actions.tex

\chapter{Group actions}
\label{ch:actions}

Historically, groups have appeared because they can ``act'' on a set
(or more general objects), that is to say, they collect some of the
symmetries of the set. This is a point of view that we will return to
many times and we give the basic theory in \cref{sec:gsets}.
This section should remind the reader of the material in \cref{cha:circle},
where we dealt with the special case of the group of integers.
More generally, connected \coverings now reappear in the guise of
``transitive $G$-sets'', and these are intimately related to
the set of subgroups of a group.

Also discussed in \cref{sec:gsets} is the notion of ``$G$-torsor''.
A $G$-torsor is a $G$-set that is merely equal to the universal \covering.
The type of $G$-torsors recovers the classifying type of the group $G$,
and this idea is used in~\cref{ch:absgroup} to build the equivalence between our definition of a group and the abstract version taught in most algebra classes.

\section{Brief overview of the chapter}

After setting things up in~\cref{sec:gsets}, and
studying subgroups in~\cref{sec:subgroups},
we introduce the important
operations of taking \emph{invariant maps} and \emph{orbits} of an action
in~\cref{sec:fixpts-orbits}.
We then construct in~\cref{sec:torsors}
the fundamental equivalence between the classifying type $\BG$ of a group $G$
and the type of $G$-torsors.

In~\cref{sec:orbit-stabilizer-theorem,sec:burnsides-lemma} we
begin the study of the combinatorics of group actions.
This allows us to count for instance how many ways there are of ``coloring''
objects acted on by groups,
and it lays the groundwork for the combinatorics of finite groups
we'll be looking at in~\cref{ch:fingp}.

\section{Group actions ($G$-sets)}
\label{sec:gsets}

One of the goals of \cref{sec:Gsetforabstract} below
is to prove that the types of groups and abstract groups are equivalent.
In doing that, we are invited to explore how elements of
abstract groups should be
thought of as symmetries and introduce the notion of a $G$-set.
However, this takes a pleasant detour where we have to explore a
most important feature of groups: they can \emph{act} on things
(giving rise to manifestations of symmetries)!

%\MB{Before we handle the more complex case of abstract groups,
%let us see what this looks like for groups. Leave out?}

\begin{definition}\label{def:Gset}
  For $G$ a group, a \emph{$G$-set} is a function
  \index{group!acting on a set}\index{group action!of $G$-set}
  \[
    X : \BG\to\Set,
  \]
  and $X(\sh_G)$ is referred to as the \emph{underlying set}.
  If $p:x\eqto y$ in $\BG$,
  then the transport function $X(x)\to X(y)$ induced
  by $X(p)\defeq\trp{X}(p) : X(x)\eqto X(y)$ is also denoted by $X(p)$.
  We denote $X(p)(a)$ by $p\cdot_X a$.
  The operation $\cdot_X$ is called the \emph{group action} of $X$.
  When $X$ is clear from the context we may leave out the
  subscript $X$.\footnote{%
    Note that in this case $\cdot: (x\eqto y) \to X(x) \to X(y)$.
    See \cref{def:principaltorsor} for a special case
    where $\cdot_X$ is indeed path composition.}
  In particular, if $g:\USymG$,
  then $X(g)$ is a permutation of the underlying set of $X$.

  The type of $G$-sets is
  \glossary{GSet}{\protect{$\GSet$}}{type of $G$-sets}
  \[
    \GSet\defequi(\BG\to\Set).\qedhere
  \]
\end{definition}
\marginnote{
  Much of what follows will work equally well for $\infty$-groups;
  if $G$ is (a group or) an infinity group,
  a \emph{$G$-type} is a function $X : \BG\to\UU$,
  with \emph{underlying type} $X(\sh_G)$.
  More generally, an action of $G$ on an element of type $A$
  is a function $X : \BG\to A$, see~\cref{sec:actions} below.}
\begin{remark}

The reader will notice that the type of $G$-sets is equivalent to the
type of \coverings over $\BG$.
The reason we have allowed ourselves two names is that our focus is different: for a $G$-set $X:\BG\to\Set$ we focus on the sets $X(z)$, whereas when talking about \coverings the first projection $\sum_{z:\BG}X(z)\to \BG$ takes center stage.  Each focus has its advantages.
\end{remark}

\begin{example}\label{def:principaltorsor}
  If $G$ is a group, then
  \[
    \princ G:\BG\to\Set,
    \qquad\princ G(z)\defequi\pathsp{\sh_G}(z)\defequi(\sh_G \eqto z)
  \]
  is a $G$-set called the \emph{principal $G$-torsor}.\footnote{%
    The term ``$G$-torsor'' will reappear several times and will mean nothing but a $G$-set in the component of $\princ G$ -- a ``twisted'' version of $\princ G$.}
  We've seen this family before in the guise of the (preimages of the) ``universal \covering'' of \cref{def:universalcover}!

  There is nothing sacred about starting the identification
  $\sh_G \eqto z$ at $\sh_G$.
  Define more generally
  \begin{equation}\label{eq:pathsp}
    \pathsp{\blank}:\BG\to\GSet,
    \qquad
    \pathsp{y} \defeq (z \mapsto (y\eqto z)),
  \end{equation}
  Applying $\pathsp{\blank}$ to a path $q:y\eqto y'$
  induces an equivalence from $\pathsp y$ to $\pathsp {y'}$ that sends $p:y \eqto z$
  to $pq^{-1}:y'\eqto z$.
  As a matter of fact, \cref{lem:BGbytorsor} will identify $\BG$ with the type of
  $G$-torsors via the map $\pathsp{\blank}$, simply denoted as $\pathsp{}$,
  using the full transport structure of the identity type $\pathsp y(z)\jdeq(y \eqto z)$.
\end{example}

Note that the underlying set of $\princ G$ is
\[
  \princ G(\sh_G) \jdeq
  \pathsp{\sh_G}(\sh_G) \jdeq
  (\sh_G \eqto \sh_G) \jdeq \USymG,
\]
the underlying symmetries of $G$.
If we vary both ends of the identifications simultaneously,
we get another $G$-set:
\begin{example}\label{def:adjointrep}
  If $G$ is a group, then
  \[
    \Ad_G:\BG\to\UU,\qquad\Ad_G(z)\defequi(z\eqto z)
  \]
  is a $G$-set (or $G$-type) called
  the \emph{adjoint $G$-set (or $G$-type)}.\footnote{%
    Note that $\Ad_G$ also makes sense for $\infty$-groups.
    The name ``adjoint'' comes from how transport works in this case; if $p:y \eqto z$,
    then $\Ad_G(p):(y\eqto y) \equivto (z\eqto z)$ is given by conjugation:
    \[
      \Ad_G(p)(q)\eqto pqp^{-1} \text{ in } z \eqto z.
    \]
    The picture
    \[
      \begin{tikzcd}[ampersand replacement=\&]
        y \ar[r,eqr,"p"]\ar[d,eql,"q"'] \& z \ar[d,eqr,"\Ad_G(p)(q)"] \\
        y \ar[r,eql,"p"'] \& z
      \end{tikzcd}
    \]
    is a mnemonic device illustrating that it couldn't have been different,
    and should be contrasted with the picture for
    $\princ G (p):(\sh_G\eqto y)\equivto (\sh_G\eqto z)$:
    \[
      \begin{tikzcd}[ampersand replacement=\&]
        \sh_G \ar[r,eqr,"{\refl{\sh_G}}"]\ar[d,eql,"q"']
          \& \sh_G \ar[d,eqr,"\princ G(p)(q)"] \\
        y \ar[r,eql,"p"'] \& z.
      \end{tikzcd}
    \]
  }\label{ft:adjoint-transport}
Notice that by the induction principle for the circle,
\[
  \sum_{z:\BG}\Ad_G(z) \jdeq \sum_{z:\BG}(z \eqto z)
\]
is equivalent to the type of (unpointed!) maps $\Sc\to\BG$,
known in other contexts as the \emph{free loop space} of $\BG$,
an apt name given that it is the type of ``all symmetries in $\BG$.''
The first projection $\sum_{z:\BG}\Ad_G(z)\to \BG$ correspond to the function $(\Sc\to\BG)\to\BG$ given by evaluating at $\base$.
\end{example}
\begin{example}
  \label{ex:HomHGasGset}
  Let $G$ and $H$ be groups. Recall that $\Hom(H,G)$ is a set
  (\cref{lem:hom-is-set}). We will define group actions on such
  sets of homomorphisms by moving the shapes of $G$ and $H$ as
  in \cref{exa:conj-concrete}. Reusing the notation
  $\Hom(H,G)$, define for any $x:\BH$ and $y:\BG$
  \[
    \Hom(H,G)(x,y)\defequi \Hom(\mkgroup(\BH_\div,x),\mkgroup(\BG_\div,y)).
  \]
  Alternatively, by \cref{def:pointedtypes} and
  \cref{def:grouphomomorphism}, we have
  \[
    \Hom(H,G)(x,y)\jdeq
%\Copy_{\mkgroup}((\BH_\div,x) \ptdto (\BG_\div,y)). 
    \Copy_{\mkgroup}\bigl(\sum_{f:\BH_\div\to \BG_\div}(y\eqto f(x))\bigr).
  \]
  Thus the type $\Hom(H,G)$ may also be considered to be a $(H\times G)$-set
  \[
    \Hom(H,G) : (\BH\times \BG)\to\Set.
  \]
  
  We shall be particularly interested in the restriction to $G$,
  giving a $G$-set for which we again reuse the notation:
  \[
    \Hom(H,G)(y)\defequi\Hom(H,G)(\sh_H,y) %\jdeq avoid overflow
%\Copy_{\mkgroup}(\sum_{f:\BH_\div\to \BG_\div}(y \eqto f(\sh_H)))
    .\qedhere
  \]
\end{example}
\begin{xca}
  \label{xca:HomZGvsAdG}
  Provide an identification between the $G$-sets
  $\Ad_G$  and $\Hom(\ZZ,G)$
  of \cref{def:adjointrep,ex:HomHGasGset}.\footnote{%
    Hint: This is similar to \cref{ex:Zinitial}:
    identify $\Hom(\ZZ,G)(y)$ with $\sum_{z:\BG}\sum_{p:z\eqto z}(y \eqto z)$
    and use~\cref{lem:contract-away}.}
\end{xca}

\begin{definition}\label{def:map-of-Gsets}
  If $G$ is a group and $X,Y$ are $G$-sets, then a
  \emph{map from $X$ to $Y$} is an element of the type\footnote{%
  Note that, if $f$ is a map from $X$ to $Y$,
  then $f_w(g\cdot_X x) = g\cdot_Y f_z(x)$ for all $z,w:\BG$,
  $x:X(z)$, $y:X(w)$, $g:z\eqto w$. \MB{Make picture.}}
  \[
  \prod_{z:\BG}(X(z)\to Y(z)).
  \]
  \index{map!of $G$-sets}\index{$G$-subset}
  
  A \emph{$G$-subset of $X$} is a map from $X$ to the $G$-set that is
  constant $\Prop$.\footnote{%
  Note that, if $P$ is a $G$-subset of $X$, 
  then $P_z(x) = P_w(g\cdot_X x)$ for all $z,w:\BG$,
  $x:X(z)$, $y:X(w)$, $g:z\eqto w$.}
  The type of all such maps is denoted by
  \[
   \Sub^G_X\defeq \prod_{z:\BG}(X(z)\to\Prop).
  \]
  %\glossary(Gsubsets){\protect{$\Sub^G_X$}}{set of $G$-subsets of $X$} ERROR
  Similarly to \cref{cor:subtype-same-level}, $\Sub^G_X$ is a set.
  If $P$ is a $G$-subset of $X$, then the \emph{underlying $G$-set of $P$},
  denoted by $X_P$, is defined by
  \[
  X_P(z) \defeq \setof{x:X(z)}{P(z)(x)},\quad\text{for all $z:\BG$}.\qedhere
  \]
  \glossary(Gset){$X_P$}{underlying $G$-set of $P$}
\end{definition}

\begin{example}\label{def:trivGset}
  If $G$ is a group and $X$ is a set, then
  \[\triv_G X(z)\defequi X\]
  is a $G$-set.
  Examples of this sort (regardless of $X$) are called \emph{trivial $G$-sets}.
\end{example}

\begin{remark}
  \label{remark:GsetsareGsets}
  A $G$-set $X$ is often presented by focusing on the underlying set $X(\sh_G)$
  and providing it with a structure relating it to $G$ determining
  the entire function $X : \BG\to\Set$.

  More precisely, since $\BG$ is connected, a $G$-set $X : \BG\to\Set$ factors
  through the component
  $\conncomp \Set {X(\sh_G)} \jdeq \sum_{Y:\Set}\Trunc{X(\sh_G) \eqto Y}$
  that contains the point $X(\sh_G)$.
  Since $\BSG_{X(\sh_G)}\jdeq(\conncomp \Set {X(\sh_G)},X(\sh_G))$,
  the $G$-set $X$ can,
  without loss of information, be considered as a homomorphism from $G$ to
  the permutation group $\SG_{X(\sh_G)}$ of $X(\sh_G)$,
  classified by a pointed map
  \[
    \BG\ptdto\BSG_{X(\sh_G)}.
  \]

  The constructions in the previous two paragraphs yields the following equivalences:
  \[
    \GSet \equivto \sum_{X:\Set} (\BG \ptdto \BSG_X)
     \equivto \sum_{X:\Set}\Hom(G,\SG_X).\qedhere
   \]
\end{remark}

\begin{xca}
Show that if $X$ is a type family with parameter type $\BG$ and $X(\sh_G)$ is a set,
then $X$ is a $G$-set.
\end{xca}

\begin{xca}\label{xca:Ad-triv-abelian}
  Prove that a group $G$ is abelian if and only if the $G$-sets $\Ad_G$ and
  $\triv_G(\USymG)$ are identical.
\end{xca}

\begin{xca}\label{xca:Ad-princ-trivial}
  Prove that a group $G$ is the trivial group if and only if the $G$-sets $\Ad_G$ and
  $\princ G$ are identical.
\end{xca}

\subsection{Transitive $G$-sets}
\label{sec:transitiveGsets}
We saw in~\cref{cha:circle} that connected \coverings play a special role:
In the case of the circle, classifying the group of integers $\ZZ$,
they correspond to cycles (\cref{thm:cycset-connS1cover}).

