $1 \neq g \in G_v$ then $\phi\circ e_v(g) = \phi(\ [(g)]\ ) = \psi(\ (g)\ ) = h_v(g)$.
Therefore $\phi \circ e_v = h_v$ for all $v \in V$. It remains to show that $\phi$ is unique. To
that end, let $\phi' \colon \Omega \rightarrow H$ be a homomorphism such that $\phi \circ e_v =
h_v$. If $1 \neq g \in G_v$ then $\phi'(\ [(g)]\ ) = \phi'\circ e_v(g) = h_v(g) = \phi\circ e_v(g)
= \phi(\ [(g)]\ )$. But $\Omega$ is generated by elements of the form $[(g)]$. Therefore $\phi =
\phi'$, and the proof is complete.
\end{proof}

\section{Normal Forms for Elements of Graph Products}

Let $(\Gamma,\{G_v\}_{v\in V})$ be a graph of groups, with graph product $G$. We will realize $G$ as the group
of equivalence classes of words from Theorem \ref{words}. Let $X$ be the set of words used in this
construction.

Choose an arbitrary linear ordering $\prec$ of $V$. Let $w = (g_1, \ldots, g_n)$ be a word, where
$1 \neq g_i \in G_{v_i}$ and $v_i \in V$ for all $i$. We say that $w$ is \emph{reduced} if it is
not equivalent to any shorter word. We say that $w$ is \emph{normal} if it satisfies the following
conditions:
\begin{enumerate}
\item $v_i \neq v_{i+1}$ for all $i$ between 1 and $n-1$, and
\item if $\{v_i,v_{i+1}\} \in E$ then $v_i \prec v_{i+1}$.
\end{enumerate}
We also consider $\lambda$ to be a normal word.

\begin{theorem} Every element of $G$ is represented by exactly one normal word.
The unique normal word representing $g$ is called the \emph{normal form} of $g$.
\end{theorem}

\begin{proof}
The following argument is modeled on the proof by van der Waerden of the normal form theorem for
free products~\cite{kurosh,vdw}. The theorem was first proved by Green~\cite{green}.

For each $v\in V$ we define $\mu_v \colon G_v \times X \rightarrow X$ by the following recursive
algorithm. Let $g \in G_v$ and let $x \in R$.
\begin{enumerate}
\item If $g=1$ then $\mu_v(g,x) = x$.
\item If $g\neq 1$ and $x=\lambda$ then $\mu_v(g,x) = (g)$.
\item Suppose that $g\neq 1$ and $x\ne\lambda$.
Let $g_1$ be the first syllable of $x$. Select $v_1 \in V$ such that $g_1 \in G_{v_1}$, and select
$y \in X$ such that $x = (g_1)\ y$.
\begin{enumerate}
     \item If $v = v_1$ and $gg_1 = 1$ then $\mu_v(g,x) = y$.
     \item If $v = v_1$ and $gg_1 \neq 1$ then $\mu_v(g,x) = (gg_1)\ y$.
     \item If $v_1 \prec v$ and $\{v,v_1\} \in E$ then $\mu_v(g,x) = (g_1)\ \mu_v(g,y)$.
     \item Otherwise $\mu_v(g,x) = (g)\ x$.
\end{enumerate}
\end{enumerate}

Let $R$ denote the set of normal words of $X$. We claim that if $x \in R$, $v \in V$ and $g \in
G_v$ then $\mu_v(g,x) \in R$. The proof is by induction on word length. Suppose that $x \in R$,
and that $\mu_v(g,y)$ is normal for every reduced word $y$ which is shorter than $x$. We need to
show that $\mu_v(g,x) \in R$. This is done by checking each of the six cases in the recursive
definition. The verification of these cases is left to the reader.

A similar case-by-case analysis shows that if $g,h \in G_v$ and $x \in R$ then
$\mu_v(g,\mu_v(h,x)) = \mu_v(gh,\mu_v(x))$. Since $\mu_v(g,\mu_v(g^{-1},x)) = x$,
it follows that $\mu_v(g,\cdot)$
is a permutation of $R$ for each $g\in G$. Therefore there is a homomorphism $h_v \colon G_v
\rightarrow \mathit{Perm}(R)$ defined by $h_v(g)=\mu_v(g,\cdot)$.

It is a routine matter to verify that $[h_u(g),h_v(k)]=1$ whenever $\{u,v\}\in E$. Therefore there
exists a homomorphism $\phi \colon G \rightarrow \mathit{Perm}(R)$ such that $\phi(g) =
\mu_v(g,\cdot)$ when $g \in G_v$.

This homomorphism allows us to compute for any word $x$ an equivalent normal word $w$. Let $x =
(x_1, \ldots, x_n)$ and let $w = \phi([x])(\lambda)$. Then $w$ is a normal word, and $w$ is
equivalent to $x$. On the other hand, $w = \phi([w])(\lambda)$, so $w$ is the only normal word
which is equivalent to $x$. For if $w'$ is another normal word equivalent to $x$, then $w' =
\phi([w'])(\lambda) = \phi([w])(\lambda) = w$.
\end{proof}

\begin{theorem}
Every normal word is reduced.
\end{theorem}

\begin{proof} Let $x$ be a normal word, and let $y$ be a reduced word which is equivalent to $x$.
If $y$ contains two successive syllables $g_i$ and $g_{i+1}$ such that $\{v_i,v_{i+1}\} \in E$ and
$v_{i+1} \prec v_i$, then swap these syllables. Repeat this until no more swaps are possible. This
must terminate because no pair of syllables can be swapped more than once. Let $z$ be the word
which results. Now $z$ cannot have two successive syllables belonging to the same vertex group,
else $y$ would not have minimal length. Therefore $z$ is normal, hence $z=x$ by the previous
theorem. Since $y$ and $z$ have the same length, it follows that $x$ is reduced. \end{proof}

\begin{cor}
A reduced word can be transformed into an equivalent normal word by swapping syllables belonging
to adjacent vertex groups. \qed
\end{cor}

\section{Graph Products as Hyperreflection Systems}

Let $(V,E)$ be a graph, and suppose that $G$ is the internal graph product of a collection
of subgroups $\{G_v\}_{v\in V}$. Also suppose that $(G_v, \Sigma_v)$ is a hyperreflection
system for each $v \in V$. The main objective of this section is to prove that
$(G, \Sigma)$ is a hyperreflection system, where $\Sigma = \bigcup_{v\in V} \Sigma_v$.

