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\begin{document}
\section*{Math 222 Fall 2022, Lecture 27: Permutations}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/22fco}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/22fco}}}

\setcounter{section}{3}

\section{\label{chap.perm}Permutations}

\setcounter{subsection}{2}

\subsection{The cycle decomposition of a permutation}

Last time (Lecture 26), we studied permutations and introduced a few ways to
denote them. The cycle digraph, in particular, has revealed a few patterns: It
always seems to consist of a bunch of cycles that are disjoint (i.e., have no
nodes in common).

Let us make this precise and draw some consequences.

\subsubsection{Orbits and the relation $\protect\overset{\sigma}{\sim}$}

\begin{definition}
\label{def.perm.sig-eq}Let $X$ be a finite set. Let $\sigma$ be a permutation
of $X$.

Consider the relation $\overset{\sigma}{\sim}$ on the set $X$ defined as
follows: For any $i,j\in X$, we set%
\[
\left(  i\overset{\sigma}{\sim}j\right)  \ \Longleftrightarrow\ \left(
i=\sigma^{k}\left(  j\right)  \text{ for some }k\in\mathbb{N}\right)  .
\]

\end{definition}

We will soon see that this relation $\overset{\sigma}{\sim}$ is an equivalence
relation.\footnote{Note that this would not be the case if $X$ were allowed to
be infinite! For example, the map
\[
\sigma:\mathbb{Z}\rightarrow\mathbb{Z},\ i\mapsto i+1
\]
is a permutation of $\mathbb{Z}$, but the relation $\overset{\sigma}{\sim}$ is
not symmetric (since $2\overset{\sigma}{\sim}1$ but not $1\overset{\sigma
}{\sim}2$) and therefore not an equivalence relation.} But first, let us
analyze its equivalence classes. Recall that if $\sim$ is a relation on some
set $X$, and if $a$ is an element of $X$, then the $\sim$-equivalence class of
$a$ is defined to be the set%
\[
\left[  a\right]  _{\sim}:=\left\{  b\in X\ \mid\ b\sim a\right\}  .
\]


If $\sim$ is the relation $\overset{\sigma}{\sim}$ from Definition
\ref{def.perm.sig-eq}, then these $\sim$-equivalence classes have a special name:

\begin{definition}
\label{def.perm.orbit}Let $X$ be a finite set. Let $\sigma$ be a permutation
of $X$. Then, the equivalence classes of the relation $\overset{\sigma}{\sim}$
are called the \textbf{orbits} of $\sigma$.
\end{definition}

\begin{example}
\label{exa.perm.cycle-dig2v2}Let $\sigma$ be the permutation of $\left[
10\right]  $ whose OLN is%
\[
5\ 4\ 3\ 2\ 6\ \left(  10\right)  \ 1\ 9\ 8\ 7.
\]
(We have put the element $10$ in parentheses to make its place clearer.) For
example, $\sigma\left(  5\right)  =6$ and $\sigma\left(  6\right)  =10$. As we
recall (Example 4.1.5 in Lecture 26), the cycle digraph of $\sigma$ is%
\[%
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%\node[main node] (5) [above left of=6] {5};
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%\node[main node] (7) [below of=1] {7};
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%(2) edge [bend right] (4)
%(4) edge [bend right] (2)
%(3) edge [loop right] (3);
%\end{tikzpicture}}}%
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(9) edge [bend right] (8)
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(2) edge [bend right] (4)
(4) edge [bend right] (2)
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\ \ .
\]


The orbits of $\sigma$ are%
\[
\left\{  1,5,6,10,7\right\}  ,\ \ \ \ \ \ \ \ \ \ \left\{  8,9\right\}
,\ \ \ \ \ \ \ \ \ \ \left\{  3\right\}  ,\ \ \ \ \ \ \ \ \ \ \left\{
2,4\right\}  .
\]
Of course, $\left\{  1,5,6,7,10\right\}  $ is an orbit as well, but that's
just the same orbit as $\left\{  1,5,6,10,7\right\}  $, so it doesn't count as
a fifth orbit.
\end{example}

As we see on this example, the orbits of a permutation $\sigma$ correspond to
the cycles on the cycle digraph of $\sigma$. More precisely, each cycle on the
cycle digraph of $\sigma$ results in an orbit, which consists of all nodes on
this cycle. However, the orbit is just a set, so it doesn't \textquotedblleft
remember\textquotedblright\ the order of the nodes on the cycle; it only
remembers which nodes are on the cycle.

This all holds in general, not just in Example \ref{exa.perm.cycle-dig2v2}. To
convince ourselves of this, we shall prove the following proposition, which
describes the structure of a single orbit $\left[  a\right]  _{\sim}$ of
$\sigma$:

\begin{proposition}
[structure of an orbit of $\sigma$]\label{prop.perm.orbit-struc}Let $X$ be a
finite set. Let $\sigma$ be a permutation of $X$.

Let $\sim$ be the relation $\overset{\sigma}{\sim}$. Let $a\in X$. Then:

\textbf{(a)} We have
\begin{align}
\left[  a\right]  _{\sim}  &  =\left\{  \sigma^{k}\left(  a\right)
\ \mid\ k\in\mathbb{N}\right\} \label{pf.prop.perm.orbit-struc.all-N}\\
&  =\left\{  \sigma^{0}\left(  a\right)  ,\ \sigma^{1}\left(  a\right)
,\ \sigma^{2}\left(  a\right)  ,\ \ldots\right\}  .
\label{pf.prop.perm.orbit-struc.all-N2}%
\end{align}
In other words, the orbit $\left[  a\right]  _{\sim}$ of $\sigma$ consists of
all elements of $X$ that can be obtained from $a$ by repeatedly applying the
map $\sigma$ some number of times. (\textquotedblleft Some
number\textquotedblright\ allows for the number $0$, so the element
$a=\sigma^{0}\left(  a\right)  $ is included as well.)