We hinted there that they are connected to subgroups, so
we now study them over a general group $G$.
As $G$-sets they are called transitive $G$-sets.
Classically, a $\abstr(G)$-set (a notion \emph{we} have yet not defined) $\mathcal X$ is said to be \emph{transitive} if there exists some $x:\mathcal X$ such that for all $y:\mathcal X$ there exists a $g:\mathcal X$ with $x=g\cdot y$.  In our world this translates to:
\begin{definition}\label{def:transitiveGset}
  A $G$-set $X:\BG\to\Set$ is \emph{transitive}\index{transitive $G$-set} if the proposition
  \[
    \istrans(X) \defequi
    \exists_{x:X(\sh_G)} \prod_{y:X(\sh_G)} \exists_{g:\USymG} x=g\cdot y
  \]
  holds.
\end{definition}
\begin{remark}
  In other words, $X$ is transitive if and only if there exists
  some $x:X(\sh_G)$ such that the map $\blank\cdot x:\USymG\to X(\sh_G)$ is
  surjective.

  Note also that by connectedness (cf.~\cref{xca:component-connected})
  it is equivalent to demand this over all $z:\BG$:
  \begin{equation}\label{eq:Gset-trans-gen}
    \prod_{z:\BG}\exists_{x:X(\sh_G)}
    \prod_{y:X(\sh_G)}\exists_{g:z\eqto z}a=g\cdot b.
  \end{equation}

  Yet another equivalent way of expressing that $X$ is transitive is to say
  that $X(\sh_G)$ is nonempty and for any $x,y:X(\sh_G)$ there
  exists some $g:\USymG$ with $x = g\cdot y$.
\end{remark}

\begin{lemma}
  \label{lem:conistrans}
  A $G$-set is transitive if and only if the associated \covering is connected.
\end{lemma}
\begin{proof}
  Consider a $G$-set $X:\BG\to\Set$ and the associated \covering
  $f:\tilde X\to\BG$ where $\tilde X\defequi\sum_{y:\BG}X(y)$ and $f$
  is the first projection.  Now, $\tilde X$ is connected if and only
  if there exists a $z:\BG$ and a $x:X(z)$ such that for
  all $w:\BG$ and $y:X(w)$ there exists some $g:z\eqto w$ such that $y=g\cdot x$.
  Since $\BG$ is connected, this is equivalent to asserting that there
  exists some $x:X(\sh_G)$ such that for all $y:X(\sh_G)$ there exists
  some $g:\USymG$ such that $x=g\cdot y$.
\end{proof}

The next lemma is an analog of~\cref{cor:ConnCycles},
but for a general group and transitive \covering
we only get injectivity, not an equivalence.
\Cref{fig:not-normal} illustrates what can go wrong.
We'll study exactly when we get surjectivity in~\cref{sec:normal}
on ``normal'' subgroups.
\begin{marginfigure}
  \noindent\begin{tikzpicture}[scale=.1]
    \node[dot,label=above:$x$] (two)   at (0,10) {};
    \node[dot] (one)   at (0, 6) {};
    \node[dot] (zero)  at (0, 2) {};
    \node[dot] (base)   at (0,-5) {};

    \pgfmathsetmacro\cc{.55228475}% = 4/3*tan(pi/8)
    \pgfmathsetmacro\cy{2*\cc}%
    \pgfmathsetmacro\cx{10*\cc}%
    \pgfmathsetmacro\intx{3.5}%
    \pgfmathsetmacro\inty{1.5}%
    \pgfmathsetmacro\ay{.35165954}%

    % right 3-cycle
    \draw (zero.center) .. controls ++(0,-\cy+\ay) and ++(-\cx,-\ay)
    .. (10,1) .. controls ++(\cx,+\ay) and ++(0,-\cy-\ay)
    .. (20,4)
    \foreach \y in {4,8} {
      .. controls ++(0,\cy + \ay) and ++(\cx,-\ay)
      .. (10,3 + \y) .. controls ++(-\cx,\ay) and ++(0,\cy-\ay)
      .. (0,2 + \y) .. controls ++(0,-\cy+\ay) and ++(-\cx,-\ay)
      .. (10,1 + \y) .. controls ++(\cx,\ay) and ++(0,-\cy-\ay)
      .. (20,4 + \y) }
    .. controls ++(0,+\cc) and ++(\cx,\ay)
    .. (10+\intx,12 + \inty) .. controls ++(-\cx,-\ay) and ++(\cx,\ay)
    .. (10-\intx,2 + \inty) .. controls ++(-\cx,-\ay) and ++(0,\cc)
    .. (zero.center);

    % left 2-cycle
    \draw (one.center) .. controls ++(0,-\cy+\ay) and ++(\cx,-\ay)
    .. (-10,5) .. controls ++(-\cx,+\ay) and ++(0,-\cy-\ay)
    .. (-20,8) .. controls ++(0,\cy + \ay) and ++(-\cx,-\ay)
    .. (-10,11) .. controls ++(+\cx,\ay) and ++(0,\cy-\ay)
    .. (two.center) .. controls ++(0,-\cy+\ay) and ++(\cx,-\ay)
    .. (-10,9) .. controls ++(-\cx,\ay) and ++(0,-\cy-\ay)
    .. (-20,12) .. controls ++(0,+\cc) and ++(-\cx,\ay)
    .. (-10-\intx,12 + \inty) .. controls ++(\cx,-\ay) and ++(-\cx,\ay)
    .. (-10+\intx,6 + \inty) .. controls ++(\cx,-\ay) and ++(0,\cc)
    .. (one.center);

    % left 1-cycle
    \draw (zero.center) .. controls ++(0,\cy) and ++(\cx,0)
    .. (-10,4) .. controls ++(-\cx,0) and ++(0,\cy)
    .. (-20,2) .. controls ++(0,-\cy) and ++(-\cx,0)
    .. (-10,0) .. controls ++(\cx,0) and ++(0,-\cy)
    .. (zero.center);

    % base right
    \draw (base.center) .. controls (0,-5+\cy) and ++(-\cx,0)
    .. (10,-3) .. controls ++(\cx,0) and ++(0,\cy)
    .. (20,-5) .. controls ++(0,-\cy) and ++(\cx,0)
    .. (10,-7) .. controls ++(-\cx,0) and ++(0,-\cy) .. (base.center);
    % base left
    \draw (base.center) .. controls (0,-5 + \cy) and (-10+\cx,-3)
    .. (-10,-3) .. controls (-10-\cx,-3) and (-20,-5 + \cy)
    .. (-20,-5) .. controls (-20,-5 - \cy) and (-10-\cx,-7)
    .. (-10,-7) .. controls (-10+\cx,-7) and (0,-5 - \cy)
    .. (base.center);
  \end{tikzpicture}
  \caption{A $\mkgroup(\Sc\vee\Sc)$-set for which $\protect\ev_x$ is not
   surjective. At the bottom the type $\Sc\vee\Sc$ is visualized as
   two circles with a common base point. }
  \label{fig:not-normal}
\end{marginfigure}

\begin{lemma}
  \label{lem:evisinjwhentransitive}
  Let $X,X':\BG\to\Set$ be $G$-sets. Let $z:\BG$ and $x:X(z)$.
  Suppose that $X$ is transitive. Then the evaluation map
  \[
    \ev_x:(X \eqto X')\to X'(z),\qquad \ev_x(f)\defequi f_z(x)
  \]
  is injective.\footnote{%
    Recall that for type families $X,X':T\to\UU$, and
    $f:\prod_{y:T}(X(y)\to X'(y))$, we may write $f_y:(X(y)\to X'(y))$
    (instead of the more correct $f(y)$) for its evaluation at $y:T$.}
\end{lemma}
\begin{proof}
  In view of function extensionality, our claim is that the evaluation
  map $\ev_x:(\prod_{s:\BG}(X(s)\eqto X'(s)))\to X'(z)$ given by the 
  same formula is injective; that is all $f$s with the same 
  value $f_z(x)$ are identical.

  Fix a value $a:X'(z)$, and consider an $f:X\eqto X'$ with $f_z(x)=a$.
  We will show that $f$ is uniquely determined by $f_z(x)=a$.
  Let $s:\BG$ and  $y:X(s)$. It suffices to show that the value 
  of $f_s(y)$ is independent of $f$. 
  For any $g:z=s$ such that $g\cdot_X x=y$ (which exists by the 
  transitivity of $X$, using \cref{lem:conistrans}) we have
  $f_s(y)=f_s(g\cdot_X x)=g \cdot_{X'} f_z(x)=g \cdot_{X'} a$,
  and the latter value does indeed not depend on $f$.
  Since we try to prove a proposition we are done.
\end{proof}

\begin{xca}\label{xca:not-normal}
Reverse engineer the $\mkgroup(\Sc\vee\Sc)$-set in \cref{fig:not-normal}.
Let's call it $X$. Show that $X\eqto X$ is contractible.
Conclude that $\ev_x$ cannot be surjective.
(Hint: the induction principle for $\Sc\vee\Sc$ is a generalization
of the induction principle for the circle to two loops.)
\end{xca}

\subsection{Actions in a type}
\label{sec:actions}
Oftentimes it is interesting not to have an action on a set, but on an element in any given type (not necessarily the type of sets).  For instance, a group can act on another, giving rise to the notion of the semidirect product in \cref{sec:Semidirect-products}.  We will return these more general types of actions many times.

\begin{definition}\label{action}
  If $G$ is any group\footnote{%
  Even an $\infty$-group in the sense of \cref{sec:inftygps}.} 
  and $A$ is any type of objects,
  then we define an \emph{action} by $G$ in %the world of elements of
  $A$ as a function
  \[
    X : \BG \to A.\qedhere
  \]
\end{definition}

The particular object of type $A$ being acted on is $X(\sh_G):A$,
and the action itself is given by transport.
This generalizes our earlier definition of $G$-sets, $X : \BG \to \Set$.

\begin{definition}\label{std-action}
  The \emph{standard action} of $G$ on its designated shape $\sh_G$ is obtained by
  taking $A \defeq \BG$ and $X \defeq \id_{\BG}$.
\end{definition}

\begin{example}
  By composing constructions we can build new actions
  starting from simple building blocks.
  For example, the standard action of symmetric group $\SG_n$
  is to permute the elements of the standard $n$-element set $\bn n$.
  Composing with the projection $\BSG_n \to \Set$,
  we get the corresponding standard $\SG_n$-set.\footnote{%
    Check that this action is transitive for $n>0$.}
  Composing further with the operation $\blank \to \bool : \Set \to \Set$,
  we get the action of $\SG_n$ on the set of decidable subsets of $\bn n$.
\end{example}

Generalizing~\cref{remark:GsetsareGsets},
notice that the type $\BG \to A$ is equivalent to the type
\[
  \sum_{a:A}\Hom(G,\Aut_A(a)),
\]
that is, the type of pairs of an element $a : A$,
and a homomorphism from $G$ to the automorphism group of $A$.
This equivalence maps an action $X:\BG\to A$
to the pair consisting of $a \defeq X(\sh_G)$
and the homomorphism represented by the pointed map
from $\BG$ to the pointed component $\conncomp A a$ given by $X$.

Because of this equivalence,
we define a \emph{$G$-action on $a:A$}
to be a homomorphism from $G$ to $\Aut_A(a)$.

\section{Subgroups}
\label{sec:subgroups}
In our discussion of the group $\ZZ\defequi\Aut_{\Sc}(\base)$ of integers
in \cref{cha:circle} we discovered that some of the symmetries of $\base$
were picked out by the degree $m$ function $\dg{m}: \Sc\to\Sc$
(for some particular natural number $m>0$, see \cref{def:mfoldS1cover}).  
On the level of the set $\base\eqto{}\base$, the symmetries picked out are
all the iterates (positive or negative or even zero-fold) of $\Sloop^m$.
The important thing is that we can compose or invert any of the iterates
of $\Sloop^m$ and get new symmetries of the same sort (because of
distributivity $mn_1+mn_2=m(n_1+n_2)$). So, while we do not get all
symmetries of $\base$ (unless $m=1$), we get what we'd like to call 
a subgroup of the group of integers.

The case of $m=0$ is special. The iterates of $\Sloop^0$, \ie of
$\refl{\base}$, can also be composed and inverted, never to give
something else than $\Sloop^0$ itself. This is what we'd like to call
the trivial subgroup of the group of integers. 
We can pick out the single symmetry $\Sloop^0$
by the constant map $\cst{\base} : \bn 1 \to \Sc$.

Both $\dg{m}$ and $\cst{\base}$ can trivially be pointed to make them
into classifying maps of homomorphisms that are injections on the
respective sets of symmetries. Using \cref{cor:dgm-conncov},
each $\dg{m}$ is a pointed connected \covering over the circle, and
$\cst{\base}$ is even the universal \covering by 
\cref{lem:univ-cover-of-groupoid}. Finally, \cref{lem:conistrans}
gives yet another equivalent view, namely the of pointed transitive
$G$-sets. This view will now be used for our first formal definition
of the notion of a subgroup of a group $G$.

\subsection{Subgroups through $G$-sets}

The idea is that a $G$-set $X$ picks out those symmetries in $G$
that keep the point of $X(\sh_G)$ in place. For this to work well
we need to point $X(\sh_G)$ and $X$ must be transitive.
%so that the set of symmetries that are picked out is closed under composition and reverse.

\begin{definition}\label{def:set-of-subgroups}
  For any group $G$, define the type of \emph{subgroups of $G$} as
  \index{type!of subgroups of a group}
%  \glossary(SubG){$\protect{\Sub_G}$}{type of subgroups of $G$} ERROR???
  $$\typesubgroup_G\defequi\sum_{X:\BG\to\Set}{\,}X(\sh_G)
  \times\istrans(X).$$
  The \emph{underlying group} of the subgroup $(X,x,!) : \Sub_G$ is
  $$\mkgroup \bigl(\sum_{z:BG}X(z),(\sh_G,x)\bigr).\qedhere$$
\end{definition}

\begin{xca}\label{xca:group-Xx!}
Show that $\sum_{z:BG}X(z)$ above is a connected groupoid.
Hint: use \cref{lem:conistrans}.
\end{xca}

As an example, recall from \cref{def:RmtoS1} the $\Sc$-set
$R_m : \Sc\to\Set$ defined by $R_m(\base) \defeq \bn m$ and
$R_m(\Sloop) \defis \etop\zs$. Here $m>0$ so that we can point
$R_m$ by $0: R_m(\base)$.\footnote{Any element of $\bn m$ would do.} 
Transitivity of $R_m$ is obvious.
Which symmetries $p: \base\eqto\base$ are picked out by $R_m$?
Those that keep the point $0: R_m(\base)$ in place, that is,
those that satisfy $R_m(p)(0)=0$, \ie $p=\Sloop^{mk}$ for some integer $k$.
Given $\alpha_m$ in \cref{con:psi-alpha-m}, it should not come as a 
surprise that these are precisely the symmetries picked out by $\dg{m}$.