Define a weight function $\wt$ on $G$ as follows. For each $v \in V$ let $\ell_v$ be the length
function associated to $(G_v, \Sigma_v)$, and let $\ell = \bigcup_{v\in V} \ell_v$.
If $(g_1, \ldots, g_n)$ is the normal form for $g$, and if
$g_i \in G_{v_i}$ for each $i$, then define $\wt(g) = \sum_{i=1}^n \ell(g_i)$.

Let $v \in V$ and $S \in \Sigma_v$ be given. Choose a linear ordering $\prec$ on $V$ such that $v$ is
minimal with respect to $\prec$. This linear ordering determines a normal form for the elements of $G$.

\begin{theorem}
If $g \in G$ then $Sg$ has a unique element of minimum weight, and $g$ is the minimum weight element of
$Sg$ if and only if $g_1 \in (G_v)^S$ or $g_1 \notin G_v$, where $g_1$ is the first syllable of the normal
form for $g$.
\end{theorem}

\begin{proof}
Let $(g_1, \ldots, g_n)$ be the normal form for $g$.  We consider three cases.
\begin{enumerate}
\item
Suppose that $g_1 \notin G_v$.
If $1\ne s \in S$ then $(s, g_1, \ldots, g_n)$ is the normal form for $sg$.
Therefore, $\wt(sg) = \wt(g) + 1$ for all $1\ne s \in S$,
so $g$ is the unique element of minimum weight in $Sg$.

\item
Suppose that $g_1 \in G_v$ and $g_1 \notin S$.
If $s \in S$, then $(sg_1, \ldots, g_n)$ is the normal form for $sg$.
Therefore, $\wt(sg)$ is minimized when $\ell(sg_1)$ is minimized. But the coset
$Sg_1$ of $G_{v}$ has a unique element of minimum length, hence
$Sg$ has a unique element of minimum weight.
Furthermore, $g$ is the minimum weight element of $Sg$ if and only if
$g_1$ is the minimum length element of $Sg_1$,
which occurs precisely when $g_1 \in (G_v)^S$.

\item
Suppose that $g_1 \in S$.
If $s = g_1^{-1}$ then $(g_2, \ldots, g_n)$ is the normal form for $sg$, otherwise
$(sg_1, g_2, \ldots, g_n)$ is the normal form for $sg$.
The weight is uniquely minimized when $s = g_1^{-1}$, hence
$Sg$ has a unique element of minimum weight.
Note that $g$ cannot be the minimum weight element of $Sg$ in this case,
since $\wt(g_1^{-1} g) = \wt(g) - 1$.
\end{enumerate}
\end{proof}

\begin{theorem}
$(G, \Sigma)$ as defined above is a hyperreflection system.
\end{theorem}

\begin{proof}
Let $C$ be the set of all elements $g \in G$ such that $g$ is the unique element of minimum
weight in $Sg$. Since every coset $Sg$ has a unique element of minimum weight, it follows
that $G$ is the disjoint union of $sC$ for $s \in S$.

Let $g \in C$ and let $(g_1, \ldots, g_n)$ be the normal form for $g$. Choose a reduced word
$\bm{s_i}$ for each $g_i$ and let $\bm{s} = (s_1, \ldots, s_N)$ be the concatenation of the $\bm{s_i}$.
Then $\bm{s}$ determines a walk $\pi$ from 1 to $g$ in $\Cay(G,\Sigma)$.

Suppose that $\pi$ crosses from $C$ to $sC$ for some $s \in S\setminus\{1\}$ ,
i.e.\ there exists $k$ such that
$$s(s_1\ldots s_{k-1}) = (s_1\cdots s_k).$$
Then $\bm{s}$ is equivalent to a new word
$$\bm{s'} = (s, s_1, \ldots, \widehat{s_k}, \ldots, s_N).$$

Let $i$ be the index such that $s_k$ lies in the subword $\bm{s_i}$, which represents the syllable $g_i$.
If $g_i = s_k$ then $g$ can be represented by $(s, g_1, \ldots, \widehat{g_i}, \ldots, g_n)$.
Since this word has $n$ syllables, it is reduced; so its normal form is obtained by swapping
adjacent syllables. The first syllable $s$ cannot be swapped,
because $v$ is the first vertex in the chosen linear ordering of V. Therefore, $s$ must be the first syllable
in the normal form of $g$, which contradicts the assumption that $g \in C$.

If $g_i \ne s_k$ then $g$ can be represented by $(s, g_1, \ldots, g'_i, \ldots, g_n)$, where $g'_i$
is obtained by deleting $s_k$ from $\bm{s_i}$.
Since this word has $n+1$ syllables, it is not normal. However, $(g_1, \ldots, g'_i, \ldots, g_n)$
is normal, and this implies (by the normal form algorithm) that $g_1 \in S$ and that
$$(sg_1, g_2, \ldots, g'_i, \ldots, g_n)$$
is the normal form for $g$. Therefore $sg_1 = g_1$ by the
uniqueness of normal forms, which is a contradiction. Therefore, every $g \in C$ can be joined to 1
by a walk that does not cross $\Fix(S)$.

It remains to prove that if $1 \ne s_0 \in S$ then every walk from 1 to $s_0$ must cross $\Fix(S)$.
Let $\pi$ be any walk from 1 to $s_0$, and let $(s_1, s_2, \ldots, s_n)$ be the corresponding word representing $s_0$.
Since $1 \in C$ and $s_0 \notin C$, there exists an index $k$ and $1 \ne s \in S$ such that
$u \ceq s_1 \cdots s_{k-1} \in C$ and $us_k \in sC$.