\textbf{(b)} There exists some positive integer $m$ such that%
\begin{align*}
\sigma^{m}\left(  a\right)   &  =a\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left[  a\right]  _{\sim}  &  =\left\{  \sigma^{0}\left(  a\right)
,\ \sigma^{1}\left(  a\right)  ,\ \ldots,\ \sigma^{m-1}\left(  a\right)
\right\}  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\left\vert \left[  a\right]  _{\sim}\right\vert  &  =m.
\end{align*}

\end{proposition}

\begin{proof}
The definition of $\left[  a\right]  _{\sim}$ yields
\begin{align*}
\left[  a\right]  _{\sim}  &  =\left\{  b\in X\ \mid\ b\sim a\right\} \\
&  =\left\{  b\in X\ \mid\ b\overset{\sigma}{\sim}a\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sim\text{ is the same as
}\overset{\sigma}{\sim}\right) \\
&  =\left\{  b\in X\ \mid\ b=\sigma^{k}\left(  a\right)  \text{ for some }%
k\in\mathbb{N}\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since \textquotedblleft}b\overset{\sigma}{\sim}a\text{\textquotedblright%
\ means \textquotedblleft}b=\sigma^{k}\left(  a\right)  \text{ for some }%
k\in\mathbb{N}\text{\textquotedblright}\\
\text{(by Definition \ref{def.perm.sig-eq})}%
\end{array}
\right) \\
&  =\left\{  \sigma^{k}\left(  a\right)  \ \mid\ k\in\mathbb{N}\right\} \\
&  =\left\{  \sigma^{0}\left(  a\right)  ,\ \sigma^{1}\left(  a\right)
,\ \sigma^{2}\left(  a\right)  ,\ \ldots\right\}  .
\end{align*}
This proves Proposition \ref{prop.perm.orbit-struc} \textbf{(a)}.

Now, consider the map%
\begin{align*}
\left\{  0,1,\ldots,\left\vert X\right\vert \right\}   &  \rightarrow X,\\
i  &  \mapsto\sigma^{i}\left(  a\right)  .
\end{align*}
If this map was injective, then we would have $\left\vert \left\{
0,1,\ldots,\left\vert X\right\vert \right\}  \right\vert \leq\left\vert
X\right\vert $ (by the Pigeonhole Principle for Injections\footnote{i.e.,
Theorem 2.4.5 \textbf{(a)} in Lecture 16}), which would contradict $\left\vert
\left\{  0,1,\ldots,\left\vert X\right\vert \right\}  \right\vert =\left\vert
X\right\vert +1>\left\vert X\right\vert $. Thus, this map cannot be injective.
In other words, there exist two elements $i$ and $j$ of $\left\{
0,1,\ldots,\left\vert X\right\vert \right\}  $ such that $i<j$ and $\sigma
^{i}\left(  a\right)  =\sigma^{j}\left(  a\right)  $. Such two elements $i$
and $j$ will be called \textbf{twins}. Let $i$ and $j$ be two twins with the
smallest possible value of $j$. Thus,%
\[
i,j\in\left\{  0,1,\ldots,\left\vert X\right\vert \right\}
\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ i<j\ \ \ \ \ \ \ \ \ \ \text{and }%
\ \ \ \ \ \ \ \ \ \ \sigma^{i}\left(  a\right)  =\sigma^{j}\left(  a\right)
.
\]
Moreover,
\begin{equation}
\text{if }u\text{ and }v\text{ are any two twins, then }v\geq j
\label{pf.prop.perm.orbit-struc.j=min}%
\end{equation}
(since $i$ and $j$ are two twins with \textbf{smallest possible }$j$).

Note that $i<j$, thus $j>i\geq0$. Hence, $j$ is a positive integer.

The map $\sigma$ is a permutation, thus a bijection. Hence, it has an inverse
$\sigma^{-1}$.

Now, we claim that $i=0$. Indeed, assume the contrary. Thus, $i\geq1$, so that
$j\geq1$ as well (since $i<j$). Applying $\sigma^{-1}$ to both sides of the
equality $\sigma^{i}\left(  a\right)  =\sigma^{j}\left(  a\right)  $, we
obtain $\sigma^{i-1}\left(  a\right)  =\sigma^{j-1}\left(  a\right)  $ (since
$\sigma^{-1}\left(  \sigma^{k}\left(  a\right)  \right)  =\sigma^{k-1}\left(
a\right)  $ for each $k\in\mathbb{Z}$). Moreover, both $i-1$ and $j-1$ belong
to $\left\{  0,1,\ldots,\left\vert X\right\vert \right\}  $ (since $i\geq1$
and $j\geq1$) and satisfy $i-1<j-1$ (since $i<j$). Hence, $i-1$ and $j-1$ are
two elements of $\left\{  0,1,\ldots,\left\vert X\right\vert \right\}  $ such
that $i-1<j-1$ and $\sigma^{i-1}\left(  a\right)  =\sigma^{j-1}\left(
a\right)  $. In other words, $i-1$ and $j-1$ are twins. Thus,
(\ref{pf.prop.perm.orbit-struc.j=min}) (applied to $u=i-1$ and $v=j-1$) yields
$j-1\geq j$. But this is absurd. This contradiction shows that our assumption
was false, and therefore $i=0$ is proved.

Hence, $\sigma^{i}\left(  a\right)  =\sigma^{0}\left(  a\right)  =a$, so that
\begin{equation}
a=\sigma^{i}\left(  a\right)  =\sigma^{j}\left(  a\right)  .
\label{pf.prop.perm.orbit-struc.asi}%
\end{equation}
In other words,
\begin{equation}
\sigma^{j}\left(  a\right)  =a. \label{pf.prop.perm.orbit-struc.1}%
\end{equation}
Hence, for each $k\in\mathbb{Z}$, we have%
\begin{equation}
\sigma^{k}\left(  a\right)  =\sigma^{k-j}\left(  \underbrace{\sigma^{j}\left(
a\right)  }_{=a}\right)  =\sigma^{k-j}\left(  a\right)  .
\label{pf.prop.perm.orbit-struc.sik}%
\end{equation}
Thus, for each $k\in\mathbb{Z}$, we have%
\begin{align*}
\sigma^{k}\left(  a\right)   &  =\sigma^{k-j}\left(  a\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.perm.orbit-struc.sik}%
)}\right) \\
&  =\sigma^{k-2j}\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.perm.orbit-struc.sik}), applied to }k-j\text{ instead of
}k\right) \\
&  =\sigma^{k-3j}\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.perm.orbit-struc.sik}), applied to }k-2j\text{ instead of
}k\right) \\
&  =\cdots.
\end{align*}
In other words, the value $\sigma^{k}\left(  a\right)  $ does not change if we
subtract $j$ from $k$ any number of times.