The case of $m=0$ connects to another old friend, the $\Sc$-set
$R : \Sc\to\Set$ defined by $R(\base) \defeq \zet$ and
$R(\Sloop) \defis \etop\zs$, see \cref{def:RtoS1}.
Again we point by $0: R(\base)$ and transitivity of $R$ is obvious.
The only symmetry that keeps $0$ in place is $\refl{\base}$,
since $R(\Sloop)= \zs$ iff $k=0$.
Again, no surprise in view of the results in \cref{sec:symcirc}
identifying $R$ as the universal \covering over $\Sc$.

The following result is analogous to the fact that $\Sub_T$ is
a set for any type $T$, see \cref{xca:subtypes-set}. It captures
that the essence of picking out symmetries (or picking out elements
of a type), is a predicate, like $R_m(p)(0)=0$ above.

\begin{lemma}
  \label{lem:SubGisset}%
  The type $\typesubgroup_G$ is a set, for any group $G$.   
\end{lemma}
\begin{proof}
Let $G$ be a group, and let $X$ and $X'$ be transitive $G$-sets with
points $x:X(\sh_G)$ and $x':X'(\sh_G)$. If $f,f': X\eqto X'$,
then both $f$ and $f'$ can be viewed as families of equivalences of
type $X(z) \equivto X'(z)$, parameterized by $z:\BG$,
with $f_{\sh_G}$ and $f'_{\sh_G}$ mapping $x$ to $x'$.
Now \cref{lem:evisinjwhentransitive} applies and we get $f=f'$.
Hence $X\eqto X'$ is a proposition and the lemma is proved.
\end{proof}

\begin{example}
  \label{exa:fix1subSGn}%
Consider the group $\SG_n$ (\cref{ex:groups}\ref{ex:permgroup}) 
for given $n>0$. For any $k:\bn n$, define
the $\SG_n$-set $X_k : \BSG_n  \to\Set$ by $X_k(A,!)\defeq A$ for any
$A:\FinSet_n$. Then $X_k$ is obviously transitive. We point $X_k$ by
$k: X_k(\sh_{\SG_n}) \jdeq \bn n$.\footnote{Here the choice of the point
does matter for the symmetries that are picked out.}
Thus we have $(X_k,k,!):\Sub_{\SG_n}$.
The symmetries that are picked out are those $\pi : \bn n \eqto \bn n$
that satisfy $(\pi \cdot_{X_k}k) = k$.\footnote{%
This uses the alternative notation for the group action of $X_k$
introduced in \cref{def:Gset}.}
In other words, $\pi$ keeps $k$ in place and can be any permutation
of the other elements of $\bn n$. From the next \cref{xca:n-is-ptd-n+1}
we get that the underlying group of each $(X_k,k,!)$ 
is isomorphic to $\SG_{n-1}$.
\end{example}

\begin{xca} \label{xca:n-is-ptd-n+1}
Give an equivalence from the type of
$n$-element sets to the type of pointed $(n{+}1)$-element sets.
Hint: use \cref{xca:finsets-decidable}.
\end{xca}

\begin{xca} \label{xca:A-is-A-1+1}
  For any set $A$ with decidable equality,
  give an equivalence from $A$ to $\sum_{B:\UU}(A\eqto(B+\bn1))$.
\end{xca}

For yet another example, consider the cyclic group $\CG_6$ of order $6$; perhaps visualized as the rotational symmetries of a regular hexagon,  \ie the rotations by $2\pi\cdot m /6$, where $m=0,1,2,3,4,5$.
The symmetries of the regular triangle (rotations by $2\pi\cdot m/3$, where $m=0,1,2$) can also be viewed as symmetries of the hexagon.
Thus there is a subgroup of $\CG_6$ which, as a group, is isomorphic to $\CG_3$.\marginnote{Make a TikZ drawing of the hexagon and triangle inscribe in it.}

\begin{example}
  \label{exa:C3subC6}%
Recall from \cref{ex:cyclicgroups} the definition 
$\CG_6\defeq\Aut_\Cyc(\bn6,\zs)$.
In order to obtain $\CG_3$ as a subgroup we can define
$F : \CG_6 \to \Set$ defined by $F(X,t) \defeq X/2$ for all $(X,t):\BCG_6$,
where $X/2$ is defined in \cref{sec:mthroot} as the quotient of $X$
modulo identifying elements that are an even power of $t$ away from each other.
Clearly, $F$ is a transitive $G$-set.
On symmetries, $F$ maps $\pi : (\bn6,\zs) \eqto (\bn6,\zs)$ to 
$([k] \mapsto [\pi(k)]) : (\bn6,\zs)/2 \eqto (\bn6,\zs)/2$.\footnote{%
The function $[k] \mapsto [\pi(k)]$ is well-defined since
permutations that commute with $\zs$ preserve distance.}
The symmetries $\pi$ satisfying $F(\pi)([0])=[0]$ are the
even powers of $\zs$.\footnote{%
In view of \cref{cor:id-m-cycle}, these symmetries can be visualized 
by the vertices of the regular triangle above.
The same is true for the symmetries picked out by $F(\pi)([1])=[1]$.
Both $F(\pi)([0])=[1]$ and $F(\pi)([1])=[0]$ give the other inscribed
regular triangle.} 
The subgroup that we have defined above is $(F,[0],!) : \Sub_{\CG_6}$.
The underlying group of $(F,[0],!)$ is 
$\mkgroup(\sum_{(X,t):\Cyc_6} X/2, ((\bn6,\zs),[0]))$.
Using $\rho_2: \BCG_3 \ptdto \BCG_6$ from \cref{lem:deg-m-on-Cyc},
and the equivalence between $X/2$ and $\inv\rho_2 (X,t)$ from
\cref{thm:fiber-cdg}, and the equivalence from \cref{lem:sum-of-fibers},
we get an equivalence of the underlying group of  $(F,[0],!)$ and $\CG_3$.
\end{example}

There are other subgroups of $\CG_6$, and in this example they are accounted for simply by the various factorizations of the number $6$.

\subsection{Subgroups as monomorphisms}

We now give a second, equivalent definition of a subgroup,
generalizing the examples $\dg{m}$ and $\cst{\base}$ from
the introduction of this chapter. Recall that both
$\USym\dg{m}$ and $\USym\cst{\base}$ are injective.
Also recall \cref{cor:fib-vs-path}\ref{set-fib-vs-path-point},
which implies that $\USymf$ is injective iff $\Bf$ is a \covering,
for any homomorphism $f$.


%\newcommand{\typemono}{Mono}
\begin{definition}
  \label{def:typeofmono}
  Let $G$ and $H$ be groups. A homomorphism $i: \Hom(H,G)$ is
  a \emph{monomorphism},\index{monomorphism} denoted by $\ismono(i)$,
  %\glossary(isMono){$\protect{\ismono(i)}$}{proposition stating that
  %$i$ is a monomorphism of groups} gives error: culprit \ismono(i)?
  if $\USymi:\USymH\to \USymG$ is an injection  
  (all preimages of $\USymi$ are propositions).
  
  The \emph{type of monomorphisms into $G$}\footnote{%
  The similarity of this type with the type of subtypes
  $\Sub_T \defeq \sum_{S:\UU}\sum_{f:S\to T}\isinj(f)$ in
  \cref{def:subtype} is not coincidental, and the remarks made 
  there in \cref{ft:incl-vs-inj} apply here as well.
  
  In particular, the identity type of $\typemono_G$
  identifies precisely the triples that define the same subgroup,
  namely when their homomorphisms differ by precomposition by an 
  identification of their underlying groups.
  
  \MB{We should add in \cref{ch:absgroup} an equivalence between
  $\typemono_G$ and the subsets of $\USymG$ with the usual closure
  properties --- the ultimate proof that we have the right 
  notion of concrete subgroup.}
  }
  \index{type! of monomorphisms into a group}
  \glossary(MonoG){$\protect{\typemono_G}$}{type of monomorphisms%
  into the group $G$} 
  is
  \[
  \typemono_G\defequi\sum_{H:\typegroup}\sum_{i:\Hom(H,G)}\ismono(i).
  \]
  
  We call $H$ the \emph{underlying group} of $(H,i,!) : \typemono_G$.
  
  A monomorphism $(H,i,!)$ into $G$ is:
      \begin{enumerate}
      \item \emph{trivial}\index{trivial monomorphism} 
      if $H$ is the trivial group;\footnote{This amounts to $Bi$
      being the universal \covering over $BG$, see \cref{def:univ-cover}}
%.contractible (or, equivalently, if $\USymH$ is contractible),
      \item \emph{proper}\index{proper monomorphism} if $i$ is 
      not an isomorphism.\qedhere
      \end{enumerate}
\end{definition}
    
\begin{example}
  \label{ex:SGninSGn+1}
   \marginnote{
     That $i:\SG_2\to\SG_3$ is a monomorphism can visualized as follows:
     if $\SG_3$ represent all symmetries of an equilateral triangle in the
     plane (with vertices $1$, $2$, $3$), then $i$ is represented by the
     inclusion of the symmetries leaving $3$ fixed; \ie reflection through
     the line marked with dots in the picture.
                         $$\xymatrix{&3\ar@{.}[dd]&\\&&\\
             1\ar@{-}[uur]\ar@{-}[rr]&            &2\ar@{-}[uul]}$$}
We will present the subgroups from \cref{exa:fix1subSGn} with
monomorphisms. For each $n:\NN$, consider the  homomorphism 
$i_n:\Sigma_n\to\SG_{n+1}$ of permutation groups
with $\Bi_n$ sending $A:\BSG_n\defequi \FinSet_n$ to $A+\true:\BSG_{n+1}$.
As pointing path we take the reflexivity path.
This is a monomorphism since $\USymi_n:\USym\SG_n\to\USym\SG_{n+1}$ 
is an injection, extending any permutation $\pi$ of $\bn{n}$ to 
a permutation of $\bn{n}+\bn1$ by adding the last element as a
fixed point.

In the picture in the margin we have taken $n=3$ and
$\set{1,2,3}$ for $\bn{3}$. How can we obtain the other proper,
non-trivial subgroups of $\SG_3$? First of all, one should not
expect to find all subgroups through monomorphisms $j:\SG_2\to\SG_3$,
see \cref{xca:C3subSG3}. Using only $\SG_2$, the
two other subgroups can be obtained by varying the pointing
path of $i_3$. These pointing paths are induced by the permutations 
of $\bn{3}$. In \cref{xca:SG2subSG3} you are asked to elaborate each case.
\end{example}

\begin{xca}\label{xca:SG2subSG3}
Calculate $\im(\USymi_3)$ for each pointing path $\pi:\bn{3}\eqto\bn{3}$.
\end{xca}

\begin{xca}\label{xca:C3subSG3}
Define monomorphisms $j,j':\CG_3\to\SG_3$ such that $\USymj\neq \USymj'$
while $(\CG_3,j,!)$ and $(\CG_3,j',!)$ can be identified.
\end{xca}

\begin{example}
  \label{ex:prodinclismono}
  If $G$ and $H$ are groups, then $i_G : G \ptdto (G\times H)$ with
  $\Bi_G(z) \defeq (z,\sh_H)$, pointed by reflexivity, is a monomorphism:
  $\USymi_G$ maps $g:\USymG$ to $(g,\sh_H)$ and is obviously injective.
  We call $i_G$ the \emph{first inclusion} and we have a similar
  \emph{second inclusion} $i_H : H \ptdto (G\times H)$.
  % More generally, if $i:\Hom(H,G)$ is a homomorphism for which there (merely) exists a homomorphism $f:\Hom(G,H)$ such that $\id_H=fi$, then $i$ is a monomorphism.
  \end{example}


\begin{lemma}\label{lem:SubG=MonoG}
  Let $G$ be a group.
  The map sending $(X,\pt,!) : \Sub_G$ to the monomorphism classified by  
  $\fst: (\sum_{z:\BG}X(z),(\sh_G,\pt))\ptdto\BG$, pointed by
  reflexivity, yields an equivalence from $\Sub_G$ to $\typemono_G$.
\end{lemma}

\begin{proof}
   The inverse equivalence is $E$ defined as follows:
 $$E:\typemono_G\to\Sub_G,\qquad 
   (H,i,!)\mapsto E(H,i,!)\defequi (\Bi_\div^{-1},(\sh_H,\Bi_\pt),!),$$
 %\glossary(E){$E$}{equivalence from $\typemono_G$ to $\\Sub_G$}
  where the monomorphism $i:\Hom(H,G)$ is given by 
  the pointed map $(\Bi_\div,\Bi_\pt):\BH\ptdto\BG$.
  The preimage function $\Bi_\div^{-1}:\BG\to\Set$ is a transitive $G$-set 
  since $i$ is a monomorphism, and $(\sh_H,\Bi_\pt):\Bi_\div^{-1}(\sh_G)
  \defequi \sum_{x:\BH}(\sh_G\eqto{}\Bi_\div(x))$.
\end{proof}

 \begin{example}\label{exa:EforSG3}
  In this example we explain how the equivalence between
  $\typemono_G$ and $\Sub_G$ works in the special case
  $G\jdeq\SG_3$ and with two versions of the same subgroup,
  both elaborated above.
  