Let $(g_1, \ldots, g_n)$ be the normal form for $u$.  Then either $g_1 \in (G_{v})^{S}$ or
$g_1 \notin G_{v}$, whereas the normal form for $us_k$ begins with
an element of $s (G_{v})^{S}$ since $us_k \in sC$. In order to affect the
first syllable of the normal form, $s_k$ must commute with $g_2, \cdots, g_n$,
hence $t_k = g_1 s_k g_1^{-1}$. If $g_1 \notin G_{v}$, then $s_k$ must commute
with $g_1$ as well. There are three cases to consider.
\begin{enumerate}
\item
If $g_1 \notin G_{v}$, then $s_k$ commutes with $u$, so $t_k = us_ku^{-1} = s_k$. Therefore, $\pi$ crosses
$\Fix(S)$ while passing from $u$ to $us_k$.
\item
If $g_1 \in G_{v}$ and $g_1 \ne s_k^{-1}$, then $(g_1s_k, g_2, \cdots, g_n)$ is the normal form for $us_k$.
Therefore $g_1s_k \not\in (G_{v})^{S}$, which implies that $\pi$ crosses $\Fix(S)$ while passing from
$u$ to $us_k$.
\item
If $g_1 = s_k^{-1}$, then $(g_2, \cdots, g_n)$ is the normal form for $us_k$. But the normal form for $us_k$
must start with an element of $G_{v}$, so this case cannot occur.
\end{enumerate}
Therefore, each component of $\Cay(W,\Sigma)\setminus\Fix(R)$ contains exactly one element of $S$,
which implies that $S$ is a hyperreflection. Since $S$ is an arbitrary element of $\Sigma$, it follows
that $(G,\Sigma)$ is a hyperreflection system.
\end{proof}