Therefore, if $k\in\mathbb{N}$, then\footnote{Here, we let $k\%j$ denote the
remainder obtained when $k$ is divided by $j$.}%
\begin{align*}
\sigma^{k}\left(  a\right)   &  =\sigma^{k\%j}\left(  a\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }k\%j\text{ can be obtained from }k\\
\text{by subtracting }j\text{ some number of times}%
\end{array}
\right) \\
&  \in\left\{  \sigma^{0}\left(  a\right)  ,\ \sigma^{1}\left(  a\right)
,\ \ldots,\ \sigma^{j-1}\left(  a\right)  \right\}
\end{align*}
(since $k\%j$ is a remainder upon division by $j$, and therefore must be one
of the $j$ numbers $0,1,\ldots,j-1$). In other words,%
\[
\left\{  \sigma^{k}\left(  a\right)  \ \mid\ k\in\mathbb{N}\right\}
\subseteq\left\{  \sigma^{0}\left(  a\right)  ,\ \sigma^{1}\left(  a\right)
,\ \ldots,\ \sigma^{j-1}\left(  a\right)  \right\}  .
\]
Combining this inclusion with the (obvious) inclusion%
\[
\left\{  \sigma^{0}\left(  a\right)  ,\ \sigma^{1}\left(  a\right)
,\ \ldots,\ \sigma^{j-1}\left(  a\right)  \right\}  \subseteq\left\{
\sigma^{k}\left(  a\right)  \ \mid\ k\in\mathbb{N}\right\}  ,
\]
we obtain%
\[
\left\{  \sigma^{k}\left(  a\right)  \ \mid\ k\in\mathbb{N}\right\}  =\left\{
\sigma^{0}\left(  a\right)  ,\ \sigma^{1}\left(  a\right)  ,\ \ldots
,\ \sigma^{j-1}\left(  a\right)  \right\}  .
\]
In view of (\ref{pf.prop.perm.orbit-struc.all-N}), we can rewrite this as%
\begin{equation}
\left[  a\right]  _{\sim}=\left\{  \sigma^{0}\left(  a\right)  ,\ \sigma
^{1}\left(  a\right)  ,\ \ldots,\ \sigma^{j-1}\left(  a\right)  \right\}  .
\label{pf.prop.perm.orbit-struc.2}%
\end{equation}


Moreover, we claim that the $j$ elements $\sigma^{0}\left(  a\right)
,\ \sigma^{1}\left(  a\right)  ,\ \ldots,\ \sigma^{j-1}\left(  a\right)  $ are
distinct. Indeed, if two of them were equal (say, $\sigma^{u}\left(  a\right)
=\sigma^{v}\left(  a\right)  $ for some $u,v\in\left\{  0,1,\ldots
,j-1\right\}  $ satisfying $u<v$), then we would get a contradiction to
(\ref{pf.prop.perm.orbit-struc.j=min}) (since $u$ and $v$ would be twins, and
thus (\ref{pf.prop.perm.orbit-struc.j=min}) would yield $v\geq j$, but this
would contradict $v\leq j-1<j$). Thus, the $j$ elements $\sigma^{0}\left(
a\right)  ,\ \sigma^{1}\left(  a\right)  ,\ \ldots,\ \sigma^{j-1}\left(
a\right)  $ are distinct. Therefore,%
\[
\left\vert \left\{  \sigma^{0}\left(  a\right)  ,\ \sigma^{1}\left(  a\right)
,\ \ldots,\ \sigma^{j-1}\left(  a\right)  \right\}  \right\vert =j.
\]
In view of (\ref{pf.prop.perm.orbit-struc.2}), we can rewrite this as
\begin{equation}
\left\vert \left[  a\right]  _{\sim}\right\vert =j.
\label{pf.prop.perm.orbit-struc.3}%
\end{equation}


We have now proved the three equalities (\ref{pf.prop.perm.orbit-struc.1}),
(\ref{pf.prop.perm.orbit-struc.2}) and (\ref{pf.prop.perm.orbit-struc.3}).
Since $j$ is a positive integer, we thus conclude that there exists some
positive integer $m$ such that
\begin{align*}
\sigma^{m}\left(  a\right)   &  =a\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left[  a\right]  _{\sim}  &  =\left\{  \sigma^{0}\left(  a\right)
,\ \sigma^{1}\left(  a\right)  ,\ \ldots,\ \sigma^{m-1}\left(  a\right)
\right\}  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\left\vert \left[  a\right]  _{\sim}\right\vert  &  =m
\end{align*}
(namely, $m=j$). This proves Proposition \ref{prop.perm.orbit-struc}
\textbf{(b)}.
\end{proof}

\begin{proposition}
\label{prop.perm.sig-eq-equiv}Let $X$ be a finite set. Let $\sigma$ be a
permutation of $X$. Then, the relation $\overset{\sigma}{\sim}$ is an
equivalence relation.
\end{proposition}

\begin{proof}
We must prove that it satisfies three axioms: reflexivity, symmetry and transitivity.

\begin{itemize}
\item Reflexivity is easy: For any $a\in X$, we have $a\overset{\sigma}{\sim
}a$, since $a=\sigma^{k}\left(  a\right)  $ for some $k\in\mathbb{N}$ (namely,
for $k=0$).

\item Transitivity is fairly easy as well: Let $a,b,c\in X$ be such that
$a\overset{\sigma}{\sim}b$ and $b\overset{\sigma}{\sim}c$. From
$a\overset{\sigma}{\sim}b$, we obtain $a=\sigma^{i}\left(  b\right)  $ for
some $i\in\mathbb{N}$. From $b\overset{\sigma}{\sim}c$, we obtain
$b=\sigma^{j}\left(  c\right)  $ for some $j\in\mathbb{N}$. Consider these $i$
and $j$. Now,%
\[
a=\sigma^{i}\left(  \underbrace{b}_{=\sigma^{j}\left(  c\right)  }\right)
=\sigma^{i}\left(  \sigma^{j}\left(  c\right)  \right)  =\sigma^{i+j}\left(
c\right)  .
\]
Hence, $a=\sigma^{k}\left(  c\right)  $ for some $k\in\mathbb{N}$ (namely, for
$k=i+j$). In other words, $a\overset{\sigma}{\sim}c$. This proves transitivity.

\item Let us now prove symmetry. Indeed, let $a,b\in X$ be such that
$a\overset{\sigma}{\sim}b$. Denoting the relation $\overset{\sigma}{\sim}$ by
$\sim$, we thus have $a\sim b$. Hence, $a\in\left[  b\right]  _{\sim}$ (since
the $\sim$-equivalence class $\left[  b\right]  _{\sim}$ is defined to consist
of all $c\in X$ that satisfy $c\sim b$).