  Recall $(\SG_2,i_3,!):\typemono_{\SG_3}$ with
  $i_3: \SG_2\ptdto\SG_3 : B\mapsto (B+\true)$ from \cref{ex:SGninSGn+1}.
  The preimage function $\inv\Bi_3$ maps any $A:\BSG_3$
  to $\sum_{B:\BSG_2}(A\eqto{}(B+\true))$.
  In particular we have $(\bn2,\refl{\bn3}):\inv\Bi_3(\bn3)$
  (recall that $i_3$ is pointed by reflexivity).
   
  We have $E(\SG_2,i_3,!)\jdeq(\inv\Bi_3,(\bn2,\refl{\bn3}),!)$.
  Going back as in \cref{lem:SubG=MonoG} we get 
  $(\sum_{A:\BSG_3}\inv\Bi_3(A),\fst,!)$. Using \cref{lem:sum-of-fibers}
  one sees that, indeed, the latter monomorphism 
  can be identified with $(\SG_2,i_3,!)$.
  
  Why do we say that $(X_3,3,!):\Sub_{\SG_3}$ from \cref{exa:fix1subSGn}
  defines the same subgroup as $(\SG_2,i_3,!):\typemono_{\SG_3}$
  from \cref{ex:SGninSGn+1}? The reason is that they pick out
  the same symmetries in $\SG_3$, as argued in these examples.
  Moreover, $(X_3,3,!)$ and $E(\SG_2,i_3,!)$ can be identified.
  Note that $X_3(A,!) \jdeq A$ and 
  $\inv\Bi_3 \jdeq \sum_{B:\BSG_2}(A\eqto{}(B+\true))$.
  Now apply \cref{xca:A-is-A-1+1} and verify that the points correspond.
  \cref{lem:E-preserves-symms} below offers a general result of this kind.
  \end{example}

\begin{xca}\label{xca:SubG=MonoG}
Complete the details of the proof of \cref{lem:SubG=MonoG} above using 
\cref{cor:fib-vs-path}\ref{set-fib-vs-path-point},
\cref{lem:sum-of-fibers},
\cref{lem:conistrans}.
\end{xca}

\begin{corollary}\label{lem:setofsubgroups}
Let $G$ be a group. Then $\typemono_G$ is a set since $\Sub_G$ is.
\end{corollary}

The following lemma states that the equivalences in \cref{lem:SubG=MonoG} preserve the subsets of symmetries that are picked out.
\begin{lemma}\label{lem:E-preserves-symms}
  Let $G$ be a group and $g:\USymG$ a symmetry.
  For all $(X,\pt,!) : \Sub_G$ and $(H,i,!) : \typemono_G$
  that correspond via the equivalences in \cref{lem:SubG=MonoG},
  we have $X(g)(\pt) = \pt$ if and only if there exists $h:\USymH$
  such that $g = \USymi(h)$.
\end{lemma}
\begin{proof}
It suffices to prove this for one of the equivalences.
We choose $\inv E$, that is, the equivalence in the statement
of \cref{lem:SubG=MonoG}, so we take 
$(H,i,!) \jdeq ((\sum_{z:\BG}X(z),(\sh_G,\pt)),\fst,!)$.
Now we have to prove: $X(g)(\pt)=\pt$ iff there exists an
$h:(\sh_G,\pt)\eqto(\sh_G,\pt)$ such that $g = \USym\fst(h)$.

If $X(g)(\pt)=\pt$, then we can simply take $h\defeq(g,\pt)$.

For the converse, assume there exists an
$h:(\sh_G,\pt)\eqto(\sh_G,\pt)$ such that $g = \USym\fst(h)$.
Then $h = (g,p)$ for some $p: X(g)(\pt)=\pt$.
\end{proof}

\marginnote{%
  Which of the equivalent sets $\typemono_G$ and $\Sub_G$ is allowed to be called ``the set of subgroups of $G$'' is, of course, a choice.  It could easily have been the other way around and we informally refer to elements in either sets as ``subgroups'' and use the given equivalence $E$ as needed.
}%
\marginnote{%
  An argument for our choice can be
  as follows.  In set-based mathematics one has two options for defining subgroup: either as a certain subset (uniquely given by its characteristic function to $\Prop$) or as an equivalence class of injections (taking care of size issues since the class of monomorphisms will not form a small set).  The former is the usual choice and is the one we model here with $\Sub_G$, whereas the other corresponds to $\typemono_G$.
  % that the identity type in $\typesubgroup_G$ seems more transparent than the one in $\typemono_G$  (``more things are equal'' in $\typemono_G$?), just as  $A\to\Prop$ gives more the intuition of picking out a subset by means of a characteristic function than what you get when considering the equivalent type of injections into $A$.
}


Through the equivalence $E$ we can translate the concepts in
\cref{def:typeofmono} to subgroups in $\Sub_G$. 
First, observe that the underlying groups correspond. 
Then, we say that a subgroup $(X,\pt,!):\Sub_G$ is:
      \begin{enumerate}
      \item \emph{trivial}\index{trivial subgroup} if the underlying subgroup
      $(\sum_{z:\BG}X(z),(\sh_G,\pt))$ is trivial;
      \item \emph{proper}\index{proper subgroup} if $X(\sh_G)$ is not
      contractible.
      \end{enumerate}

      \begin{remark}
      \label{rem:notationsubgroup}
      A note on classical notation is in order.
If $(X,\pt,!)$ is a subgroup corresponding to a monomorphism $(H,i,!)$ into a group $G$, tradition would permit us to relax the burden of notation and we could write ``a subgroup $i:H\subseteq G$'', or, if we didn't need the name of $i:\Hom(H,G)$, simply ``a subgroup $H\subseteq G$'' or ``a subgroup $H$ of $G$''.
    \end{remark}
    

    {\large cursor 1}
    
    \subsection{Move to better place}
    
    % the ``subsymmetries'' formed a very organized structure.
% For each natural number $n$ we obtained a set of subsymmetries in the identity type $\base=\base$, namely the set of all the iterates $(\Sloop^{n})^m$ where $m$ varies over the integers.
% When $n$ was positive this was realized as the $n$-fold \covering of $S^1$, when $n=0$ this was given by the universal \covering.


The other extreme of the idea of a subgroup was exposed in \cref{sec:groupssubperm} in the form of the slogan ``any symmetry is a symmetry in $\Set$''.
By this we meant that, if $G \defequi \Aut_A(a)$ is a group, we produced a monomorphism $\rho_G:\Hom(G,\Aut_{\USymG}(\Set))$,
\ie any symmetry of $a$ is uniquely given by a symmetry (``permutation'') of the set $\USymG\defequi (a\eqto{}a)$.

For many purposes it is useful to define ``subgroups'' slightly differently.
A monomorphism into $G$ is given by a pointed connected groupoid  $\BH=(\BH_\div,\pt_H)$, a function $F:\BH_\div\to\BG_\div$ whose fibers are sets (a \covering) and an identification $p_f:\sh_G\eqto{}F(\sh_H)$.  There is really no need to specify that $\BH_\div$ is a groupoid: if $F:T\to \BG$ is a \covering, then $T$ is automatically a groupoid.

On the other hand,  the type of \coverings over $\BG$ is equivalent to the type of $G$-sets: if $X:\BG\to\Set$ is a $G$-set, then the \covering is given by the first projection $\tilde X\to \BG$ where $\tilde X\defequi\sum_{y:\BG}X(y)$ and the inverse is obtained by considering the fibers of a \covering.  Furthermore, we saw in \cref{lem:conistrans} that $\tilde X$ being connected is equivalent to the condition $\istrans(X)$ of \cref{def:transitiveGset} claiming that the $G$-set $X$ is transitive.

Hence, the type (set, really) $\typemono_G$ of monomorphisms into $G$ is equivalent to the type of pointed connected \coverings over $\BG$, which again is equivalent to the type $\typesubgroup_G$ of transitive $G$-sets $X:\BG\to\Set$ together with a point in $X(\sh_G)$.

and the \emph{type of epimorphisms from $G$}\index{type! of epimorphisms from a groups}\glossary(EpiG){$\protect{\typeepi_G}$}{type of epimorphisms from the group $G$} is
  $$\typeepi_G\defequi\sum_{H:\typegroup}\sum_{f:\Hom(G,G')}\isepi(f).$$
  
  
  \begin{exercise}
      \begin{enumerate}
      \item Show that $i:\Hom(H,G)$ is a monomorphism if and only if $\USymi$ is an injection of sets and that $i$ is proper if and only $Ui$ is not a bijection.
      \item Show that $f:\Hom(G,G')$ is an epimorphism if and only if $\USymf$ is an surjection of sets.
      \item Consider a composite $f=f_0f_2$ of homomorphisms.  Show that $f_0$ is an epimorphism if $f$ is and $f_2$ is a monomorphism if $f$ is.\qedhere
      \end{enumerate}
    \end{exercise}
    
    If $G_1$ and $G_2$ are groups, then the first projection from $G_1\times G_2$ is an epimorphism 

\begin{xca}
  \label{xca:SG2=SG2contractible} \MB{Move or shorten?}
  Show that $\Iso(\SG_2,\SG_2)$ is contractible. Hint: use
  \cref{xca:CG2isSG2}. Here comes another interesting solution.
  Let $f:\Iso(\SG_2,\SG_2)$. Then $\USym f: (\bn2\eqto\bn2)\equivto_*
  (\bn2\eqto\bn2)$. There is exactly one such pointed equivalence,
  namely the identity on $\bn2\eqto\bn2$. Now show that 
  \[
  \sum_{p:A\eqto \Bf(A)}\,\prod_{q: A\eqto\bn{2}} 
  (\Bf^{-1}_\pt\cdot \ap{\Bf}(q)\cdot p) = q
  \]
  is contractible for all $A:\FinSet_2$, which induces an
  identification of type $\id_{\SG_2}\eqto f$.
  
  This argument is a special case of a technique that we
  will develop in \cref{sec:homabsisconcr} and which is called `delooping'.
\end{xca}


\begin{example}
  \label{ex:prodinclisGset}
  We saw in \cref{ex:prodinclismono} that the first inclusion $i_1:G\to G\times G'$ is a monomorphism.
  The corresponding $G\times G'$-set is the composite of the first projection $\mathrm{proj}_1:\BG_\div\times\BG'_\div\to \BG_\div$ followed by the principal $G$-torsor $\princ G:\BG\to\Set$.\MB{Later!}

  More generally, if $i:\Hom(H,G)$ and $f:\Hom(G,H)$, and $fi\eqto{}\id_H$, then $(H,i,!):\typemono_G$, corresponding to the subgroup with $G$-set given by the composite of $\Bf$ with the princial $H$-torsor $\princ H$.
\end{example}



{\large cursor 2}

    % commented out by BID 211117 Some examples and references should be included when the cyclic subgroups are fully developed
    % \subsection{The geometry of subgroups: some small examples}\footnote{this subsection is not touched: it needs attention}
% \label{smallsubgpex}

% As a teaser, and in order to get a geometric feel for the subgroups and their intricate interplay, it can be useful to have some fairly manageable examples to stare at.
% Some of the main tools for analyzing the geometry of subgroups are collected in \cref{sec:fingp} on finite groups, and we hope the reader will be intrigued by our mysterious claims and go on to study \cref{sec:fingp}.
% That said, the examples we'll present are possible to muddle through by hand without any fancy machinery, but brute force is generally not an option and even for the present examples it is not something you want to show publicly.

% When presenting the subgroups of a group $G$, three types are especially revealing: the set of subgroups $\typesubgroup_G(\sh_G)$, the \emph{groupoid of subgroups} $\typesubgroup(G)\defequi\sum_{y:\BG}\typesubgroup_G(y)$ and what we for now call the ``set of normal subgroups'' $\prod_{y:\BG}\typesubgroup_G(y)$.   Our local use of ``normal subgroup'' is equivalent to the official definition to come.

% The first projection $\typesubgroup(G)\to \BG$ is referred to as the \emph{\covering of subgroups}.

% \footnote{Write out and fix the concrete examples (cyclic groups and $\Sigma_3$) commented out}
% % \begin{remark}
% % In  \cref{cha:circle} we studied the subgroups of the group of integers $G\eqto{}\ZZ$ through \coverings over the circle $S^1$ (which we showed was equivalent to $B\ZZ$).
% % We discovered a subgroup $n\ZZ$ for each natural number $n:\NN$ and in the groupoid $\typesubgroup({\ZZ})$ these sit as elements in separate components.  Each of these components are contractible (because addition is commutative: $\ZZ$ is an abelian group).

% % In general, a component $K$ of the groupoid $\sum_{y:\BG}\typesubgroup_G(y)$ of subgroups of a group $G$ may be much more interesting. For one thing the, $K$ can contain many subgroups in the sense that the preimage of the first projection $K\to \BG$ is a set that may have many different elements; each representing a subgroup.  However, this set of subgroup will be a \emph{conjugacy class} of subgroups: the different subgroups are related by the conjugation action of $G$.

% % If $G$ is abelian this action is trivial, and $\sum_{y:\BG}\typesubgroup_G(y)$ consists of contractible components indexed over the subgroups of $G$.  Otherwise different subgroups may live in the same component of the groupoid of subgroups -- we'll see examples in a moment.

% % In addition, the components will not in general be contractible, revealing the symmetries of the subgroups under the conjugation action.
% % \end{remark}


% % \begin{example}
% %   The trivial group only has itself as a subgroup; the groupoid of subgroups and the set of normal subgroups are singletons.
% % \end{example}
% % \begin{example}
% %   The cyclic group $C_p$ of prime order $p$ has only two subgroups, the trivial and the full subgroup itself and both are normal.  In fact, all subgroups of abelian groups are normal.