\begin{cor} If $G$ is the internal graph product of $\{G_v\}_{v\in V}$ then $(G, \{G_v\}_{v\in V})$
is a hyperreflection system.
\end{cor}
\begin{proof}
Each $G_v$ has a trivial hyperreflection system $(G_v, \{G_v\})$, so the previous theorem implies
that $(G, \{G_v\}_{v\in V})$ is a hyperreflection system.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Dancing with sea lions, walking with giants: Unmissable Galapagos highlights
"The boat moved noiselessly over the calm surface, the water so clear it was almost imperceptible, as if we were gliding over emerald green air. The captain steered expertly, close enough to the stony islands to look pelicans in the eye, far enough to never scratch the boat on the surreal lava bridges. We passed over turtles and parrot fish, by blue-footed boobies and palo santo trees. None of it felt real…"
Galapagos highlights are never just things to do or see on the islands; they are visions, emotions and enduring memories. Below are some of the most unmissable Galapagos experiences, and the favorite moments of those who have encountered them.
Hike an active volcano
From the moment you touch down in Galapagos you are treading on the land of volcanoes: the 13 main islands and six islets were spluttered out from the depths of the Earth itself over the course of millions of years of eruptions. And although the results are abundantly obvious in the strange lunar scenery and spiny rock formations, a true immersion in the archipelago's geologic history comes when you hike the Sierra Negra caldera on Isabela, one of the most eye-opening, adventurous things to do in Galapagos. This island is something of a work in process; formed from an eruption just one million years ago, the volcano remains active. Walking along the rim of the caldera, the now-blackened lava seems to have frozen in time, rivers and gorges of molten magma from its last activity in 2005 cooled into eerie rock forms. If you haven't tired once you've reached the highest point, continuing on to Volcán Chico holds further insight into the volcanic past, with eruptions from different eras cooled into a palette of rich shades. Still in evolution, the landscape huffs and puffs with fumaroles, telling of the awesome power and heat below the Earth's surface.
Stumble upon marine iguanas
When Spanish ships first approached the shores of the Galapagos Islands, the sailors were terrified of the nightmarish, blackened monsters crawling along the rocks and darting through the seas. Marine iguanas, the creatures that Charles Darwin named "imps of darkness", are a mainstay of the Galapagos scenery, apparent on every island in different colors, shapes and forms – stumbling over a scaly tail or being startled when a perfectly motionless one suddenly creeps into action once it's warmed under the equatorial sun are part of any Galapagos day. The only sea-diving lizard on the planet, these creatures are wonders of evolution, their mouths, teeth, skin and respiratory system adapted to feast on vegetation in the ocean. They can hold their breath for more than a minute, using their tails to propel themselves through the waves. Though two-a-penny on the islands, marine iguanas are truly a marvelous thing to see in Galapagos.
My Galapagos Highlight
"My most amazing Galapagos moment was when we went snorkeling round Los Tuneles in Isabela. My guide beckoned me over to a spot and when I put my head underwater, I was face to face with a white-tipped shark. It darted out of the cave where it had been hiding in, and suddenly, loads of them followed it, maybe 10 or 12 sharks streaming out of the hole, swimming all around me. Of course I knew that this kind of shark is totally harmless, but it was still an incredible thrill."
Christian Trepaud, regular visitor to Galapagos since his teens.
Attend the spectacle of the dancing birds
We often talk of the theatre of nature, the elements and wildlife conspiring to put on a show to beat anything on Broadway. Never is this truer than with the mating rituals of Galapagos birds, too elaborate, too choreographed and too entertaining to be a coincidence of nature. There are the blue-footed boobies and their solemn stomp, a left to right two-step of a routine that can be seen wherever there are colonies around the archipelago. The deal is sealed when the female accepts a twig that the male passes to her, a signal that they will build a home together. The waved albatross, which breeds on Española Island from January to March, has its own courtship jig, a dance of circling and bowing while rhythmically tapping beaks. It is a delightful spectacle.
Witness the plight of the bird that can't take off
It seems like a cruel joke: a bird that can't fly, trapped on the lava shorelines of Isabela and Fernandina by its shriveled, deformed wings, as if they'd foolishly swapped theirs with a finch. But the flightless cormorant is anything but hopeless. In fact, what they lack in flight, they make up for in remarkable swimming ability, and those stunted wings come in handy for balance when jumping from rock to rock. Their unique skills allow them a diet of kings: octopus, eel and fish are all on the menu. Witnessing their swimming prowess is a memorable thing to do in Galapagos and watching their courtship ritual even more so, an intricate process of intertwining their necks while spinning in circles.
Stroll among giants
Some things to see in Galapagos are pure, dreamlike surrealism; cycling past a lumbering giant tortoise or spotting their great shells from the main highway is one of them. More humbling still is visiting the great Galapagos tortoises in their natural habitat, in the sanctuaries set up around the national park to preserve these endangered species. To stroll among these gentle giants is to stroll among the most dignified of prehistoric beasts, their expression condescending as they munch rhythmically through greenery up in the lush highlands. Look out for the 14 species divided into two main shell groups: the domed and the saddle backed, shapes that dictate how much they can raise their heads. A trip to see the tortoises in the Santa Cruz highlands or on San Cristóbal (among other places) is a stark reminder of the fragility of our ecosystems and the necessity of our protection: it is due to human interference that the Pinta tortoise died out completely with the death of the species' last known survivor, Lonesome George, in 2012.
See evolution in action
Finches, the fist-sized birds with a knack for staring at you with an air of deeply judgmental curiosity, are absolutely key to the significance of the Galapagos Islands and allow us to see the process of evolution working in real time. It was these birds that inspired the seminal theories of Charles Darwin, as he noted the extraordinary differences in the sizes of beak in 13 different species of finch around the archipelago. And the evolutionary process is far from over: evolutionary biologists Peter and Rosemary Grants have continued to observe the birds over a period of two decades, noting the incredible rate at which they "evolve" and "adapt". Of all the things to see in Galapagos, the finches provide the greatest insight into how the natural world works.
Share oceans with extraordinary beasts
If the terrestrial part of the Galapagos National Park is a living nature documentary, what lurks beneath the waves is pure science fiction. The underwater world (and a 'world' it truly is) is home to a party of rare beasts: massive whale sharks and endemic Galapagos sharks, graceful Pacific turtles, manta rays soaring like aquatic eagles, orcas, eels, and the absurd sunfish, otherwise known as the mola mola. This last monster from the blue has no tail, and instead propels itself along with a fin above and below its body. Weighing up to 1,000 kg and big and rounded like a rock, the sunfish is a bizarre thing to see in Galapagos, around Punta Vicente Roca on Isabela Island. Scuba divers reach their Mecca when they take to the Galapagos waters, but even the most inexperienced snorkelers can be rewarded with rainbow-colored fish, golden rays, turtles, white-tipped sharks and speeding Galapagos penguins, to name but a few of the ocean's eclectic community.
Make friends with Galapagos sea lions
Just like marine iguanas, sea lions provide much of the backdrop – and soundtrack, and scent – to a Galapagos adventure. Whether you're stepping over them as they sprawl and groan like old drunks on the wooden walkways, narrowly avoiding sitting on them as recline on benches, or swimming with their curious pups in the shallows of beaches, these endemic species are entwined into any Galapagos experience. See them in hundred-strong groups with their pups resting in coves, or throngs of unfortunate, lonely bachelors sticking together for company. The trick is to learn how to tell a Galapagos sea lion from a fur seal: both slick and sleek when wet, it's all in the ears and tail…
"Every time I go to Galapagos I always look forward to swimming with the sea lions. With nothing but a mask, flippers and snorkel, you can dance with them, swirl around with them, even blow bubbles at them. That connection with an innocent, wild animal, behaving almost like a dog, who wants to play, looking you in the eye. When we go out to Las Tintoreras, seeing the joy that other visitors get, including my family, is always an exciting, memorable moment."
Dominic Hamilton, Director of Ecuadorian travel communications agency, Ñan Comunicación.
Bike through an ever-changing natural wonderland
The cycle route beginning in Isabela Island's Puerto Villamil and ending at the eerie Wall of Tears feels like a Galapagos nature museum, as if it were a landscaped installation engineered to display the archipelago's diverse ecosystems and environments. But, save for the very much manmade wall of lava rocks built by inmates of the former penal colony in the middle of the 20th century, this spectacle is all Mother Nature's doing. Pass through the fabulous Humedales Reserve with its smattering of pink flamingos, past roaming giant tortoises, stopping off to marvel at the delicate sands of La Playita, the coralline Playa del Amor with its blood-red crabs, the lava tunnels and mysterious mangroves, the surreal lagoons surrounded by cacti and palo santo, and the Orchilla viewpoint, from which you make out the five volcanoes that created the stunning island. It's as if all of the things to see in Galapagos were condensed into one, stunning path.
Kayak and paddleboard where no one has been before
The idea of exploring the unknown, touching the untouched and seeing things never seen before just like the great adventurers of the past is an allure for many but a reality for few. And though with the flow of tourists in the Galapagos Islands this can seem more unattainable than ever, there are moments when kayaking or paddle-boarding around the northern coast of San Cristobal when you feel like, just maybe, you're the first human on Earth to have ever to have set foot, or oar, there. Paddling almost noiselessly through tunnels and islets, on your kayak or board you're just another marine creature, treated with nonchalance by frigate birds, penguins, sea lions and turtles. It's a thrilling thing to do in Galapagos: to feel like you've been where no one has before.
"I love the Galapagos; every corner of the archipelago has its magic. La Ratonera is a beach that I am really fond of because it is near my office in the National Park in Santa Cruz: it's my escape and relaxation place where I love taking photos after a day of work."
Diego Bermeo, The Galapagos Photographer
Take it all in on the world's best beaches
With so many animals to see and adventures to be had, you may think that downtime is off the agenda here. Yet the Galapagos Islands offer some of the most magical locations on the planet in which to relax: their utterly perfect beaches. There's Tortuga Bay, a gentle walk across Santa Cruz leading to the gorgeous stretch of white sand, easy surf and a cove backing onto a peaceful lagoon. There's the great expanse of creamy sand dotted with the black of iguanas all along Puerto Villamil, the main settlement on Isabela. There's postcard-perfect Carola Beach on San Cristobal, where you'll share the turquoise waters with sea lions, sea turtles and marine iguanas. And there are even more to discover when you hike, bike, kayak or paddleboard around the coastline.
Stargaze like nowhere else on Earth
The great majority of the Galapagos Islands remains untouched by man; the same goes for its night skies. With so few settlements, light pollution is kept to a bare minimum, meaning that stars are amazingly radiant and clear. As the islands straddle the equator it is possible to pick out constellations from both hemispheres; the Big Dipper, Tres Marias and the Southern Cross blinking on the velvety sky. Isabela Island is one of the best spots for stargazing, as is anywhere where you can get away from streetlights. Download the free apps Star Chart, SkyView Free or Night Sky Lite to help you identify the stars, constellations and planets. Of all the things to see in Galapagos, the stars are the most magical.
Learn from Galapagos locals
The Galapagos Islands would not be the extraordinary natural sanctuary that they are were it not for the incredible people who live there, working each day to protect it and find green solutions to the challenges of a developing community. These are people from far and wide, Ecuador and beyond, who moved to the archipelago as pioneers and have built their homes here, each one with incredible stories and projects. There are the fishermen who sell their catch in Puerto Ayora, whom you can accompany on trips on the open waters. There are the organic coffee and cacao farmers who explain their unusual agricultural experiences. There are chefs performing magic with the fresh seafood. And there are the dozens of local guides and people working in the Galapagos National Park who are passionate about conservation. All these people are waiting to share their experiences and island tales. All you have to do is ask.
The Galapagos Islands are full of wonder, with so many incredible things to do and see that however long you stay it never seems like enough. As Diego Bermeo points out: "every corner of the archipelago has its magic."