However, Proposition \ref{prop.perm.orbit-struc} \textbf{(b)} (applied to $b$
instead of $a$) yields that there exists some positive integer $m$ such that%
\begin{align*}
\sigma^{m}\left(  b\right)   &  =b\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left[  b\right]  _{\sim}  &  =\left\{  \sigma^{0}\left(  b\right)
,\ \sigma^{1}\left(  b\right)  ,\ \ldots,\ \sigma^{m-1}\left(  b\right)
\right\}  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\left\vert \left[  b\right]  _{\sim}\right\vert  &  =m.
\end{align*}
Consider this $m$. From $a\in\left[  b\right]  _{\sim}=\left\{  \sigma
^{0}\left(  b\right)  ,\ \sigma^{1}\left(  b\right)  ,\ \ldots,\ \sigma
^{m-1}\left(  b\right)  \right\}  $, we obtain that $a=\sigma^{i}\left(
b\right)  $ for some $i\in\left\{  0,1,\ldots,m-1\right\}  $. Consider this
$i$. Then, $i\leq m-1<m$, so that $m-i\in\mathbb{N}$. Now, $\sigma
^{m-i}\left(  \underbrace{a}_{=\sigma^{i}\left(  b\right)  }\right)
=\sigma^{m-i}\left(  \sigma^{i}\left(  b\right)  \right)  =\sigma^{m}\left(
b\right)  =b$, so that $b=\sigma^{m-i}\left(  a\right)  $. Hence,
$b=\sigma^{k}\left(  a\right)  $ for some $k\in\mathbb{N}$ (namely, for
$k=m-i$). In other words, $b\overset{\sigma}{\sim}a$. This proves symmetry.
\end{itemize}

Altogether, we have now proved all three axioms, so that Proposition
\ref{prop.perm.sig-eq-equiv} is proven.
\end{proof}

Proposition \ref{prop.perm.sig-eq-equiv} tells us that $\overset{\sigma}{\sim
}$ is an equivalence relation, so that the orbits of $\sigma$
\textquotedblleft behave well\textquotedblright:

\begin{corollary}
\label{cor.perm.orbits-sp}Let $X$ be a finite set. Let $\sigma$ be a
permutation of $X$. Then, the set of all orbits of $\sigma$ is a set partition
of $X$. In particular, each element of $X$ belongs to precisely one orbit of
$\sigma$.
\end{corollary}

\begin{proof}
The orbits of $\sigma$ are the $\overset{\sigma}{\sim}$-equivalence classes
(by their definition). Since $\overset{\sigma}{\sim}$ is an equivalence
relation, each element of $X$ belongs to exactly one $\overset{\sigma}{\sim}%
$-equivalence class (namely, its own)\footnote{Here, we have used Theorem
3.3.11 from Lecture 24.}. In other words, each element of $X$ belongs to
precisely one orbit of $\sigma$. Hence, the orbits of $\sigma$ are disjoint,
and their union is $X$. Moreover, each orbit of $\sigma$ is nonempty (since
the relation $\overset{\sigma}{\sim}$ satisfies the reflexivity axiom, and
thus each of its equivalence classes $\left[  a\right]  _{\overset{\sigma
}{\sim}}$ contains at least the element $a$). Therefore, the set of all orbits
of $\sigma$ is a set partition of $X$. Thus, Corollary
\ref{cor.perm.orbits-sp} is proved.
\end{proof}

For future use, we note the following:

\begin{remark}
\label{rmk.perm.orbit-1elt}Let $X$ be a finite set. Let $\sigma$ be a
permutation of $X$. Let $a\in X$. Then, $a$ is a fixed point of $\sigma$ if
and only if the orbit of $\sigma$ containing $a$ is a $1$-element set.
\end{remark}

\begin{proof}
Let $\sim$ be the relation $\overset{\sigma}{\sim}$. Then, the orbit of
$\sigma$ containing $a$ is the $\sim$-equivalence class $\left[  a\right]
_{\sim}$. Thus, we need to prove that $a$ is a fixed point of $\sigma$ if and
only if $\left[  a\right]  _{\sim}$ is a $1$-element set. Let us prove the
\textquotedblleft$\Longrightarrow$\textquotedblright\ and \textquotedblleft%
$\Longleftarrow$\textquotedblright\ directions of this statement separately:

$\Longrightarrow:$ Assume that $a$ is a fixed point of $\sigma$. Then,
$\sigma\left(  a\right)  =a$. Hence, by induction, we see that $\sigma
^{k}\left(  a\right)  =a$ for each $k\in\mathbb{N}$. However,
(\ref{pf.prop.perm.orbit-struc.all-N}) yields
\[
\left[  a\right]  _{\sim}=\left\{  \underbrace{\sigma^{k}\left(  a\right)
}_{=a}\ \mid\ k\in\mathbb{N}\right\}  =\left\{  a\ \mid\ k\in\mathbb{N}%
\right\}  =\left\{  a\right\}  .
\]
Thus, $\left[  a\right]  _{\sim}$ is a $1$-element set. This proves the
\textquotedblleft$\Longrightarrow$\textquotedblright\ direction.

$\Longleftarrow:$ Assume that $\left[  a\right]  _{\sim}$ is a $1$-element
set. Thus, $\left[  a\right]  _{\sim}=\left\{  a\right\}  $ (since $\left[
a\right]  _{\sim}$ contains $a$).

Now, $\sigma\left(  a\right)  \overset{\sigma}{\sim}a$, since we have
$\sigma\left(  a\right)  =\sigma^{k}\left(  a\right)  $ for some
$k\in\mathbb{N}$ (namely, for $k=1$). Hence, $\sigma\left(  a\right)
\in\left[  a\right]  _{\sim}=\left\{  a\right\}  $. In other words,
$\sigma\left(  a\right)  =a$. In other words, $a$ is a fixed point of $\sigma
$. This proves the \textquotedblleft$\Longleftarrow$\textquotedblright\ direction.
\end{proof}

\subsubsection{The DCD (= disjoint cycle decomposition) of a permutation}

The orbits of a permutation $\sigma$ do not (in general) determine $\sigma$.
For example, the $3$-cycles $\operatorname*{cyc}\nolimits_{1,2,3}$ and
$\operatorname*{cyc}\nolimits_{1,3,2}$ have the same orbits (since an orbit is
just a set, and we have $\left\{  1,2,3\right\}  =\left\{  1,3,2\right\}  $).
In order to properly encode the information from the cycle digraph, we need to
also remember how a permutation acts on each orbit.

Looking back at Example \ref{exa.perm.cycle-dig2v2}, we see that if we
restrict a permutation $\sigma$ to a given orbit of $\sigma$, then it becomes
a cycle (more precisely, an $m$-cycle, where $m$ is the size of the orbit).
This follows almost immediately from Proposition \ref{prop.perm.orbit-struc}
\textbf{(b)}:

\begin{proposition}
[structure of an orbit of $\sigma$, part 2]\label{prop.perm.orbit-struc-2}Let
$X$ be a finite set. Let $\sigma$ be a permutation of $X$.