% % In general, the cyclic group $C_n$ of order $n$ has exactly one subgroup for each divisor $i$ of $n$.
% % \end{example}


% % \begin{example}
% %   The group $C_2\times C_2$ has has no less than five subgroups: the trivial one, three subgroups that as groups (as opposed as \emph{sub}groups) are equivalent to $C_2$ and the full group $C_4$ itself.
% % \end{example}
% % \begin{remark}
% %   The permutation group $\Sigma_3$ has four nontrivial proper subgroups.  Three conjugate subgroups isomorphic as groups to $C_2$ and one normal one which is as a group is isomorphic to $C_3$.  The component containing the copies of $C_2$ is equivalent to a circle.
% % \end{remark}

\section{Invariant maps and orbits}
\label{sec:fixpts-orbits}
We now return to some important constructions involving $G$-sets for a group $G$.
However, since they make equally good sense for \emph{$G$-types} for \aninftygp
$G$, we'll mostly work in that generality.\footnote{\MB{MB is for
moving all $\infty$ stuff to the margin, not having to worry
about $\USymG$, $=$, \cref{lem:conistrans}, ...}}
\begin{definition}
  \label{def:actiontype}
  If $X : \BG\to\UU$, then the \emph{action type}\index{action type}
  is the total type of $X$, denoted\footnote{%
    The superscripts and subscripts are decorated with ``$hG$'',
    following a convention in homotopy theory.
    This helps to distinguish them from other uses, such as powers.
    The action type is sometimes denoted $X \dblslash G$.}
\[
  X_{hG} \defeq \sum_{z:\BG} X(z).
\]
The type of \emph{invariant maps}\footnote{%
These are dependent functions $f$ and the reason for the new name
in this context is that $(g \cdot f(z))\eqto f(z)$ for any $z:\BG$
and $g:z\eqto z$.}
\index{invariant map type} is
\[
  X^{hG} \defeq \prod_{z:\BG} X(z).
\]
The \emph{set of orbits}\footnote{\MB{Where: terminology homotopy orbit spaces?}}
\index{set!of orbits}\index{orbit set} is the subset of $\Sub^G_X$ consisting
of all $G$-subtypes $P$ of $X$ such that the underlying $G$-subtype $X_P$
is transitive:
\[
  X / G \defeq \sum_{P:\Sub^G_X}\istrans(X_P).
\]
We say that the action is \emph{transitive}\index{transitive group action}
if $X / G$ is contractible.
\end{definition}

\begin{xca}
  Show that the above notion of transitive coincides with the one we introduced
  in \cref{def:transitiveGset} for a $G$-set $X$ for an ordinary group $G$:
  that $X/G$ is contractible exactly encodes that there is just one ``orbit'':
  there is some $x:X(\sh_G)$ so that for any $y:X(\sh_G)$
  there is a $g:\USymG$ such that $x=g\cdot y$.
\end{xca}

We have seen many instances of action types before:
When $G$-sets are considered as \coverings $f : A \to \BG$,
they are the domains $A$.
Recall for example~\cref{fig:two-comp-S1-cover},
showing an action of $\ZZ$ on $\set{1,2,3,4,5}$ with no invariant maps
and an action type equivalent to a sum of two circles.
In~\cref{fig:ZZ-set-orbits}, we show a similar $\ZZ$-set,
with underlying set $\set{0,1,2,3,4,5}$, three orbits,
and $5$ corresponding to the only invariant map.\footnote{%
Sending $\base$ to $5$ and $\Sloop$ to $\refl{5}$.}

\begin{marginfigure}
  \begin{tikzpicture}[scale=.15]
    \node (Sc) at (0,-5) {$\B\ZZ$};
    \node[dot,label=left:$5$] (five)   at (-10,30) {};
    \node[dot,label=left:$4$] (four)   at (-10,22) {};
    \node[dot,label=left:$3$] (three)  at (-10,18) {};
    \node[dot]                (base)   at (-10,-5) {};
    \node[label=left:$\Sloop$] (Sloop) at (10,-5) {};

    \pgfmathsetmacro\cc{.55228475}% = 4/3*tan(pi/8)
    \pgfmathsetmacro\cy{2*\cc}%
    \pgfmathsetmacro\cx{10*\cc}%
    \pgfmathsetmacro\intx{3.5}%
    \pgfmathsetmacro\inty{1.5}%
    \pgfmathsetmacro\ay{.35165954}%

    \draw (-10,18) .. controls (-10,18 - \cy + \ay) and (-\cx,17 - \ay)
    .. (0,17) .. controls (\cx,17 + \ay) and (10,20 - \cy - \ay) .. (10,20)
    .. controls (10,20 + \cy + \ay) and (\cx,23 - \ay)
    .. (0,23) .. controls (-\cx,23 + \ay) and (-10,22 + \cy - \ay)
    .. (-10,22) .. controls (-10,22 - \cy + \ay) and (-\cx,21 - \ay)
    .. (0,21) .. controls (\cx,21 + \ay) and (10,24 - \cy - \ay)
    .. (10,24)
    .. controls (10,24 + \cc) and (\intx + \cx, 24 + \inty + \ay)
    .. (\intx,24 + \inty) .. controls (\intx - \cx,24 + \inty - \ay)
    and (-\intx + \cx,20 + \ay)
    .. (-\intx,18 + \inty) .. controls (-\intx - \cx,18 + \inty - \ay)
    and (-10,18 + \cc) .. (-10,18);
    \draw[casblue] (-10,2) .. controls (-10,2 - \cy + \ay) and (-\cx,1 - \ay)
    .. (0,1) .. controls (\cx,1 + \ay) and (10,4 - \cy - \ay)
    .. (10,4)
    \foreach \y in {4,8} {
      .. controls (10,\y + \cy + \ay) and (\cx,3 + \y - \ay)
      .. (0,3 + \y) .. controls (-\cx,3 + \y + \ay) and (-10,2 + \y + \cy - \ay)
      .. (-10,2 + \y) .. controls (-10,2 + \y - \cy + \ay) and (-\cx,1 + \y - \ay)
      .. (0,1 + \y) .. controls (\cx,1 + \y + \ay) and (10,4 + \y - \cy - \ay)
      .. (10,4 + \y) }
    .. controls (10,12 + \cc) and (\intx + \cx, 12 + \inty + \ay)
    .. (\intx,12 + \inty) .. controls (\intx - \cx,12 + \inty - \ay)
    and (-\intx + \cx,4 + \ay)
    .. (-\intx,2 + \inty) .. controls (-\intx - \cx,2 + \inty - \ay)
    and (-10,2 + \cc) .. (-10,2);
    \draw (10,-5) .. controls ++(0,\cy) and ++(\cx,0)
    .. (0,-3) .. controls ++(-\cx,0) and ++(0,\cy)
    .. (-10,-5) .. controls ++(0,-\cy) and ++(-\cx,0)
    .. (0,-7) .. controls ++(\cx,0) and ++(0,-\cy) .. (10,-5);

    \draw (10,30) .. controls ++(0,\cy) and ++(\cx,0)
    .. (0,32) .. controls ++(-\cx,0) and ++(0,\cy)
    .. (-10,30) .. controls ++(0,-\cy) and ++(-\cx,0)
    .. (0,28) .. controls ++(\cx,0) and ++(0,-\cy) .. (10,30);

    \node[dot,label=left:$2$,casred] (two)   at (-10,10) {};
    \node[dot,label=left:$1$,casred] (one)   at (-10, 6) {};
    \node[dot,label=left:$0$,casred] (zero)  at (-10, 2) {};
  \end{tikzpicture}
  \caption{A $\ZZ$-set with three orbits and one invariant map.}
  \label{fig:ZZ-set-orbits}
\end{marginfigure}

In~\cref{fig:ZZ-set-orbits} we have highlighted a single component of
the action type in blue (\ie corresponding to an element of the set of orbits),
and we see that it contains a subset of the underlying set,
the three red elements $\set{0,1,2}$.
Such a set is what is traditionally called an orbit.
This connection is emphasized in the following result,
and in particular in \cref{cor:orbit-equiv}.

\begin{lemma}\label{lem:surj-set-orbit}
  Let $G$ be a group and $X : \BG\to\UU$. \MB{$\UU$ or $\Set$?} 
  The map $[\blank] : X(\sh_G) \to X/G$, defined by
  \[
    [x](z,y)\defeq \Trunc{(\sh_G,x)\eqto(z,y)}\quad
    \text{for all $x:X(\sh_G)$ and $(z,y):X_{hG}$},
  \]
  is surjective, and $[x] = [y]$ is equivalent to
  $\exists_{g:\USymG}(g\cdot x = y)$.
\end{lemma}
\begin{proof}\MB{New:}
  By definition, $[x]$ is a $G$-subtype of $X$. Clearly,
  $\Trunc{(z,x)\eqto(w,y)}$ iff $\exists_{g:z\eqto w}(g\cdot x = y)$.
  It follows that the underlying $G$-type $X_{[x]}\jdeq
  (z\mapsto \sum_{y:X(\sh_G)}\Trunc{(\sh_G,x)\eqto(z,y)})$ is transitive,
  so that the map $[\blank]$ is well-defined.
  
  For proving surjectivity, assume an orbit $O:X/G$, \ie
  $O$ is a $G$-subtype of $X$ such that $X_O$ is transitive. 
  We have to show that there exists an $x:X(\sh_G)$ such that $O=[x]$.
  By the connectivity of $\BG$ it suffices to show 
  $O(\sh_G)=_{X(\sh_G)\to\Prop}[x](\sh_G)$ for some $x$.
  Transitivity of $X_O$ means that there exists an $x:X(\sh_G)$
  such that $O(\sh_G,x)$ and for all $y:X(\sh_G)$ such that
  $O(\sh_G,y)$ there
  exists a $g:\USymG$ such that $g\cdot x = y$, \ie $[x](\sh_G,y)$.
  Thus transitivity of $X_O$ warrants the existence of an $x$ that we need.
  
  The last part of the lemma follows using the connectivity of $\BG$.
\end{proof}

\begin{remark}\label{rem:equivalents-of-[x]=[y]}\MB{New:}
Given a $G$-set $X$ and $x,y:X(\sh_G)$,
the following propositions are all equivalent and we may pass from
one to another without mention:
$[x]=_{X/G}[y]$;\quad 
$[x](\sh_G)=_{X(\sh_G)\to\Prop}[y](\sh_G)$;\quad
%$G\cdot x =_{\Sub_{X(\sh_G)}} G\cdot y$;\quad
$\exists_{g:\USymG}(g\cdot x = y)$;\quad
$\Trunc{(\sh_G,x)\eqto(\sh_G,y)}$;\quad 
$[x](\sh_G,y)$;\quad
$[y](\sh_G,x)$. As functions of $x$ and $y$ they all define the equivalence
relation on $X(\sh_G)$ induced be the surjection $[\blank]$. 

\end{remark}

\begin{corollary}\label{cor:orbit-equiv}\MB{New:}
Being surjective, the map $[\blank] : X(\sh_G) \to X/G$ defines\footnote{%
\MB{To do in Chapter 2, then refer to it.}}
an equivalence between the induced quotient of $X(\sh_G)$ and $X/G$.
\end{corollary}

\begin{xca}\label{xca:orbit-set-as-quotient}\MB{New:}
Let conditions be as in \cref{def:actiontype}.
Construct an equivalence between $X/G$ and $\setTrunc{X_{hG}}$.
Hint: use \cref{rem:set-trunc-as-quotient}.
\end{xca}

\begin{definition}\label{def:orbit-stabilizer}
  Let $G$ be a group, $X : \BG \to \Set$ a $G$-set, 
  and $x : X(\sh_G)$ an element.
  \begin{enumerate}
  \item Define the group $G_x \defeq \Aut_{X_{hG}}(\sh_G,x)$.
  Clearly, $\fst:\BG_x \to \BG$ is a set bundle:
  each fiber at $z:\BG$ is a subset\footnote{\label{ft:fib-fst}%
  $\sum_{y:X(z)}\Trunc{(\sh_G,x)\eqto(z,y)}$} of $X(z)$.
  Hence $(G_x,\fst,!):\typemono_G$ is monomorphism into $G$.
  We call the subgroup $G_x$ of $G$ the 
  \emph{stabilizer (sub)group}\index{stabilizer}%
    \index{group!stabilizer} at $x$. The inclusion $\fst$ of $\BG_x$ in
    $\BG$ classifies a monomorphism denoted by $i_x : \Hom(G_x,G)$.
  \item Define $G\cdot x \defeq \setof{y : X(\sh_G)}{[x] =_{X/G} [y]}$
    to be the \emph{orbit}\index{orbit!of $x$} of $x$.\footnote{%
    This terminology relies on the equivalence in \cref{cor:orbit-equiv}.}
    \qedhere
  \end{enumerate}
\end{definition}

\begin{xca}\label{xca:Gx-in-Sub_G}
Find $(X_x,x,!):\Sub_G$ corresponding to $(G_x,\fst,!):\typemono_G$.
\MB{Steer away from $\inv E$??}
\end{xca}



The following lemma states that the orbits of a $G$-set $X$
sum up to its underlying set, with the sum taken over the set
of orbits $X/G$.

\begin{lemma}\footnote{\MB{Replaces \cref{old:splitting into orbits}.
Do we need it?}}
  \label{lem:splitting into orbits}
  The inclusions of the orbits form an equivalence
\[
  (O,p)\mapsto\fst(p) \quad:\quad
  \bigl(\sum_{O:X/G} \inv{[O]} \bigr) \we X(\sh_G).
\]
\end{lemma}
\begin{proof}
Recall that $\inv{[O]}\jdeq \sum_{x:X(\sh_G)}(O=[x])$, 
and then abstract
away the $O$ using \cref{lem:contract-away}.\footnote{%
In fact, for every set $A$ and every equivalence relation on $A$,
the equivalence classes sum up to $A$.}
\end{proof}

There are two possible extreme cases for $G_x$ that are important:
\begin{definition}\label{def:fixed-free}
  Let $X$ be a $G$-set and $x:X(\sh_G)$ an element of the underlying set.
  We say that
  \begin{enumerate}
  \item $x$ is \emph{fixed}\index{fixed}
    if $i_x$ is an isomorphism (so $G_x$ is all of $G$),
  \item and $x$ is \emph{free}\index{free}\index{action!free}
    is $G_x$ is trivial.
  \end{enumerate}
  We say that $X$ itself is \emph{free} if each $x:X(\sh_G)$ is free.
\end{definition}

\begin{lemma}\label{lem:fixed-char}
  Given a $G$-set $X$, an element $x:X(\sh_G)$ is
 fixed if and only if the orbit $G\cdot x$ is contractible,
  \ie $x = g\cdot x$ for all $g:\USymG$.
\end{lemma}
\begin{proof}
  The orbit $G\cdot x$ is the fiber of $\Bi_x : \BG_x \ptdto \BG$
  at $\sh_G$. Since $\BG$ is connected,
  this is contractible if and only if all fibers of $\Bi_x$ are contractible,
  \ie $\Bi_x$ is an equivalence, which in turn is equivalent to $i_x$
  being an isomorphism.
\end{proof}