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Finding and Printing all Links within a DIV I am trying to find all links in a div and then printing those links.
I am using the Simple HTML Dom to parse the HTML file. Here is what I have so far, please read the inline comments and let me know where I am going wrong.
include('simple_html_dom.php');

$html = file_get_html('tester.html');

$articles = array();

//find the div the div with the id abcde
foreach($html->find('#abcde') as $article) {

    //find all a tags that have a href in the div abcde
    foreach($article->find('a[href]') as $link){

        //if the href contains singer then echo this link
        if(strstr($link, 'singer')){

            echo $link;

        }

    }

}

What currently happens is that the above takes a long time to load (never got it to finish). I printed what it was doing in each loop since it was too long to wait and I find that its going through things I don't need it to! This suggests my code is wrong.
The HTML is basically something like this:

Thanks all for any help

A: The correct way to select a div (or whatever) by ID using that API is:
$html->find('div[id=abcde]');

Also, since IDs are supposed to be unique, the following should suffice:
//find all a tags that have a href in the div abcde
$article = $html->find('div[id=abcde]', 0);

foreach($article->find('a[href]') as $link){

    //if the href contains singer then echo this link
    if(strstr($link, 'singer')){
        echo $link;
    }
}

A: Why don't you use the built-in DOM extension instead?
<?php

$cont = file_get_contents("http://stackoverflow.com/") or die("1");

$doc = new DOMDocument();
@$doc->loadHTML($cont) or die("2");

$nodes = $doc->getElementsByTagName("a");

for ($i = 0; $i < $nodes->length; $i++) {
    $el = $nodes->item($i);
    if ($el->hasAttribute("href"))
        echo "- {$el->getAttribute("href")}\n";
}

gives

... (lots of links before) ...
- http://careers.stackoverflow.com
- http://serverfault.com
- http://superuser.com
- http://meta.stackoverflow.com
- http://www.howtogeek.com
- http://doctype.com
- http://creativecommons.org/licenses/by-sa/2.5/
- http://www.peakinternet.com/business/hosting/colocation-dedicated#
- http://creativecommons.org/licenses/by-sa/2.5/
- http://blog.stackoverflow.com/2009/06/attribution-required/

redpajama_set_name: RedPajamaStackExchange


text:
AFP-FL to host summer town hall series with Ron DeSantis
Published on August 9, 2017 in 2017/Top Headlines by Staff Reports
Americans for Prosperity-Florida is hitting the road to energize activists about tax reform.
The grassroots organization is scheduled to hold "Un-Rig the Economy" town hall meetings in Miami, Fort Lauderdale and Orlando in August and September to talk with activists about the current situation in Washington, D.C. and energize them for what's to come. The AFP-FL team will be joined by local leaders and members of the federal affairs team to discuss the need to enact comprehensive tax reform.
Rep. Ron DeSantis, a Ponte Verde Republican, is scheduled to join AFP-FL at each of the town hall events. DeSantis, who is believed to be mulling a 2018 gubernatorial bid, is expected to discuss how Congress can act to fix the tax code.
"AFP has been leading the charge on calling for Congress to un-rig the American economy," said Chris Hudson, the state director for AFP-FL, in a statement "We hope the rest of the Florida delegation will join Congressman DeSantis in fighting back against the current rigged tax system by joining our effort to pass pro-growth tax reform. Americans want a system that's based on simplicity, efficiency, equitability, predictability, and creates no new burden on taxpayers. We want to speak directly to Floridians who want to help fix our broken tax code."
The summer town hall series kicks off on Aug. 24 with a town hall in Miami, followed by a town hall scheduled on Sept. 19 in Fort Lauderdale and Sept. 28 in Orlando.
Tags: AFP-FLAmericans For Prosperity – FloridaChris HudsonRon Desantis
Poll: Adam Putnam is front-runner in governor's race
It's clear that Agriculture Commissioner Adam Putnam has been running a careful race for Florida governor. And
Paulson's principles: Money, money, money!
It has been said that money is the lifeblood of politics. If so, many members of
Chris Latvala says his dad's not the moderate in governor's race
Chris Latvala predicts the race for governor will be a campaign unlike any ever seen before
Previous Story Previous post: Immigrant groups urge Florida GOP lawmakers to oppose Trump
Next Story Next post: The Delegation for 8.9.17 — Insights from the Beltway to the Sunshine State
Ballard Partners, which has close ties with the Trump White House, has
It's clear that Agriculture Commissioner Adam Putnam has been running a
It has been said that money is the lifeblood of