Let $\sim$ be the relation $\overset{\sigma}{\sim}$. Let $a\in X$. Let $m$ be
the positive integer whose existence is claimed in Proposition
\ref{prop.perm.orbit-struc} \textbf{(b)}. Then, the permutation $\sigma$ and
the $m$-cycle%
\[
\operatorname*{cyc}\nolimits_{\sigma^{0}\left(  a\right)  ,\ \sigma^{1}\left(
a\right)  ,\ \ldots,\ \sigma^{m-1}\left(  a\right)  }\in S_{X}%
\]
are equal on the orbit $\left[  a\right]  _{\sim}$. In other words, for any
$b\in\left[  a\right]  _{\sim}$, we have $\sigma\left(  b\right)
=\operatorname*{cyc}\nolimits_{\sigma^{0}\left(  a\right)  ,\ \sigma
^{1}\left(  a\right)  ,\ \ldots,\ \sigma^{m-1}\left(  a\right)  }\left(
b\right)  $.
\end{proposition}

\begin{proof}
Let $b\in\left[  a\right]  _{\sim}$. We must prove that $\sigma\left(
b\right)  =\operatorname*{cyc}\nolimits_{\sigma^{0}\left(  a\right)
,\ \sigma^{1}\left(  a\right)  ,\ \ldots,\ \sigma^{m-1}\left(  a\right)
}\left(  b\right)  $.

The properties of $m$ stated in Proposition \ref{prop.perm.orbit-struc}
\textbf{(b)} yield $\sigma^{m}\left(  a\right)  =a$ and $\left[  a\right]
_{\sim}=\left\{  \sigma^{0}\left(  a\right)  ,\ \sigma^{1}\left(  a\right)
,\ \ldots,\ \sigma^{m-1}\left(  a\right)  \right\}  $.

We have $b\in\left[  a\right]  _{\sim}=\left\{  \sigma^{0}\left(  a\right)
,\ \sigma^{1}\left(  a\right)  ,\ \ldots,\ \sigma^{m-1}\left(  a\right)
\right\}  $ and therefore $b=\sigma^{k}\left(  a\right)  $ for some
$k\in\left\{  0,1,\ldots,m-1\right\}  $. Consider this $k$. From $b=\sigma
^{k}\left(  a\right)  $, we obtain
\begin{equation}
\sigma\left(  b\right)  =\sigma\left(  \sigma^{k}\left(  a\right)  \right)
=\sigma^{k+1}\left(  a\right)  . \label{pf.prop.perm.orbit-struc-2.0}%
\end{equation}
On the other hand,
\begin{align}
&  \operatorname*{cyc}\nolimits_{\sigma^{0}\left(  a\right)  ,\ \sigma
^{1}\left(  a\right)  ,\ \ldots,\ \sigma^{m-1}\left(  a\right)  }\left(
\underbrace{b}_{=\sigma^{k}\left(  a\right)  }\right) \nonumber\\
&  =\operatorname*{cyc}\nolimits_{\sigma^{0}\left(  a\right)  ,\ \sigma
^{1}\left(  a\right)  ,\ \ldots,\ \sigma^{m-1}\left(  a\right)  }\left(
\sigma^{k}\left(  a\right)  \right) \nonumber\\
&  =%
\begin{cases}
\sigma^{k+1}\left(  a\right)  , & \text{if }k<m-1;\\
\sigma^{0}\left(  a\right)  , & \text{if }k=m-1
\end{cases}
\label{pf.prop.perm.orbit-struc-2.1}%
\end{align}
(by the definition of the $m$-cycle $\operatorname*{cyc}\nolimits_{\sigma
^{0}\left(  a\right)  ,\ \sigma^{1}\left(  a\right)  ,\ \ldots,\ \sigma
^{m-1}\left(  a\right)  }$).

We claim that the right hand sides of the equalities
(\ref{pf.prop.perm.orbit-struc-2.1}) and (\ref{pf.prop.perm.orbit-struc-2.0})
are equal. Indeed, they are clearly equal in the case when $k<m-1$ (since both
of them are $\sigma^{k+1}\left(  a\right)  $ in this case). However, they are
also equal in the case when $k=m-1$ (because in this case, we have $k+1=m$ and
thus $\sigma^{k+1}\left(  a\right)  =\sigma^{m}\left(  a\right)  =a=\sigma
^{0}\left(  a\right)  $, so that $\sigma^{0}\left(  a\right)  =\sigma
^{k+1}\left(  a\right)  $). Thus, we know that the right hand sides of the
equalities (\ref{pf.prop.perm.orbit-struc-2.1}) and
(\ref{pf.prop.perm.orbit-struc-2.0}) are always equal. Thus, so are the left
hand sides. In other words,%
\[
\operatorname*{cyc}\nolimits_{\sigma^{0}\left(  a\right)  ,\ \sigma^{1}\left(
a\right)  ,\ \ldots,\ \sigma^{m-1}\left(  a\right)  }\left(  b\right)
=\sigma\left(  b\right)  .
\]
This completes the proof of Proposition \ref{prop.perm.orbit-struc-2}.
\end{proof}

\bigskip

Proposition \ref{prop.perm.orbit-struc-2} describes how a permutation $\sigma$
of a finite set $X$ acts on a given orbit of $\sigma$. However, by
understanding a permutation $\sigma$ on each of its orbits, we gain an
understanding of the whole permutation $\sigma$ on the entire set $X$. This is
best illustrated on an example:

\begin{example}
\label{exa.perm.dcd.1}Let $X=\left[  10\right]  $, and let $\sigma$ be the
permutation in Example \ref{exa.perm.cycle-dig2v2}. Its orbits are $\left\{
1,5,6,10,7\right\}  $, $\left\{  8,9\right\}  $, $\left\{  3\right\}  $ and
$\left\{  2,4\right\}  $. From Proposition \ref{prop.perm.orbit-struc-2}, we
know that $\sigma$ equals the cycles $\operatorname*{cyc}%
\nolimits_{1,5,6,10,7}$, $\operatorname*{cyc}\nolimits_{8,9}$,
$\operatorname*{cyc}\nolimits_{3}$ and $\operatorname*{cyc}\nolimits_{2,4}$ on
these respective orbits. We claim that the whole permutation $\sigma$ can
therefore be written as%
\begin{equation}
\sigma=\operatorname*{cyc}\nolimits_{1,5,6,10,7}\circ\operatorname*{cyc}%
\nolimits_{8,9}\circ\operatorname*{cyc}\nolimits_{3}\circ\operatorname*{cyc}%
\nolimits_{2,4}. \label{eq.exa.perm.dcd.1.claim}%
\end{equation}
Why is this true?