When $X : \BG \to \Set$ is a $G$-set for an ordinary group $G$, the subset
\[
  \setof{x : X(\sh_G)}{\text{$x$ is fixed}}
\]
is closely related to the type $X^{hG}$ of invariant maps.
If we evaluate an invariant map $f : \prod_{z:\BG}X(z)$ at $\sh_G$
we do indeed land in this subset:
Letting $x\defeq f(\sh_G)$,
and taking the dependent action on paths,
$\apd{f}(g) : \pathover x X g x$,
we can use~\cref{def:pathover-trp} to conclude
$\trp[X] g(x)\jdeq g\cdot x = x$, for all $g:\USymG$.
The following lemma states that, conversely, each fixed $x$ uniquely
determines an invariant map.

\begin{lemma}\label{lem:fixpts-are-fixed}
  Let $G$ be a group\footnote{\MB{We only use that $\BG_\div$ is connected,
  so the lemma holds for $\infty$-groups as well.}}, 
  $X$ a $G$-set, with $X^{hG} \jdeq \prod_{z:\BG}X(z)$
  the set of invariant maps. 
  Evaluation $\ev\defeq(f:X^{hG})\mapsto f(\sh_G)$ at $\sh_G$ gives
  \begin{enumerate}
  \item\label{it:ev-is-inj} 
  an injection of type $\bigl(\prod_{z:\BG}X(z)\bigr) \to X(\sh_G)$, which is
  \item\label{it:ev-is-eq-on-inv} an equivalence of type 
  $\bigl(\prod_{z:\BG}X(z)\bigr) \equivto
  \setof{x : X(\sh_G)}{\text{$x$ is fixed}}.$
  \end{enumerate}
\end{lemma}
\begin{proof}
  Let $x : X(\sh_G)$. We prove that the fiber of $\ev$ at $x$,
  \[
    \inv\ev(x) \defeq \sum_{f : \prod_{z:\BG}X(z)} x=f(\sh_G),
  \]
  is a proposition. Let $(f,!),(g,!) : \inv\ev(x)$. 
  Then it suffices to prove $f=g$, which follows by extensionality
  from $f(\sh_G)=x=g(\sh_G)$ since $\BG$ is connected. This proves
  \ref{it:ev-is-inj}.
  
  For \ref{it:ev-is-eq-on-inv}, assume that
  $x : X(\sh_G)$ is fixed, so $i_x\jdeq\fst:\BG_x \to \BG$
  is an equivalence. This means that $\inv\fst(z)$ is contractible
  for all $z:\BG$. Spelling out $\inv\fst(z)$, using \cref{lem:contract-away},
  identifies each fiber $\inv\fst(z)$ with
  $\sum_{y:X(z)}\Trunc{(\sh_G,x)\eqto(z,y)}$.
  Projecting on the first component of each center of contraction
  gives a invariant map $f$ such that $\Trunc{(\sh_G,x)\eqto(\sh_G,f(\sh_G))}$,
  from which the proposition $x=f(\sh_G)$ follows. This proves that
  $\ev$ is surjective, so an equivalence by \ref{it:ev-is-inj}.
\end{proof}

\begin{xca}\label{xca:Gset-A->B}
Let $G$ be the group $\SG_2\times\SG_2$ and $X$ the $G$-set mapping
any pair $(A,B)$ of $2$-element sets to the set $A\to B$.
Elaborate the action of $G$ on $X(\sh_G)$ and determine the
set of orbits and the set of invariant maps. You can
do the same exercise for the following easier cases first:
the $G$-set that is constant $\bn2\times\bn2$, and
the $\SG_2$-sets $X(\blank,\bn2)$ and $X(\bn2,\blank)$.
\end{xca}

\subsection{The Orbit-stabilizer theorem and Lagrange's theorem}

Recall \cref{def:orbit-stabilizer}.
The classifying type of the stabilizer group $G_x$
is the component of $X_{hG}$ containing the shape $(\sh_G,x)$.
The first projection of a symmetry of $(\sh_G,x)$ is a symmetry of 
$\sh_G$, and the second projection is just a true proposition.
This suggest a simple way for $G_x$ to act on the
symmetries of $\sh_G$, by just ignoring the second projection. 
Then one gets a $G_x$-set whose action type can, as expected,
easily be identified with the orbit of $x$. This is spelled
out in the following construction.

\begin{construction}[Orbit-stabilizer theorem]
\footnote{\MB{Replaces \cref{old:orbitstab}.}}
  \label{thm:orbitstab} Let $G$ be a group.\footnote{%
  \MB{This construction works for $\infty$-groups as well.}}
  Fix a $G$-type $X$ and a point $x : X(\sh_G)$. Define the $G_x$-type
  $\tilde G_x : \BG_x \to \UU$ by $\tilde G_x(z,\blank,!)\defeq(\sh_G\eqto z)$,
  acting on $\tilde G_x(\sh_{G_x})\defeq (\sh_G\eqto\sh_G)$.
  Then we have an equivalence from the action type $(\tilde G_x)_{hG_x}$
  to the orbit $(G\cdot_X x)$ of $x$.
\end{construction}
\begin{implementation}{thm:orbitstab}
  The desired equivalence is the composition of elementary equivalences
  for sums and products, followed by contracting away the variable $z$:
  \begin{align*}
    (\tilde G_x)_{hG_x}
    &\defeq \sum_{u : \BG_x}\tilde G_x(u) \\
    &\equivto \sum_{z:\BG}\sum_{y:X(z)}
    (\settrunc{(\sh_G,x)} =_{X/G} \settrunc{(z,y)})\times (\sh_G \eqto z) \\
    &\equivto \sum_{y:X(\sh_G)}[x] =_{X/G} [y]\quad\defeq\quad(G\cdot_X x).\qedhere
  \end{align*}
\end{implementation}

The above theorem holds for $\infty$-groups $G$ and $G$-types $X$,
but if $\BG$ is a groupoid and $X$ is a $G$-set,
then we can draw some interesting additional conclusions. 
In that case also $\BG_x$ is a groupoid and $(G\cdot_X x)$ is a set.
\cref{thm:orbitstab} then tells us that $(\tilde G_x)_{hG_x}$ is a set.
By the following exercise we get that all stabilizer subgroups $(G_x)_g$
of the $G_x$-set $\tilde G_x$ are trivial,
\ie the action $\tilde G_x$ is free.
\cref{lem:free-pt-char} below will allow us to conclude that
all orbits $(G_x \cdot_{\tilde G_x} g)$ are equivalent.

\begin{xca}\label{xca:set-symms-contractible}\MB{New:}
Let $G$ be a \aninftygp and $X$ a $G$-type. Then the
action type $X_{hG}$ is a set if and only if each
stabilizer group $G_x$ is trivial.
\end{xca}
%solution\begin{proof}
%Any type $T$ is a set iff all identity types $t\eqto_T t$ are contractible.
%Since $\BG$ is connected we get that $X_{hG}$ is a set iff each
%identity type $(\sh_G,x)\eqto (\sh_G,x)$ is contractible, \ie
%each stabilizer group $G_x$ is trivial.\end{proof}

\begin{lemma}\label{lem:free-pt-char}
  Let $G$ be a group.
  Given a $G$-set $X$, an element $x:X(\sh_G)$ is
  free if and only if the (surjective) map
  $(\blank \cdot x) : \USymG \to (G\cdot x)$ is injective
  (and hence a bijection).
\end{lemma}
\begin{proof}
  Consider two elements of the orbit, say $g\cdot x,g'\cdot x$ for $g,g':\USymG$.
  We have $g\cdot x=g' \cdot x$ if and only if $x = \inv{g} g'\cdot x$
  if and only if $\inv{g} g'$ lies in $\USymG_x$.
  \MB{Now apply \cref{xca:connected-trivia} yielding that that 
  $G_x$ is trivial iff $\USymG_x$ is contractible.}
\end{proof}

The above results culminate in the following theorem.

\begin{theorem}[Lagrange's Theorem]\label{thm:lagrange}
\footnote{\MB{Replaces \cref{old:lagrange}.}}
  %\cref{def:set-of-subgroups}
  For all subgroups $(X,x,!):\Sub_G$, the $G_x$-set $\tilde G_x$ acts 
  freely on the set $\USymG$, so all orbits under this action are
  equivalent to $\USymG_x$ via 
  $(\blank \cdot_{\tilde G_x} g): \USymG_x \equivto (G_x\cdot_{\tilde G_x} g)$
  by \cref{lem:free-pt-char}.
\end{theorem}

\MB{Extra (exercise?).}
The subgroup is given by $(X,x,!):\Sub_G$ with $X$ transitive,
so the one and only orbit is $X(\sh_G)$. Using 
\cref{lem:splitting into orbits} (equivalence $\settrunc{\blank}$),
\cref{thm:orbitstab} (equivalence $(y,p)\mapsto(\sh_G,y,p,\refl{\sh_G})$),
we get the following chain of equivalences:
\[
\USymG \equivto \bigl(\sum_{o:{\tilde G_x}/G_x} \inv{[o]} \bigr)
\equivto \bigl(\sum_{p:(\tilde G_x)_{hG_x}} \inv{[\settrunc{p}]} \bigr)
\equivto \bigl(\sum_{y:X(\sh_G)} \inv{[\settrunc{(\sh_G,y,!,\refl{\sh_G})}]} \bigr).
\]
Here $!$ is the proof of the proposition $[x] =_{X/G} [y]$,
which exists since $X$ has only one orbit.
Furthermore, $[\blank]: \USymG \to {\tilde G_x}/G_x$ is defined by
$[g]\defeq \settrunc{(\sh_G,x,\trunc{\refl{(\sh_G,x)}},g)}$.
Now, $\inv{[\settrunc{(\sh_G,y,!,\refl{\sh_G})}]}$ is easily
identified with $\sum_{g:\USymG}((\sh_G,x)\eqto(\sh_G,y))$.
For $y\jdeq x$ the type $(\sh_G,x)\eqto(\sh_G,y)$ is $\USymG_x$, 
but in general this is problematic. \MB{(Finite case? Only merely? AC?)}. 

\subsection{Not used}
Thus, both the underlying set $X(\sh_G)$ and the action type
$X_{hG}$ have equivalence relations with quotient set $X/G$.\footnote{%
  This also justifies the notation $X/G$.
  We have a diagram of surjective maps:
  \[
    \begin{tikzcd}[ampersand replacement=\&]
      X(\sh_G) \ar[rr,"{x\mapsto(\sh_G,x)}"]\ar[dr,"{[\blank]}"']
      \& \& X_{hG}\ar[dl,"{\settrunc\blank}"] \\
      \& X/G \&
    \end{tikzcd}
  \]}
The equivalence classes of both are important:



Note that the classifying type $\BG_x$ of $G_x$
is identified with the fiber of $\settrunc{\blank} : X_{hG} \to X/G$,
and $G\cdot x$ (pointed at $x$)
is identified with the fiber of $[\blank] : X(\sh_G) \to X/G$,
both taken at $[x]$, the orbit containing $x$.

Also, the base point of $\BG_x$ depends on the choice of $x$,
but the underlying type $(\BG_x)_\div$, \MB{being a connected component,}
only depends on $[x]:X/G$.
Thus, we can decompose our diagram by writing $X(\sh_G)$ and $X_{hG}$
as sums of the respective fibers.\footnote{%
  Yielding the diagram
  \[
    \begin{tikzcd}[ampersand replacement=\&,column sep=tiny]
      \displaystyle\sum_{o:X/G} \inv{[o]} \ar[rr]\ar[dr,"\fst"']
      \& \& \displaystyle\sum_{o:X/G} \inv{\settrunc{o}}\ar[dl,"\fst"] \\
      \& X/G, \&
    \end{tikzcd}
  \]
  Indeed, $\inv{[[x]]}\jdeq(G\cdot x)$
  and $\inv{\settrunc{[x]}}\equivto \BG_x$ via identity on $X_{hG}$.
  \MB{Do we still need this?}}
  




\section{The classifying type is the type of torsors}
\label{sec:torsors}
Recall the definition of the principal $G$-torsor 
$\princ G \jdeq (\sh_G \eqto \blank)$ from \cref{def:principaltorsor}.
In this section we elaborate the concept of torsor and give one example
of its use.
In \cref{sec:Gsetforabstract} we'll use torsors to prove that the type of groups and the type of abstract groups are equivalent by classifying abstract groups via their pointed connected groupoid of torsors.  To see how this might work it is good to start with the case of a (concrete) group $G$.
In the end we want the torsors of $\abstr(G)$ to be equivalent to $\BG$, so to get the right definition we should first explore what the torsors of $G$ look like and prove~\cref{lem:BGbytorsor}, showing that $\BG$ is equivalent to the type of $G$-torsors.
\begin{definition}\label{def:Gtorsor}
  Given a group $G$, the type of \emph{$G$-torsors}%
  \index{torsor@torsor}%
  \glossary(TorsorG){$\protect\typetorsor_G$}{the type of $G$-torsors}%
  \footnote{This works equally well with $\infty$-groups: $G$-torsors are in that case $G$-types in the component of the principal torsor $\princ G:\BG\to\UU$. There is no conflict with the case when the $\infty$-group $G$ is actually a group since then any $G$-type in the component of the principal $G$-torsor will be a $G$-set.}
  is
  \[
    \typetorsor_G\defequi\sum_{X:\GSet}\Trunc{\princ G \eqto X},
  \]
  where $\princ G \jdeq (\sh_G \eqto \blank)$ is the 
  principal $G$-torsor of \cref{def:principaltorsor}.
\end{definition}

\begin{xca}\label{xca:torsor=free+transitive}
  Show that a $G$-set is a $G$-torsor if and only if it is free and transitive.
\end{xca}

\begin{remark}
  For $G$ a group, the type of $G$-torsors is just another name for the component of the type of \coverings over $\BG$ containing the universal \covering.