text:
Blakeslee House, located at 132 Curtis Street on the Medford campus, was constructed in 1925 and purchased by the university in 1939 to serve as a dormitory for women students at the Fletcher School. It was named in honor of Professor George W. Blakeslee, one of the original faculty members of the Fletcher School.
During World War II, Blakeslee house was shared by the Fletcher School with Jackson College and the Bouve-Boston School. Blakeslee Hall was relinquished by the Fletcher School upon the completion of Mugar Hall in 1964.
As of 2005, Blakeslee House is again used by the Fletcher School of Law and Diplomacy.
Source: LOH1; LOH2
Blakeslee, George H.
Jackson College
Bouve-Boston School
School of Nutrition Science and Policy
Blakeslee House
Medford Campus


text:
Q: How to programmatically erase an optional TextField value in SwiftUI? In this sample I want to clear the value of a TextField with a custom button. However, the button action (value = nil) seems to be ignored. Why?
Sample video
Tested on Xcode 14 beta 6 (iOS 16).
struct ContentView: View {
    @State private var value: Int?
    @FocusState var focused: Bool

    var buttonOpacity: Double {
        focused && value != nil ? 1 : 0
    }

    var body: some View {
        Form {
            HStack {
                TextField("Value", value: $value, format: .number.grouping(.never))
                    .focused($focused)
                    .keyboardType(.numberPad)
                Spacer()
                Button {
                    value = nil
                } label: {
                    Label("Erase", systemImage: "multiply")
                        .labelStyle(.iconOnly)
                        .imageScale(.medium)
                        .symbolVariant(.circle.fill)
                        .symbolRenderingMode(.hierarchical)
                        .foregroundStyle(.secondary)
                }
                .buttonStyle(.plain)
                .opacity(buttonOpacity)
            }
        }
    }
}


Corinne's bungalow is a 2-bedroom ground-floor apartment facing the beach within a large beach house, which is part of a small family beach complex located right on the beach in Trou aux Biches – the most sought-after beach of Mauritius.
Its covered terrace (alfresco eating area) offers 180° full sea view and leads to a tastefully furnished living-dining room with flat-screen TV with Satellite programmes.
Its fully equipped Kitchen has a new large Fridge-freezer, Microwave oven and Washing machine.
Secure parking space for 1 car is available in the gated parking area with electronic opening system.
Free WIFI Internet connection and Security alarm system with response.
Both bedrooms are air-conditioned & have double beds and ceiling fans. They share a bathroom with shower unit and toilet.
2 hour maid housekeeping service in the morning is included (except Sundays & Public holidays). Cooking service is not included but Yolande the maid who is an excellent cook would be pleased to earn a little extra by cooking some local dish for you.
2 sun loungers for your personal us.
Check-in as from 14h00 (2pm) & Check-out at 10h00.
Residents are further reminded that they should behave in an orderly manner so as not disrupt the enjoyment of other residents staying in other units of the residential complex.


Discussion in 'Buying Tips and Advice' started by direzz, Feb 5, 2006.
i currently have four products ready to buy using my education discount in my cart.
i heard apple only lets u use one a year? is this true..
and can i still get all my stuff?
**The above purchases can be made either online or through your Authorized Campus Reseller. Total product quantities remain the same regardless of where you purchase is made.
Straight from the Apple website.
did you get this from the American apple site? could you tell me where?
It was from the Aus site, but I'm pretty sure I've seen it on the US store also. To find it, go to the Edustore. When you first go to it, it will ask you to select your institution. Once you've done this, a screen comes up displaying this info.
US site has the same limitations as the Aussie one. I purchased 2 notebooks though and way more than 2 copies of software last year edu priced and they never blinked an eye.
I've bought a total of three ipods within a year using Edu discount.


There's a developing "Rising Wedge", so the price faced resistance on the 34 Moving Average. In this case, the pair is likely going to decline towards support at 1.2738 - 1.2705. At the same time, if we see a pullback from these levels, there'll be an open door for another upward price movement towards resistance at 1.2768 - 1.2793.
The price is consolidating along the Moving Averages. It seems like bulls are likely going to test the 89 Moving Average. If we see a pullback from this line, bears will probably try to achieve the next support at 1.2738 - 1.2705.


This six-story building houses students in single and double room suites, sharing adjacent or adjoining bathrooms. Kitchens, study rooms, and lounges are on each floor and a laundry facility is on the first floor. Collaboration lounges are located on the third and fifth floors. Munroe Hall is a predominantly first-year student residence hall. Munroe Hall shares a grassy courtyard with Belden-Racine Hall, which is accessible to all on-campus residents between 9 a.m. and 9 p.m. via DePaul ID.
Four students (two in each room) share a bathroom in a suite-double.
Two students in a deluxe-double share a bathroom.
One student in suite-single shares a bathroom with an adjoining room.
Want to get a better idea of what this hall looks like? Jump right to Munroe Hall on our virtual tour for a look inside the building.
Munroe Hall also has converted housing, in which regular double residential units are converted to triple units. 360 tours of converted housing units are not yet available for Munroe Hall, but you can view a 360 tour of a lived-in converted housing unit in Clifton-Fullerton Hall, with an occupancy of three residents.


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Andrew Black, Eternity Knocker (2019)
Courtesy the artist
Andrew Black
Andrew Black is an artist who lives and works in Glasgow. He was on the committee of Transmission gallery in 2016 and 2017, and in 2018 took part in the Experimental Film & Moving Image Residency at Cove Park, and the Autumn Residency at Hospitalfield. He is currently working with Atlas Arts on the Plural Futures Community Film commission on the Isle of Skye, and was selected as the recipient of the 2021 Margaret Tait Award.
Solo exhibitions include: Eternity Knocker, Intermedia, CCA, Glasgow (2019); Submerged Village, Market Gallery, Glasgow (2017); Our Andrew of the Flowers, Embassy, Edinburgh (2016).
Group exhibitions include: Seized by the Left Hand, DCA, Dundee (2019); Class Trip, Casino, Luxembourg (2018).
Private: Andrew Black and Aman Sandhu
The Magic Roundabout and The Naked Man
GI online 2021
11 Jun — 01 Jan 1970


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Published : Mon, Oct 8 2018 :7 AM.
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GAN ACCESSORIES. The Perfect Accessories.
GAN, the indoor brand belonging to GANDIABLASCO, offers a range of accessories connected with the design of interior spaces; elements that breathe, enhance and compliment any ambience through their design and functionality.
Inspired by everyday items and made using traditional techniques and superior quality materials, these accessories can be integrated naturally in the layout of any interior, together with carpets, modules and other furniture.
With this initiative, GAN expands upon its inventory with accessories that make life a little easier and indoors spaces more pleasant.
The carpet becomes a reference point in SILAÏ space, it can be coordinated with elements: cushions, poufs and wooden side tables, that provide continuity and complement the interior of residential, public or commercial spaces.