For example, let us apply both sides of (\ref{eq.exa.perm.dcd.1.claim}) to the
element $8\in X$. The left hand side clearly yields $\sigma\left(  8\right)
=9$. Applying the right hand side, we obtain%
\begin{align*}
&  \left(  \operatorname*{cyc}\nolimits_{1,5,6,10,7}\circ\operatorname*{cyc}%
\nolimits_{8,9}\circ\operatorname*{cyc}\nolimits_{3}\circ\operatorname*{cyc}%
\nolimits_{2,4}\right)  \left(  8\right) \\
&  =\operatorname*{cyc}\nolimits_{1,5,6,10,7}\left(  \operatorname*{cyc}%
\nolimits_{8,9}\left(  \operatorname*{cyc}\nolimits_{3}\left(
\underbrace{\operatorname*{cyc}\nolimits_{2,4}\left(  8\right)  }_{=8}\right)
\right)  \right) \\
&  =\operatorname*{cyc}\nolimits_{1,5,6,10,7}\left(  \operatorname*{cyc}%
\nolimits_{8,9}\left(  \underbrace{\operatorname*{cyc}\nolimits_{3}\left(
8\right)  }_{=8}\right)  \right) \\
&  =\operatorname*{cyc}\nolimits_{1,5,6,10,7}\left(
\underbrace{\operatorname*{cyc}\nolimits_{8,9}\left(  8\right)  }_{=9}\right)
\\
&  =\operatorname*{cyc}\nolimits_{1,5,6,10,7}\left(  9\right) \\
&  =9.
\end{align*}
We note that this calculation followed a very simple pattern: We started with
the element $8$ and successively applied the cycles $\operatorname*{cyc}%
\nolimits_{2,4}$, $\operatorname*{cyc}\nolimits_{3}$, $\operatorname*{cyc}%
\nolimits_{8,9}$ and $\operatorname*{cyc}\nolimits_{1,5,6,10,7}$ to it. The
first two of these cycles left $8$ unchanged (since $8\notin\left\{
2,4\right\}  $ and $8\notin\left\{  3\right\}  $). The next cycle (that is,
$\operatorname*{cyc}\nolimits_{8,9}$) sent $8$ to $9$. Finally, the last cycle
($\operatorname*{cyc}\nolimits_{1,5,6,10,7}$) left $9$ unchanged again (since
$9\notin\left\{  1,5,6,10,7\right\}  $). The reason for this simple behavior
is that each of the four cycles comes from an orbit of $\sigma$, but our
element $8$ lies in only one orbit, and thus is moved by only one cycle,
whereas the remaining cycles (whether they are applied before or after its
move) leave it unchanged.

Of course, there was nothing special about the element $8$ that we used here.
We can likewise see that for any $x\in X$, we have%
\[
\sigma\left(  x\right)  =\left(  \operatorname*{cyc}\nolimits_{1,5,6,10,7}%
\circ\operatorname*{cyc}\nolimits_{8,9}\circ\operatorname*{cyc}\nolimits_{3}%
\circ\operatorname*{cyc}\nolimits_{2,4}\right)  \left(  x\right)  .
\]
In other words, $\sigma=\operatorname*{cyc}\nolimits_{1,5,6,10,7}%
\circ\operatorname*{cyc}\nolimits_{8,9}\circ\operatorname*{cyc}\nolimits_{3}%
\circ\operatorname*{cyc}\nolimits_{2,4}$. Thus, (\ref{eq.exa.perm.dcd.1.claim}%
) is proved.

Note that each of the four cycles on the right hand side of
(\ref{eq.exa.perm.dcd.1.claim}) can be rewritten by cyclically rotating its
subscripts: For instance, $\operatorname*{cyc}\nolimits_{1,5,6,10,7}$ can be
rewritten as $\operatorname*{cyc}\nolimits_{5,6,10,7,1}$ or as
$\operatorname*{cyc}\nolimits_{6,10,7,1,5}$ or in two more ways. Furthermore,
the four cycles on the right hand side of (\ref{eq.exa.perm.dcd.1.claim}) can
be swapped at will: For instance, we have%
\begin{align*}
\sigma &  =\underbrace{\operatorname*{cyc}\nolimits_{1,5,6,10,7}%
\circ\operatorname*{cyc}\nolimits_{8,9}}_{=\operatorname*{cyc}\nolimits_{8,9}%
\circ\operatorname*{cyc}\nolimits_{1,5,6,10,7}}\circ\operatorname*{cyc}%
\nolimits_{3}\circ\operatorname*{cyc}\nolimits_{2,4}\\
&  =\operatorname*{cyc}\nolimits_{8,9}\circ\underbrace{\operatorname*{cyc}%
\nolimits_{1,5,6,10,7}\circ\operatorname*{cyc}\nolimits_{3}}%
_{=\operatorname*{cyc}\nolimits_{3}\circ\operatorname*{cyc}%
\nolimits_{1,5,6,10,7}}\circ\operatorname*{cyc}\nolimits_{2,4}\\
&  =\operatorname*{cyc}\nolimits_{8,9}\circ\operatorname*{cyc}\nolimits_{3}%
\circ\underbrace{\operatorname*{cyc}\nolimits_{1,5,6,10,7}\circ
\operatorname*{cyc}\nolimits_{2,4}}_{=\operatorname*{cyc}\nolimits_{2,4}%
\circ\operatorname*{cyc}\nolimits_{1,5,6,10,7}}\\
&  =\operatorname*{cyc}\nolimits_{8,9}\circ\operatorname*{cyc}\nolimits_{3}%
\circ\operatorname*{cyc}\nolimits_{2,4}\circ\operatorname*{cyc}%
\nolimits_{1,5,6,10,7}=\cdots.
\end{align*}
The reason for this is the following (very easily proved) fact:
\end{example}

\begin{proposition}
\label{prop.perm.cycles-commute}Let $X$ be a set. Let $b_{1},b_{2}%
,\ldots,b_{i},c_{1},c_{2},\ldots,c_{j}$ be distinct elements of $X$. Then,%
\[
\operatorname*{cyc}\nolimits_{b_{1},b_{2},\ldots,b_{i}}\circ
\operatorname*{cyc}\nolimits_{c_{1},c_{2},\ldots,c_{j}}=\operatorname*{cyc}%
\nolimits_{c_{1},c_{2},\ldots,c_{j}}\circ\operatorname*{cyc}\nolimits_{b_{1}%
,b_{2},\ldots,b_{i}}.
\]

\end{proposition}

\begin{proof}
Easy and left to the reader.
\end{proof}

Generalizing the reasoning from Example \ref{exa.perm.dcd.1}, we find the following:

\begin{theorem}
[disjoint cycle decomposition of a permutation]\label{thm.perm.dcd}Let $X$ be
a finite set. Let $\sigma$ be a permutation of $X$. Then:

\textbf{(a)} There is a list%
\begin{align*}
&  \Big(\ \left(  a_{1,1},a_{1,2},\ldots,a_{1,m_{1}}\right)  ,\\
&  \ \ \ \left(  a_{2,1},a_{2,2},\ldots,a_{2,m_{2}}\right)  ,\\
&  \ \ \ \ldots,\\
&  \ \ \ \left(  a_{k,1},a_{k,2},\ldots,a_{k,m_{k}}\right)  \ \Big)
\end{align*}
of nonempty lists of elements of $X$ such that:

\begin{itemize}
\item Each element of $X$ appears exactly once in the composite list%
\begin{align*}
&  (\ a_{1,1},a_{1,2},\ldots,a_{1,m_{1}},\\
&  \ \ \ a_{2,1},a_{2,2},\ldots,a_{2,m_{2}},\\
&  \ \ \ \ldots,\\
&  \ \ \ a_{k,1},a_{k,2},\ldots,a_{k,m_{k}}\ ).
\end{align*}


\item We have%
\[
\sigma=\operatorname*{cyc}\nolimits_{a_{1,1},a_{1,2},\ldots,a_{1,m_{1}}}%
\circ\operatorname*{cyc}\nolimits_{a_{2,1},a_{2,2},\ldots,a_{2,m_{2}}}%
\circ\cdots\circ\operatorname*{cyc}\nolimits_{a_{k,1},a_{k,2},\ldots
,a_{k,m_{k}}}.
\]

\end{itemize}

Such a list is called a \textbf{disjoint cycle decomposition} (short:
\textbf{DCD}) of $\sigma$. Its entries (which themselves are lists of elements
of $X$) are called the \textbf{cycles} of $\sigma$.

\textbf{(b)} Any two DCDs of $\sigma$ can be obtained from one another by
(repeatedly) swapping the cycles with each other and rotating each cycle
(i.e., replacing $\left(  a_{i,1},a_{i,2},\ldots,a_{i,m_{i}}\right)  $ by
$\left(  a_{i,2},a_{i,3},\ldots,a_{i,m_{i}},a_{i,1}\right)  $).

\textbf{(c)} Now assume that $X$ is a set of integers (or, more generally, any
totally ordered set). Then, there is a unique DCD%
\begin{align*}
&  \Big(\ \left(  a_{1,1},a_{1,2},\ldots,a_{1,m_{1}}\right)  ,\\
&  \ \ \ \left(  a_{2,1},a_{2,2},\ldots,a_{2,m_{2}}\right)  ,\\
&  \ \ \ \ldots,\\
&  \ \ \ \left(  a_{k,1},a_{k,2},\ldots,a_{k,m_{k}}\right)  \ \Big)
\end{align*}
of $\sigma$ that satisfies the following two additional requirements:

\begin{itemize}
\item We have $a_{i,1}\leq a_{i,p}$ for each $i\in\left[  k\right]  $ and each
$p\in\left[  m_{i}\right]  $ (that is, each cycle in the DCD is written with
its smallest entry first).

\item We have $a_{1,1}>a_{2,1}>\cdots>a_{k,1}$ (that is, the cycles appear in
the DCD in the order of decreasing first entries).
\end{itemize}
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.perm.dcd} (sketched).]\textbf{(a)} Let $\sim$ be
the relation $\overset{\sigma}{\sim}$.

Let $X_{1},X_{2},\ldots,X_{k}$ be the orbits of $\sigma$ (listed with no repetition).

Let $i\in\left[  k\right]  $. Then, the set $X_{i}$ is an orbit of $\sigma$,
that is, a $\sim$-equivalence class (by the definition of an orbit). Hence, it
can be written in the form $X_{i}=\left[  x_{i}\right]  _{\sim}$ for some
$x_{i}\in X$. Consider this $x_{i}$. Proposition \ref{prop.perm.orbit-struc}
\textbf{(b)} (applied to $a=x_{i}$) yields that there exists some positive
integer $m_{i}$ such that%
\begin{align}
\sigma^{m_{i}}\left(  x_{i}\right)   &  =x_{i}\ \ \ \ \ \ \ \ \ \ \text{and}%
\label{pf.thm.perm.dcd.a.1}\\
\left[  x_{i}\right]  _{\sim}  &  =\left\{  \sigma^{0}\left(  x_{i}\right)
,\ \sigma^{1}\left(  x_{i}\right)  ,\ \ldots,\ \sigma^{m_{i}-1}\left(
x_{i}\right)  \right\}  \ \ \ \ \ \ \ \ \ \ \text{and}%
\label{pf.thm.perm.dcd.a.2}\\
\left\vert \left[  x_{i}\right]  _{\sim}\right\vert  &  =m_{i}.
\label{pf.thm.perm.dcd.a.3}%
\end{align}
Consider this $m_{i}$.

Proposition \ref{prop.perm.orbit-struc-2} (applied to $a=x_{i}$ and $m=m_{i}$)
yields that the permutation $\sigma$ and the $m_{i}$-cycle%
\[
\operatorname*{cyc}\nolimits_{\sigma^{0}\left(  x_{i}\right)  ,\ \sigma
^{1}\left(  x_{i}\right)  ,\ \ldots,\ \sigma^{m_{i}-1}\left(  x_{i}\right)
}\in S_{X}%
\]
are equal on the orbit $\left[  x_{i}\right]  _{\sim}=X_{i}$.