  Observe that for a group $G$, $\typetorsor_G$ is a connected groupoid\footnote{Admittedly in a higher universe, but we can use the
    Replacement~\cref{pri:replacement} to see that $\typetorsor_G$ is equivalent
    to a type in the same universe as $G$ -- even before we
    have~\cref{lem:BGbytorsor} showing we can take $\BG$.}
  and so -- by specifying the base point $\princ G$ -- it classifies a group.
  Guess which one!\footnote{%
    By the way, the name ``torsor'' is a translation from the French \emph{torseur},
    introduced by \citeauthor{giraud1971},\footnotemark{} who
    related them to “twisting” operations on bundles.
    Since $BG$ is equivalent to the type of $G$-torsors,
    we can also think of shapes $t:\BG$ as giving rise to “twists”.
    Indeed, for a $G$-set $X$,
    we can think of $X(x)$ as a ``twisted'' version of the underlying set,
    $X(\sh_G)$.}\footcitetext{giraud1971}.
\end{remark}


\begin{definition}
  \label{def:BG2TorsG}
Recall from~\cref{def:principaltorsor}\eqref{eq:pathsp}
the definition, for all $y:\BG$, of $\pathsp y:\BG\to\Set$
as the $G$-set with $\pathsp y(z)\jdeq(y\eqto z)$
(so that in particular $\princ G\jdeq\pathsp{\sh_G}$).
Note that $\pathsp y$ is a $G$-torsor, so we can define
  \[
    \pathsp{\blank}:\BG\ptdto(\typetorsor_G,\princ G): y\mapsto P_y,
  \]
  with pointing path $\refl{\princ G}:\princ G\eqto\pathsp{\sh_G}$.\footnote{%
    That is, we have classified a homomorphism from $G$
    to $\Aut_{\GSet}(\princ G)$. It'll turn out to be an isomorphism.}
If $G$ is not clear from the context, we may choose to write $\pathsp{\blank}^G$ instead of $\pathsp{\blank}$.
\end{definition}

\begin{remark}\label{rem:pathsptransport}
  We will use several variants of $\pathsp{\blank}$, in combination with
  some of the conventions introduced back in \cref{ch:univalent-mathematics}.
  In this remark, to avoid confusion, we explain these variants.
  
  First, we also use $\pathsp{\blank}$ to denote its induced action on paths:
  for $y,z:\BG$ we have
  \[
    \pathsp{\blank}:(y\eqto z)\to (\pathsp y\eqto \pathsp z),
  \]
  defined by path induction as in \cref{def:ap}.
  
  Then, as $\pathsp y\eqto \pathsp z$ is an identity between
  families of types, function extensionality (\cref{def:funext}) applies.
  For $q:y\eqto z$, we may also use $\pathsp q$ to denote the corresponding 
  function of type $\prod_{x:\BG}(\pathsp y(x)\eqto \pathsp z(x))$.
  
  Finally, as $\pathsp y(x)$ and $\pathsp z(x)$ are types,
  univalence (\cref{def:univalence}) applies.
  Therefore we may use $\pathsp q(x)$ to denote the corresponding 
  equivalence, \ie transport in the type family $\pathsp{\blank}(x)$,
  sending $p:\pathsp y(x)\jdeq (y\eqto x)$ to
  $pq^{-1}:\pathsp z(x)\jdeq(z\eqto x)$.\footnote{In a commutative diagram,
    \[
      \begin{tikzcd}[ampersand replacement=\&]
        y \ar[rr,eqr,"q"]\ar[dr,eql,"p"'] \& \& z \ar[dl,eqr,"{\pathsp q(p)}"] \\
        \& x.
      \end{tikzcd}
    \]}
\end{remark}

\begin{lemma}\label{lem:pathsptransportiseq}
  Let $G$ be a group. For all $y,z:\BG$ the induced map of identity types
  \[
    \pathsp{\blank}:(y\eqto z)\to (\pathsp y\eqto \pathsp z)
  \]
  is an equivalence.\footnote{%
    For connoisseurs of category theory,
    this is also a corollary of a \emph{type-theoretic Yoneda lemma},
    stating that transport gives an equivalence
    \[
      X(a) \equivto \prod_{b:A}\bigl((a \eqto b) \to X(b)\bigr)
    \]
    for any pointed type $(A,a)$ and type family $X: A \to \UU$.
    Try to prove this yourself!}
\end{lemma}
\begin{proof}
  We craft an inverse $Q:(\pathsp y\eqto \pathsp z) \to (y\eqto z)$ for
  $\pathsp{\blank}$. Given an identity $f:\pathsp y \eqto \pathsp z$, the map
  $f_y: (y\eqto y) \to (z\eqto y)$ maps the reflexivity path $\refl y$ to a path
  $f_y(\refl y):z\eqto y$, and we define
  \[
    Q(f) \defequi \inv{f_y(\refl y)} : (y\eqto z).
  \]
  First we construct an identification of $\pathsp {Q(f)}$ and $f$:
  for any $x:\BG$, 
  $\pathsp {Q(f)}(x)$ maps any $p:\pathsp{y}(x)\jdeq(y\eqto x)$ to
  $p f_y(\refl y)(x):\pathsp{z}(x)\jdeq(z\eqto x)$. Hence we must
  construct an identification of type $p f_y(\refl y)(x)=f_x(p)$,
  which is immediate by induction on $p:y\eqto x$, setting $p\jdeq \refl y$.
  
  Next, we prove the equality of $Q(\pathsp q)$ and $q$
  for every $q:y\eqto z$. Indeed, 
  $Q(\pathsp q)\jdeq\inv{(\pathsp{q})_y(\refl y)} = \inv{(\refl y \inv q)} = q$.
  
  Now apply \cref{lem:weq-iso} to complete the proof of the lemma.
\end{proof}

The following theorem justifies the title of this section, stating 
that the classifying type of a group is the type of its torsors.

\begin{theorem}\label{lem:BGbytorsor}
  If $G$ is a group, then the function
  $\pathsp{\blank}:\BG\to\typetorsor_G$ from~\cref{def:BG2TorsG}
  is an equivalence.\footnote{A similar results holds for $\infty$-groups.}
\end{theorem}

\begin{proof}
  Since both $\typetorsor_G$ and $\BG$ are pointed and connected,
  it suffices by
  \cref{cor:fib-vs-path}\ref{conn-fib-vs-path-point} to show that 
  $\pathsp{\blank}:(\sh_G\eqto\sh_G)\to(\pathsp{\sh_G}\eqto \pathsp{\sh_G})$
  is an equivalence.\footnote{%
  This holds for all variants of $\ap{\pathsp{\blank}}$.}
  This follows directly from \cref{lem:pathsptransportiseq}.
\end{proof}

\subsection{Homomorphisms and torsors}
\label{sec:homotor}
In view of the equivalence $\pathsp{}^G$ between $\BG$ and 
$(\typetorsor_G,\princ G)$ of \cref{lem:BGbytorsor} one might 
ask what a group homomorphism  $f:\Hom(G,H)$ translates to on 
the level of torsors.  Off-hand, the answer is the round-trip 
$(\pathsp{}^H)\Bf(\pathsp{}^G)^{-1}$, but we can be more concrete than that.
We do know that for $x:\BG$ the $G$-torsor $\pathsp x^G$ should be sent to
$\pathsp {\Bf(x)}^H$, but how do we express this for an arbitrary $G$-torsor?
\begin{definition}
  \label{def:restrictandinduce}
  Let $f:\Hom(G,H)$ be a group homomorphism.  If $Y:\BH\to\Set$ is an $H$-set,
  then the \emph{restriction}\index{action!restricted}\index{restriction}
  $f^*Y$ of $Y$ to $G$ is the $G$-set given by precomposition
  \[
    f^*Y\defequi (Y\circ\Bf) :\BG\to\Set.
  \]

  If $X:\BG\to\Set$ is a $G$-set, we define
  the \emph{induced $H$-set}\index{action!induced}\index{induced action}
  $f_*X : \BH\to\UU$ by setting, for $y:\BH$,
  \[
    f_*X(y)\defeq\myTrunc{\sum_{z:\BG}(\Bf(z) \eqto y)\times X(z)}{0}.\qedhere
  \]
\end{definition}
The following exercise motivates the set-truncation in the definition
of $f_*$ above.\footnote{%
    This situation is common in algebra and is often referred to by saying
    that some construction, in this case the untruncated
    definiens of $f_*X$, is not ``exact''. See also \cref{xca:why-setTrunc_f_*}.}

\begin{xca}\label{xca:why-setTrunc_f_*}
Find groups $G,H$, $f:\Hom(G,H)$ and $G$-set $X$ such that
$\sum_{z:\BG}(\Bf(z) \eqto y)\times X(z)$ is not an $H$-set (but an $H$-type).
\end{xca}
% Solution: $G=\ZZ$, $H=\TG$, $f$ the unique homomorphism $f: \Hom(G,H)$,
% $X$ constant $\bn 1$. Then 
% $\sum_{z:\Sc}(0=0)\times \bn 1$ is a circle, so not a set.

\begin{xca}\label{xca:id_*-is-id}
\MB{New:} Give an equivalence from $f_*\,X$ to $X\circ\inv\Bf$
    if $f$ is an isomorphism. Give an equivalence between the identity
    types $f_*\,X \eqto Y$ and $X \eqto f^*\,Y$, for all $G$-sets $X$
    and $H$-sets $Y$.
\end{xca}


Note that the type $f_*X(y)$ is also the action type $(H^y \times X)_{hG}$,
of the $G$-set $H^y\times X$,
where $(H^y\times X)(x) \defeq (\Bf(x) \eqto y)\times X(x)$ for $x:\BG$,
and whose underlying set is equivalent to $(\sh_H\eqto y)\times X(\sh_G)$.

\begin{remark}
  Dually, there is also a \emph{coinduced $H$-set}\index{action!coinduced}
  $f_!:\BH\to\Set$ given by
  \[
    f_!X(y)\defeq\prod_{z:\BG}\bigl((\Bf(z)\eqto y) \to X(z)\bigr).
  \]
  Note that this always lands in sets when $X$ does.
\end{remark}

When $X$ is the $G$-torsor $\pathsp x^G$, for some $x:\BG$,
the contraction (recall \cref{lem:contract-away})
of $\sum_{z:\BG}(x\eqto z)$ induces an equivalence
\[
  \eta_y:f_*\pathsp x^G(y) \jdeq 
  \myTrunc{\sum_{z:\BG}(\Bf(z) \eqto y)\times(x \eqto z)}{0}
  \equivto (\Bf(x)\eqto y)\jdeq\pathsp{\Bf(x)}^H(y).
\]
Taking $x\jdeq\sh_G$, so $\pathsp x^G\jdeq\princ G$, we get a
path $\eta:f_*\,\princ G\eqto \pathsp{\Bf(\sh_G)}^H$.
We also have the path $\Bf_\pt : \sh_H\eqto\Bf(\sh_G)$,
so that the action of $\pathsp{\blank}^H$ gives us a path
$\pi : \princ H \jdeq\pathsp{\sh_H}^H \eqto \pathsp{\Bf(\sh_G)}^H$.
Combining we get $\inv\eta\pi:\princ H \eqto f_*\,\princ G$.

If $X$ is a $G$-set such that $\Trunc{\princ G \eqto X}$, then $f_*X$
is an $H$-set such that $\Trunc{\princ H \eqto f_*X}$, so that
$f_* : \typetorsor_G \ptdto \typetorsor_H$, pointed by $\inv\eta\pi$.

Summing up, we have implemented the following:
\begin{construction}
  \label{lem:inducedtorsor}
   Let $f:\Hom(G,H)$ be a group homomorphism. Then $f$ induces a 
   pointed map $f_*:\typetorsor_G\ptdto\typetorsor_H$,
   and we have a path of type 
   $f_*\,\pathsp{\blank}^G = \pathsp{\blank}^H\,\Bf \jdeq 
   f^*\,\pathsp{\blank}^H$,
   all represented by the following diagram:
   \[
     \begin{tikzcd}
     \sh_G \ar[ddd,mapsto] \ar[rrr,mapsto] &
     && \Bf(\sh_G)  & \ar[l,eql,"{\Bf_\pt}"'] \sh_H \ar[ddd,mapsto]\\
       &\BG \ar[r,"\Bf"]\ar[d,"{\pathsp{\blank}^G}"'] &
        \BH\ar[d,"{\pathsp{\blank}^H}"] \\
       &\typetorsor_G \ar[r,"f_*"'] & \typetorsor_H \\
     \princ G \ar[rrr,mapsto] &
     && f_*\,\princ G&  \ar[l,eqr,"{\inv\eta\pi}"] \princ H.
     \end{tikzcd}
   \]
\end{construction}

\section{Any symmetry is a symmetry in $\Set$}
\label{sec:groupssubperm}


For abstract groups there is a result, attributed to Cayley,
which is often stated as ``any group is a permutation group''. 
In our parlance this translates to ``any symmetry is a symmetry in $\Set$''.
The aim of this section is to give a precise formulation of the latter
and prove it.
% \footnote{which reminds me of the following: my lecturer in cosmology once tried to publish a paper about rotating black holes, only to have it rejected because it turned out that it was his universe, not the black hole, that was rotating}