Suppose, the last date of paying your electricity bill is the day after tomorrow and you are out of money? Now, what will you do, because it is almost impossible to arrange credit at such a short notice? But, you do not have to worry at all because we have a way out for you. For all such urgencies, you can easily go for the same day cash loans option. As is clear from the name itself, it can provide you the necessary cash support instantly on just the submission of a filled online application form. The lack of unwanted faxing hassles helps you in getting fast approval and the desired amount within 24 hours of time. Not only electricity bills, but you can also manage your all sorts of unexpected needs well on time.
Instant cash loans can allow you finance from $100 to $1500 for a short term duration of 30 days that comes out to be by your next payday. The borrowed sum is totally collateral free and is also meant for the bad creditors due to the lack of credit checking process. But, as the lenders bear all the risk, so they charge slightly higher interest rates on this scheme.
At times, need for money makes you so helpless that you are not even able to fulfill your critical demands, such as medical emergencies, immediate house or vehicle repairs, credit card bills, electricity bills, child's school or examination fees, etc. But, now you can easily tackle all such things through the support of the payday loans option. It can make you financially capable in managing your urgencies well on time by providing you credit instantly till your next payday. Thus, in this way, it can mend the unusual gap between your two consecutive salaries. For this, you just have to fill a free application form available on the internet with all the mandatory details and submit it. There are no faxing hassles involved that delay the process unnecessarily.
Paycheck cash loans can assist you with funds anywhere in between $ 100-5000 for a short duration of about 30 days. This amount is totally unsecured and free from any kind of collateral. There is no credit checking process at all for any borrower which makes these finances accessible for the bad creditors also. But, the lenders charge a bit higher rate of interests on this scheme as they bear all the risk.
In today's competitive world, life of a person has become so busy that it can hardly spare some time for anything. So, to remember each and everything also becomes impossible. Suppose, you forgot to pay your electricity bill and at the nick of time you came to know that there is just one day left for the last date of making payment. So, now what will you do? Don't worry, because we have a solution and that is the sameday cash advance loans. It can very well help you in managing your various urgencies, like medical care bills, monthly household rent, grocery bills, unexpected travelling expenses, electricity bills, etc well on time. For this, you just have to submit a filled online application form without any faxing hassles and you will get the desired amount in less than 24 hours of time.
Same day loans can provide you cash support up to a maximum of $ 1,500, whose repayment can be made by your next payday. There is neither any credit checking process for the bad creditors nor any need of depositing your valuable asset as a collateral. Instead it is a bit high on the interest rates.


The Dudley Clinic Group Day Program is unique in Orange and the Central West of NSW. Since its establishment in 2007 participants have reported great benefits from both the content of the program as well as from the camaraderie and support that comes from being with others who suffer similar illnesses.
Referrals can be made to this service by contacting our Psychologist on 02 6361 6714.


School will either be delayed or closed on days when weather and driving/road conditions are hazardous and dangerous for students and staff. The forecasting and predicting of hazardous weather in Ohio is always challenging, and the decision to cancel or delay school must be made very early in the morning. Delays and closings will almost always be announced by 6:30 AM.
If there is no announcement to cancel or delay school by 6:30 AM, you may safely assume that school will be in session.
Parents will be notified by the "Alert Now" Emergency Telephone Notification System in the event school is delayed or cancelled. The "Alert Now" phone call and parent notification may take several minutes to reach your home if there are a number of area school delays or closings. Please verify any "Alert Now" school delay or school closing telephone calls to the information provided by one of the radio stations, television stations or website addresses listed below.
If school is delayed, Louisville Schools will only delay the opening of school for two hours. Rarely, if ever, will school be delayed and then cancelled. All bus runs and school starting times will simply be delayed by two hours. There will be no AM or PM pre-school when school is delayed for two hours, and no breakfast will be served in any of the school cafeterias.
A two hour delay simply means that school will be delayed by two hours in the morning. Students should report to school or to their bus stop two hours later than their normal times. Buses will "pick-up" students two hours later than the normal pick-up times. School buildings will operate on an adjusted schedule, and students will be dismissed from school at their regular times and at their regular bus return trip times.
Once school is in session, Louisville Schools will very rarely, if ever, send children home from school early. We will make every effort to keep classes in session until the regular release times.
Parents have the right to keep their children at home if they believe weather conditions are hazardous and school is not cancelled. In those situations, please contact the school your child attends and report the absence, in addition to sending a written note with your child on the day they return to school.
This year, school districts in Ohio are limited to five "snow days or calamity days." Any days missed due to hazardous weather conditions or "calamity" reasons beyond five days must be made up.
Please do not call the school buildings for delay or closing information. Any scheduled or evening events, athletic practices, game cancellations, etc. will be announced later that day on the same communication networks listed above (radio and television stations) and will also be posted on the school district and other websites.
If you have any questions or concerns, or if you need additional information, please contact the school your child attends and speak with the building principal or contact the district offices at (330) 875-1666.


