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Rem. 4.2.21 replaced by mn 8
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marcbezem committed Aug 25, 2024
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Expand Up @@ -436,33 +436,10 @@ \subsection{First examples}
so we can also identify the trivial group with
$\mkgroup (\true,\triv)$, or with $\mkgroup (C,c)$ for
any contractible type $C$ and element $c:C$, or
with $\Aut_{S}(x)$ for any set $S$ and element $x:S$.

\item\label{ex:permgroup}
If $n:\NN$, then the \emph{permutation group of $n$ letters}
(also known as the \emph{symmetric group of order $n$}) is%
\glossary(918Sigma2){$\protect\SG_n$}{symmetric group of order $n$,
\cref{ex:groups}\ref{ex:permgroup}}\index{symmetric group}%
\index{group!symmetric group}
\[
\SG_n\defequi \Aut_{\Set}(\bn n). %\mkgroup(\FinSet_n,\bn{n}),
\]
The classifying type is thus $\BSG_n\jdeq (\FinSet_n,\bn{n})$,
where $\FinSet_n \jdeq \Set_{(\bn{n})}$ is the groupoid of
sets of cardinality $n$ (\cf \ref{def:groupoidFin}).

Again, we can also identify the group $\SG_n$ with
$\Aut_\FinSet(\bn{n})$ (by \cref{xca:group-example-details}), with
$\Aut_{\FinSet_n}(\bn n)$ (by \cref{rem:symmetriesofnonconnectedgroupoids}),
or even with $\Aut_{\UU}(\bn n)$ (by stretching the definition of $\Aut$,
using that $\UU_{(\bn n)}$ is a connected groupoid, see~\cref{rem:autinfgp}).


If the reader starts worrying about size issues,
see~\cref{rem:groupsarebig}.\footnote{%
This footnote is for those who worry about size issues -- a
with $\Aut_{S}(x)$ for any set $S$ and element $x:S$.\footnote{%
This note is for those who worry about size issues -- a
theme we usually ignore in our exposition.
Recall from \MB{updated} \cref{sec:universes} the chain of
Recall from \cref{sec:universes} the chain of
universes $\UU_0 : \UU_1 : \UU_2 : \dots$ such that for each $i$
all types in $\UU_i$ are also in $\UU_j$ for all $j>i$.
Let $\Set_0$ be the type of sets in $\UU_0$,
Expand Down Expand Up @@ -494,9 +471,30 @@ \subsection{First examples}
However, even with this principle there are groups that only belong
to $\typegroup_i$ for $i>0$ large enough.

These matters concerning universes are nontrivial and important,
Issues concerning universes are nontrivial and important,
but in this text we have chosen to focus on other matters.
}

\item\label{ex:permgroup}
If $n:\NN$, then the \emph{permutation group of $n$ letters}
(also known as the \emph{symmetric group of order $n$}) is%
\glossary(918Sigma2){$\protect\SG_n$}{symmetric group of order $n$,
\cref{ex:groups}\ref{ex:permgroup}}\index{symmetric group}%
\index{group!symmetric group}
\[
\SG_n\defequi \Aut_{\Set}(\bn n). %\mkgroup(\FinSet_n,\bn{n}),
\]
The classifying type is thus $\BSG_n\jdeq (\FinSet_n,\bn{n})$,
where $\FinSet_n \jdeq \Set_{(\bn{n})}$ is the groupoid of
sets of cardinality $n$ (\cf \ref{def:groupoidFin}).

Again, we can also identify the group $\SG_n$ with
$\Aut_\FinSet(\bn{n})$ (by \cref{xca:group-example-details}), with
$\Aut_{\FinSet_n}(\bn n)$ (by \cref{rem:symmetriesofnonconnectedgroupoids}),
or even with $\Aut_{\UU}(\bn n)$ (by stretching the definition of $\Aut$,
using that $\UU_{(\bn n)}$ is a connected groupoid, see~\cref{rem:autinfgp}).



\item\label{ex:genpermgroup}
More generally, if $S$ is a set, is there a pointed connected groupoid $(A,a)$ so that $a\eqto_Aa$ models all the ``permutations'' $S\eqto_{\Set}S$ of $S$?
Expand Down Expand Up @@ -524,37 +522,6 @@ \subsection{First examples}
Also, give an identification of type $\Aut_\Set(\NN)\eqto\Aut_\Set(\zet)$.
\end{xca}

\begin{remark}
\label{rem:groupsarebig}
This remark is for those who worry about size issues -- a theme we usually ignore in our exposition. If we
start with a base universe $\UU_0$, the groupoid $\FinSet_n$ of sets of cardinality $n$ is the $\Sigma$-type
$\sum_{A:\UU_0}\Trunc{A\eqto\bn{n}}$ over $\UU_0$ and so (without any modification) will lie in any bigger
universe $\UU_1$. In order to accommodate the permutation groups of sets in $\UU_0$, the universe ``$\UU$'' appearing as
a subscript of the first $\Sigma$-type in \cref{def:pt-conn-groupoid}, appearing later in the definition of
``group'', needs to be at least as big as $\UU_1$. If $\UU$ is taken to be $\UU_1$, then the type
$\typegroup$ of groups will not be in $\UU_1$, but in some bigger universe $\UU_2$.
If we then choose some group $G:\typegroup$
and look at its group of automorphisms, $\Aut_\Group(G)$,\footnote{%
\Cref{xca:typegroupisgroupoid} below asks you to verify that $\Group$
is a (large) groupoid.}
based on the identity type $G \eqto_\Group G$, this will be an element of $\typegroup$ only if the universe $\UU$ in the definition of
$\typegroup$ is at least as big as $\UU_2$. Our convention from \cref{sec:universes} is
that the universes form an ascending chain $\UU_0\subseteq\UU_1\subseteq\UU_2\subseteq\dots$, corresponding
to which there will an ascending chain of types of groups,
\[
\typegroup_i \defequi \Copy_{\mkgroup}\bigl( (\UU_i)_*^{=1} \bigr) : \UU_{i+1},
\]
and any group we encounter will be an element of $\typegroup_i$ for $i$
large enough.

These matters concerning universes are nontrivial and important,
but in this text we have chosen to focus on other matters.\footnote{%
We will note, however, that the Replacement~\cref{pri:replacement}
often allows us to conclude that a group $G$ belongs to $\typegroup_0$.
This is the case for $\SG_S$, for $S:\Set_0$, and for $\Aut_\Group(G)$,
for $G:\Group_0$, as we invite the reader to check.
Hint: use \cref{xca:comp-loc-small-ess-small}.}
\end{remark}

\begin{example}\label{ex:cyclicgroups}
In \cref{cor:id-m-cycle} we studied the symmetries of the
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