%which is equivalent to saying that $X$ is the universal \covering
Let $G$ be a group.
Recall from \cref{def:principaltorsor} the principal torsor
$\princ G:\BG\to\Set: z\mapsto (\sh_G\eqto z)$.
Since $\princ G (\sh_G)\defequi \USymG$, $\princ G$ restricts to
a pointed function $\BG\ptdto \BSG_{\USymG}$, \ie 
classifies a homomorphism from $G$ to the permutation group 
$\SG_{\USymG}\jdeq \Aut_{\Set}(\USymG)$,
denoted by\footnote{The letter $\rho$ commemorates the word ``regular''} 
$$\rho_G:\Hom(G,\SG_{\USymG}).$$

\begin{theorem}[Cayley]
  \label{lem:allgpsarepermutationgps}
  For all groups $G$, $\rho_G$ is a monomorphism.\footnote{By
  \cref{def:typeofmono}, $\rho_G$ is a monomorphism means 
  that the induced map $\USym\rho_G$ from the symmetries of $\sh_G$ in 
  $\BG_\div$ to the symmetries of $\USymG$ in $\Set$ is an injection, 
  \ie ``any symmetry is a symmetry in $\Set$''.}  
\end{theorem}

\begin{proof}
  In view of \cref{def:typeofmono} we need to show that 
  $\B\rho_G\jdeq\princ G :\BG \to \BSG_{\USymG}$ is a \covering.
  Under the pointed equivalence
  $$\pathsp{\blank}:\BG\ptdto (\typetorsor_G,\princ G)$$ of
  \cref{lem:BGbytorsor}, $\princ G$ is transported to\footnote{
  See \cref{xca:evP_isPrG}.} to the
  evaluation map
  $$\mathrm{ev}_{\sh_G}:\conncomp{(\BG\to\Set)}{\princ G}\ptdto
  \conncomp{\Set}{\USymG},\qquad
  \mathrm{ev}_{\sh_G}(E)\defeq E(\sh_G).$$
  We must show that the preimages
  $\inv{\ev_{\sh_G}}(X)$ for $X:\Sigma_{\USymG}$ are sets.  This
  fiber is equivalent to
  $\sum_{E:\conncomp{(\BG\to\Set)}{\princ G}}(X\eqto E(\sh_G))$ which,
  being a subtype, is a
  set precisely when $\sum_{E:\BG\to\Set}(X\eqto E(\sh_G))$ is a set.
  The latter is the type of pointed maps from $\BG$ to $(\Set,X)$
  and hence a set by \cref{lem:hom-is-set}, 
  in particular \cref{ft:ptd-decr-h-lev}.
\end{proof}
  Note that the above theorem yields that 
  $(G,\rho_G,!)$ is a monomorphism into $\SG_{\USymG}$.
  In other words, $G$ is a subgroup of $\SG_{\USymG}$.

\begin{xca}\label{xca:evP_isPrG}
\MB{New:} Show that $\princ G$ and ${\ev_{\sh_G}}\circ{\pathsp{\blank}}$ are
equal as pointed maps.
\end{xca}

\begin{remark}\label{rem:CayleyOversize}
\MB{New:} In many cases, the set $\USymG$ used in \cref{lem:allgpsarepermutationgps} is larger than necessary for
obtaining the symmetries in $G$ as symmetries of a set.
A case in point is the group $\SG_3$, where the symmetries \emph{are}
already symmetries of a set, namely of the set $\bn3$. However,
$\USG_3\jdeq(\bn3\eqto\bn3)$ is a $6$-element set. 
Let's take a closer look at where and how this happens in the proof.

As stated in \cref{xca:evP_isPrG}, the map $\princ G: 
\BG\ptdto\conncomp{\Set}{\USymG}$ classifying the monomorphism $\rho_G$ 
is decomposed as an equivalence $\pathsp{\blank}$
followed by the evaluation map $\ev_{\sh_G}$.
This is depicted in the following diagram, where the
second line shows the induced maps on the symmetries.
   \[
     \begin{tikzcd}
     \BG \ar[r,equivr,"\pathsp{\blank}"] & 
     (\typetorsor_G,\princ G) \ar[r,"\ev_{\sh_G}"]&
     \conncomp{\Set}{\USymG}\\
     \USymG \ar[r,equivr,"\pathsp{\blank}"] &   %\loops?
     (\princ G \eqto \princ G) \ar[r,"\ev_{\sh_G}"]&  %\loops?
     (\USymG \eqto \USymG)  
     \end{tikzcd}
   \]
Let $\PP$ be the $G$-set given by 
$\PP(z)\defeq(\princ G(z) \eqto \princ G(z))$ for all $z:\BG$.
By function extensionality, $\princ G \eqto \princ G$ is
equivalent to $\prod_{z:\BG}\PP(z)$, the type of
invariant maps of $\PP$.
By \cref{lem:fixpts-are-fixed}\ref{it:ev-is-inj}, such invariant maps,
and hence the corresponding symmetries of $\princ G$, are uniquely
determined by their value at $\sh_G$.\footnote{%
This is an alternative way to understand that $\ev_{\sh_G}$,
and hence $\princ G$, classifies a monomorphism.}

Note that the underlying set of $\PP$ is 
$\PP(\sh_G)\jdeq(\USymG \eqto \USymG)$.
However, $\PP$ has more structure than its underlying set.
\cref{lem:fixpts-are-fixed}\ref{it:ev-is-eq-on-inv} characterizes
exactly the invariant maps of $\PP$ as corresponding via $\ev_{\sh_G}$
with fixed elements of $\USymG \eqto \USymG$. In other words,
$\ev_{\sh_G}$ forgets about the extra structure of $\PP$
and sends invariant maps of $\PP$ to fixed permutations of $\USymG$.
For example, in the case of $\SG_3$, we go in total from permutations
of $\bn3$ to fixed permutations of the $6$-element set $\bn3\eqto\bn3$.

In \cref{xca:PP-fixed-permutations} you are asked to explore
the abstract group of fixed permutations of $\USymG$.
\end{remark}

\begin{xca}\label{xca:PP-fixed-permutations} \MB{New:}
Let conditions be as in \cref{rem:CayleyOversize}.
By analyzing transport in the type family $\princ G(\blank)$,
show that a permutation $\pi$ of $\USymG$
is fixed if and only if $\pi(gg')=g\pi(g')$ for all $g,g':\USymG$.
Show that the fixed permutations of $\USymG$ form an abstract group
and that evaluation of such a permutation at $\refl{\sh_G}$
yields an abstract isomorphism from this group to $\abstr(G)$.
\end{xca}
  

\begin{xca} \MB{MB doesn't understand:}
    Given a group $G$ we defined in \cref{sec:groupssubperm} a monomorphism from $G$ to the permutation group $\Aut_{\USymG}(\Set)$. Write out the corresponding subgroup of $\Aut_{\USymG}(\Set)$.
  \end{xca}

\section{The orbit-stabilizer theorem}
\label{sec:orbit-stabilizer-theorem}


Let $G$ be a group (or $\infty$-group) and  $X : \BG \to \UU$  a $G$-type.
Recall the action type of \cref{def:actiontype} $X_{hG}\defequi
\sum_{z:\BG}X(z)$ and its truncation, the set of orbits $X / G \defeq \setTrunc{X_{hG}}$ and the map $X(\sh_G)\to X_{hG}$ sending $x:X(\sh_G)$ to $[x]\defequi (\sh_G,x)$.


For an element $x:X/G$ consider the associated subtype of $X(\sh_G)$ consisting of
 all $y : X(\sh_G)$ so that $\settrunc{[y]}$ is equal to $x$ in the set of orbits:
\[
  \mathcal O_{x} \defeq \sum_{y : X(\sh_G)} x =_{X/G} \settrunc{[y]}.
\]
 For a point $x : X(\sh_G)$, we call $G\cdot x \defeq \mathcal O_{\settrunc{[x]}}$ 
 the \emph{orbit through $x$}\index{orbit through $x$}.
 % as the subtype of $X(\sh_G)$ consisting of
%  all $y : X(\sh_G)$ so that $[y]$ is equal to $[x]$ in the set of orbits:
% \[
%   G\cdot x \defeq \mathcal O_{[x]} \defeq \sum_{y : X(\sh_G)} [x] =_{X/G} [y].
% \]
Note the (perhaps) unfortunate terminology: an ``orbit'' is not an element in the
action type, but rather consists of all the elements in $X(\sh_G)$ sent to a given element in the orbit set.
% Also note that the orbit $G\cdot x\defeq \mathcal O_x$ only depends on the image of $x$ in the set $X/G$ of orbits,
% thus justifying the names.

In this way, the (abstract) $G$-type $X(\sh_G)$ splits as a disjoint union (i.e., a $\Sigma$-type over a set) of orbits,
parametrized by the set of orbits:
\begin{lemma}
  \label{old:splitting into orbits}
  The inclusions of the orbits induce an equivalence 
\[
  \sum_{z : X / G} \mathcal O_z  \we X(\sh_G).
\]
\end{lemma}

The \emph{stabilizer group}\index{stabilizer group}
(also known as the \emph{isotropy group}\index{isotropy group}) of
$x : X(\sh_G)$ is the automorphism group of $[x]$ in the action type
$$G_x\defequi\Aut_{X_{hG}}([x]),$$
so that
$X_{hG}$ is equivalent to the disjoint union $\sum_{z:X/G}(\BG_x)_\div$.

%The \emph{stabilizer group}\index{stabilizer group}
%(also konwn as the \emph{isotropy group}\index{isotropy group}) of
%$x : X(\sh_G)$ is the automorphism group of $[x]$ in the action type
%$$G_x\defequi\Aut_{X_{hG}}([x]).$$

  For $x:X(\sh_G)$ the restriction of the first projection $\fst:X_{hG}\to\BG$ to the component of $[x]$ gives a homomorphism $i_x:\Hom(G_x,G)$ if we point it by $\refl{\sh_G}:\sh_G=\fst([x])$.

  Recall that if $H$ and $G$ are groups, the set of homomorphisms $\hom(H,G)$ is a $G$-set.  In particular, if $g:\sh_G=\sh_G$ and $f:\hom(H,G)$, then we can denote the result of letting $g$ act on $f$ by $f^g:\hom(H,G)$ which has $\B(f^g)_\div\defequi\Bf$, but is pointed by $p_fg^{-1}:\sh_G=\Bf_\div(\sh_H)$ (i.e., precomposing the witness $p_f:\sh_G=f(\sh_H)$ with the inverse of $g$).  In the case where $f$ was a monomorphism so that it defines a subgroup, we called $f^g$ the conjugate of the original subgroup.
  
 If $g:\sh_G=\sh_G$ the identity map of $X_{hG}$ can be pointed by the identity between $(\sh_G,x) $ and $(\sh_G,gx)$ given by $g:\sh_G=\sh_G$ and $\refl{gx}$ to give an isomorphism from $G_x$ to $G_{gx}$ identifying (by univalence) the homomorphism $i_{gx}:\hom(G_{gx},G)$ with $i^{g^{-1}}_x:\hom(G_x,G)$.


In the case where $X$ is a $G$-\emph{set}, the homomorphism $i_x:\hom(G_x,G)$ is a monomorphism, and with this extra information we consider $G_x$ as a subgroup of $G$.
Thus, different points in the same orbit of $X(\sh_G)$ have conjugate stabilizer subgroups.

Another way of formulating these ideas is to consider the $G$-type
\[
  \mathcal O_x\colon\BG\to\UU,\qquad
  \mathcal O_x(z)\defequi\sum_{y:X(z)}\trunc{[x]}=_{X/G}\trunc{(z,y)},
\]
and obtain the alternative definition of $\BG_x$ as $\sum_{z:\BG}\mathcal O_x(z)$, which we see is a subgroup exactly when $\mathcal O_x$ is a $G$-set.

We say that the action is \emph{free}\index{free $G$-type} if all stabilizer groups are trivial.

\begin{theorem}[Orbit-stabilizer theorem]
  \label{old:orbitstab}
  Fix a $G$-type $X$ and a point $x : X(\sh_G)$.
  There is a canonical action $\tilde G : \BG_x \to \UU$,
  acting on $\tilde G(\sh_G)\equiv G$
  with action type $\tilde G_{hG_x} \equiv \mathcal O_x$.
\end{theorem}
\begin{proof}
  Define $\tilde G(t,y,!) \defeq (\sh_G \eqto t)$.
  Then we calculate:
  \begin{align*}
    \tilde G_{hG_x}
    &\defeq \sum_{u : \BG_x}\tilde G(u) \\
    &\eqto \sum_{t:\BG}\sum_{y:X(t)}\trunc{(\sh_G,x)} =_{X/G} \trunc{(t,y)}
      \times (\sh_G \eqto t) \\
    &\equivto \sum_{y:X(\sh_G)}[x] =_{X/G} [y] \defeq \mathcal O_x.\qedhere
  \end{align*}
\end{proof}

Now suppose that $G$ is a $1$-group acting on a set.
We see that the action type is a set
(and is thus equivalent to the set of orbits)
if and only if
all stabilizer groups are trivial,
\ie if and only if the action is free.

If $G$ is a $1$-group,
then so is each stabilizer-group,
and in this case (of a set-action),
the orbit-stabilizer theorem
tells us that

\begin{theorem}[Lagrange's Theorem]\footnote{insert that the action is free (referred to)}
\label{old:lagrange}
  If $H \to G$ is a subgroup, then $H$ has a natural action on $G$,
  and all the orbits under this action are equivalent.
\end{theorem}

% Interaction with cardinality: if G acts freely on S, then card(S/G) = card(S)/card(G)

\section{The lemma that is not Burnside's}
\label{sec:burnsides-lemma}
\label{lem:burnsides-lemma}

\begin{lemma}
  \label{lem:burnside}
  Let $G$ be a finite group acting on a finite set $X$.
  Then the set of orbits is finite with cardinality
  \[
    \Card(X/G) = \frac1{\Card(G)}\sum_{g:\El G}\Card(X^g),
  \]
  where $X^g = \setof{x:X}{gx=x}$ is the set of elements
  that are fixed by $g$.
\end{lemma}
\begin{proof}
  Since $X$ and $G$ is finite, we can decide equality of their elements.
  Hence each $X^g$ is a finite set, and since $G$ is finite,
  we can decide whether $x,y$ are in the same orbit by searching
  for a $g : \El G$ with $gx = y$.
  Hence the set of orbits is a finite set as well.

  Consider now the set $R \defeq \sum_{g:\El G} X^g$,
  and the function $q : R \to X$
  defined by $q(g,x) \defeq x$.
  The map $q^{-1}(x) \to G_x$ that sends $(g,x)$ to $g$ is a bijection.
  Thus, we get the equivalences
  \[
    R \jdeq \sum_{g:\El G} X^g \equiv \sum_{x:X} G_x
    \equiv \sum_{z:X/G} \sum_{x : \mathcal O_z} G_x,
  \]
  where the last step writes $X$ as a union of orbits.
  Within each orbit $\mathcal O_z$,
  the stabilizer groups are conjugate,
  and thus have the same finite cardinality,
  which from the orbit-stabilizer theorem (\cref{thm:orbitstab}),
  is the cardinality of $G$ divided by the cardinality of $\mathcal O_z$.
  We conclude that $\Card(R) = \Card(X/G)\Card(G)$, as desired.
\end{proof}

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