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HS2 GETS GREEN LIGHT FOR NEXT MAJOR PHASE OF DESIGN AND CONSTRUCTION WORKS
Industry leaders have warmly welcomed the green light given to the next major phase of HS2, describing it as a huge boost for both the construction industry and the wider supply chain, and the recovery of the economy post-coronavirus.
The department for transport (DfT) has given approval for HS2 Ltd to issue Notice to Proceed (NtP) to the four Main Works Civils Contractors working on the project – SCS Railways, Align JV, EKFB JV, and BBV JV – to commence full detailed design and construction of Phase One of HS2 (the route from London – West Midlands).
To date, the contracts between HS2 Ltd and the Main Works Civils Contractors (MWCC) for Phase One have focused on undertaking scheme design and site preparation (including pricing and scoping) for the civils work. The issuing of NtP marks the point in the MWCC contracts where the work transitions from scheme design and preparatory work, to full detailed design and construction.
Preparatory works are due to start imminently with main construction to commence in earnest in the second quarter of 2021. All sites will be fully compliant with UK government guidance for safe working practices in relation to social distancing.
HS2 Minister Andrew Stephenson said: "While the government's top priority is rightly to combat the spread of coronavirus, protect the NHS and save lives, we cannot delay work on our long-term plan to level up the country. HS2 will be the spine of the country's transport network, boosting capacity and connectivity while also rebalancing opportunity fairly across our towns and cities.
"Following the decision earlier this year to proceed with the project, this next step provides thousands of construction workers and businesses across the country with certainty at a time when they need it and means that work can truly begin on delivering this transformational project."
Hannah Vickers, chief executive of the Association for Consultancy and Engineering (ACE), said: "Given the uncertain times we live in, the scale and scope of the major project means that this decision by the government will bring clarity and relief to thousands of companies while allowing work to begin in earnest. We have not had much to celebrate recently as an industry, but the renewed support for HS2 is proof that construction projects will be vital to kickstarting the economy post-coronavirus."
The four work packages are for full detailed design and construction of Phase One of the HS2 railway. Through these contracts, small and medium businesses have the guarantee of a pipeline of activity for the future, helping to protect jobs and boost certainty for them in the current climate.
The joint ventures, originally awarded contracts by HS2 Ltd in July 2017, are:
SCS Railways (Skanska Construction UK Ltd, Costain Ltd, STRABAG AG);
Align JV (Bouygues Travaux Publics SAS, a subsidiary of Bouygues Construction, Sir Robert McAlpine and VolkerFitzpatrick, a subsidiary of VolkerWessels UK);
EKBF JV (Eiffage Genie Civil SA, Kier Infrastructure and Overseas Ltd, BAM Nuttall, Ferrovial Agroman);
BBV JV (Balfour Beatty Group Ltd, VINCI Construction Grands Projets, VINCI Construction UK Ltd, VINCI Construction Terrassement).
Mark Thurston, CEO of HS2 Ltd, said: "In these difficult times, today's announcement represents both an immediate boost to the construction industry - and the many millions of UK jobs that the industry supports - and an important investment in Britain's future: levelling up the country, improving our transport network and changing the way we travel to help bring down carbon emissions and improve air quality for the next generation."
Leo Quinn, Balfour Beatty Group chief executive, said: "HS2 is a transformational infrastructure scheme, underpinning both the economic resilience and future growth of the nation, during these unprecedented and challenging times. Notice to Proceed provides certainty for many businesses up and down the UK, and will drive investment in skills and capability for current and future generations."
Civil engineering contractors welcomed the news, and CECA director of external affairs, Marie-Claude Hemming, said: "The scheme will act as the backbone of our recovery and will not only deliver transformational improvements to capacity and connectivity across the UK, but will support many thousands of jobs. HS2 is vital to this country's future and will act as a springboard for the infrastructure revolution the government has promised.
Maria Machancoses, director of Midlands Connect, said: "Today's confirmation that detailed design and construction of Phase One of HS2 will begin as soon as this summer offers certainty to the construction industry when it needs it the most. By integrating the network, we will encourage millions more people to make environmentally friendly travel choices, stimulate huge regeneration plans and create tens of thousands of new jobs."
An estimated 400,000 supply chain contract opportunities for UK businesses will be created during Phase One of HS2, supporting thousands of jobs on site and many more around the country. It is estimated that around 95% of those contract opportunities will be won by UK based businesses and around two thirds of those will be small and medium sized businesses.


text:
Q: Web service response compression i'm trying understand what are consequences of enabling compression for web service response.
Web service is implemented as IIS hosted WCF service with http binding. I would like to add compression on http level, so it should be seamless for clients. Soap response size is in 100kb - 1Mb range. We do this to improve user experience on low-bandwidth network. What are  drawbacks?

A: 100kb - 1Mb : Are you sending some form of binary data in your response such as files? In this case you can consider using MTOM which will optimize responses.
If it is only text then you could gain pretty much bandwidth by compressing the response. The only drawback I can think of is that it will not be interoperable (both server and clients need to implement it) and it will probably incur some CPU overhead.

A: In my reading on this topic, what you are proposing is only seamless when the WCF client is Silverlight, because the browser then implements the client-side portion of the compression handshake.  With a standard .net client, you would need additional code in place to make this work.


We are a client-centered and quality-conscious organization that offers a complete portfolio of IT Technology services and solutions.
We guarantee a high degree of confidentiality and security in every project. All client information remains strictly confidential. Acquire Consulting Group will never sell or disclose any client information under any unauthorized circumstances. We have policies and procedures that enable us to do so.


On 25 June The Danish Working Environment Authority (WEA) has made the annual inspection of our company´s working environment.
WEA looks at the safety and health at work in the company and we have received their Green Smiley as we have no issues with the Working Environment Authority. The Green Smiley applies for the next 5 years.
We are pleased that our safety organization is working to continuously improve both the physical and mental working environment.
We send out newsletters every 2 months containing information about new products, new customer cases and general information about the company.


The essential concept or parts of economics are: desires, scale of desire, choice, and opportunity cost. No matter they find yourself deciding to build the fee wont simply be represented by the ultimate price ticket.
At occasions, opportunity nonetheless knocks on my door and I get tempted to present myself some slack.
On this case, its virtue is to remind us that the price of utilizing a useful resource arises from the worth of what it could possibly be used for instead.
We suggest avoiding the meals in the America and Chinese pavilions as you'll find better examples of these cuisines wherever.
An opportunity to do sth There will probably be an opportunity to discuss the matter again next week.