Forget that we fixed $i$. Thus, for each $i\in\left[  k\right]  $, we have
found an $x_{i}\in X$ and a positive integer $m_{i}$ satisfying
(\ref{pf.thm.perm.dcd.a.1}), (\ref{pf.thm.perm.dcd.a.2}) and
(\ref{pf.thm.perm.dcd.a.3}) and with the property that the permutation
$\sigma$ and the $m_{i}$-cycle%
\[
\operatorname*{cyc}\nolimits_{\sigma^{0}\left(  x_{i}\right)  ,\ \sigma
^{1}\left(  x_{i}\right)  ,\ \ldots,\ \sigma^{m_{i}-1}\left(  x_{i}\right)
}\in S_{X}%
\]
are equal on the orbit $\left[  x_{i}\right]  _{\sim}=X_{i}$. Hence, each
element of $X$ appears exactly once in the list%
\begin{align*}
&  (\ \sigma^{0}\left(  x_{1}\right)  ,\ \sigma^{1}\left(  x_{1}\right)
,\ \ldots,\ \sigma^{m_{1}-1}\left(  x_{1}\right)  ,\\
&  \ \ \ \sigma^{0}\left(  x_{2}\right)  ,\ \sigma^{1}\left(  x_{2}\right)
,\ \ldots,\ \sigma^{m_{2}-1}\left(  x_{2}\right)  ,\\
&  \ \ \ \ldots,\\
&  \ \ \ \sigma^{0}\left(  x_{k}\right)  ,\ \sigma^{1}\left(  x_{k}\right)
,\ \ldots,\ \sigma^{m_{k}-1}\left(  x_{k}\right)  \ )
\end{align*}
(by (\ref{pf.thm.perm.dcd.a.1}) and (\ref{pf.thm.perm.dcd.a.2}), since each
element of $X$ belongs to exactly one of the orbits $X_{1},X_{2},\ldots,X_{k}%
$), and furthermore, we have
\begin{align*}
\sigma &  =\operatorname*{cyc}\nolimits_{\sigma^{0}\left(  x_{1}\right)
,\ \sigma^{1}\left(  x_{1}\right)  ,\ \ldots,\ \sigma^{m_{1}-1}\left(
x_{1}\right)  }\circ\operatorname*{cyc}\nolimits_{\sigma^{0}\left(
x_{2}\right)  ,\ \sigma^{1}\left(  x_{2}\right)  ,\ \ldots,\ \sigma^{m_{2}%
-1}\left(  x_{2}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \circ\cdots\circ\operatorname*{cyc}%
\nolimits_{\sigma^{0}\left(  x_{k}\right)  ,\ \sigma^{1}\left(  x_{k}\right)
,\ \ldots,\ \sigma^{m_{k}-1}\left(  x_{k}\right)  }%
\end{align*}
(this can be proved by the same reasoning as we used in Example
\ref{exa.perm.dcd.1} to prove (\ref{eq.exa.perm.dcd.1.claim})). Thus, the list%
\begin{align*}
&  \Big(\ \left(  \sigma^{0}\left(  x_{1}\right)  ,\ \sigma^{1}\left(
x_{1}\right)  ,\ \ldots,\ \sigma^{m_{1}-1}\left(  x_{1}\right)  \right)  ,\\
&  \ \ \ \left(  \sigma^{0}\left(  x_{2}\right)  ,\ \sigma^{1}\left(
x_{2}\right)  ,\ \ldots,\ \sigma^{m_{2}-1}\left(  x_{2}\right)  \right)  ,\\
&  \ \ \ \ldots,\\
&  \ \ \ \left(  \sigma^{0}\left(  x_{k}\right)  ,\ \sigma^{1}\left(
x_{k}\right)  ,\ \ldots,\ \sigma^{m_{k}-1}\left(  x_{k}\right)  \right)
\ \Big)
\end{align*}
is a DCD of $\sigma$. This proves that a DCD of $\sigma$ exists. In other
words, Theorem \ref{thm.perm.dcd} \textbf{(a)} is proven.

(For alternative proofs of Theorem \ref{thm.perm.dcd} \textbf{(a)}, see (e.g.)
\cite[proof of Theorem 5.5.2 \textbf{(a)}]{21s}, \cite[Exercise 7 \textbf{(e)}
and \textbf{(d)}]{17f-hw7s}, \cite[Theorem 1.5.3]{Goodman}, \cite[Chapter I,
\S 5.7, Proposition 7]{Bourba74}, or
\url{https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Cycle_Decomposition}
(but see Remark \ref{rmk.perm.dcd.1} further below for a certain disagreement
about the definition of a DCD).)\medskip

\textbf{(b)} See \cite[Theorem 1.5.3]{Goodman} or \cite[Chapter I, \S 5.7,
Proposition 7]{Bourba74}. The idea is fairly simple: Let%
\begin{align*}
&  \Big(\left(  a_{1,1},a_{1,2},\ldots,a_{1,m_{1}}\right)  ,\\
&  \ \ \ \left(  a_{2,1},a_{2,2},\ldots,a_{2,m_{2}}\right)  ,\\
&  \ \ \ \ldots,\\
&  \ \ \ \left(  a_{k,1},a_{k,2},\ldots,a_{k,m_{k}}\right)  \Big)
\end{align*}
be a DCD of $\sigma$. Then, for each $i\in X$, the cycle of this DCD that
contains $i$ is uniquely determined by $\sigma$ and $i$ up to cyclic rotation
(indeed, it is a rotated version of the list $\left(  i,\sigma\left(
i\right)  ,\sigma^{2}\left(  i\right)  ,\ldots,\sigma^{r-1}\left(  i\right)
\right)  $, where $r$ is the smallest positive integer satisfying $\sigma
^{r}\left(  i\right)  =i$). Therefore, all cycles of this DCD are uniquely
determined by $\sigma$ up to cyclic rotation and up to the relative order in
which these cycles appear in the DCD. But this is precisely the claim of
Theorem \ref{thm.perm.dcd} \textbf{(b)}. \medskip

\begin{noncompile}
First of all, it is clear that if you start with a DCD of $\sigma$ and
cyclically rotate its cycles (any number of times) and swap the cycles around,
the result will still be a DCD of $\sigma$. (Indeed, the cycles in a DCD have
no elements in common, and thus the permutation $\operatorname*{cyc}%
\nolimits_{a_{1,1},a_{1,2},\ldots,a_{1,m_{1}}}\circ\operatorname*{cyc}%
\nolimits_{a_{2,1},a_{2,2},\ldots,a_{2,m_{2}}}\circ\cdots\circ
\operatorname*{cyc}\nolimits_{a_{k,1},a_{k,2},\ldots,a_{k,m_{k}}}$ does not
change if they are swapped\footnote{Here we are using the following (very
easily proved) fact: If $b_{1},b_{2},\ldots,b_{i},c_{1},c_{2},\ldots,c_{j}$
are distinct elements of $X$, then $\operatorname*{cyc}\nolimits_{b_{1}%
,b_{2},\ldots,b_{i}}\circ\operatorname*{cyc}\nolimits_{c_{1},c_{2}%
,\ldots,c_{j}}=\operatorname*{cyc}\nolimits_{c_{1},c_{2},\ldots,c_{j}}%
\circ\operatorname*{cyc}\nolimits_{b_{1},b_{2},\ldots,b_{i}}$.}.)
\end{noncompile}

\textbf{(c)} Clearly, such a DCD of $\sigma$ exists: Indeed, we can start with
an arbitrary DCD of $\sigma$ (this exists because of Theorem
\ref{thm.perm.dcd} \textbf{(a)}), and then transform it using cyclic rotations
and swaps into a form that satisfies the two additional requirements of
Theorem \ref{thm.perm.dcd} \textbf{(c).} Namely, we

\begin{itemize}
\item first cyclically rotate each cycle to ensure that it begins with its
smallest entry, and

\item then we swap the cycles appropriately to ensure that they appear in the
order of decreasing first entries.
\end{itemize}

It remains to prove that a DCD of $\sigma$ satisfying the two additional
requirements of Theorem \ref{thm.perm.dcd} \textbf{(c)} is unique. But this
follows easily from Theorem \ref{thm.perm.dcd} \textbf{(b)}: Indeed, Theorem
\ref{thm.perm.dcd} \textbf{(b)} shows that any two such DCDs can be
transformed into one another by rotating each cycle and swapping the cycles;
however, the two additional requirements uniquely determine the first entry of
each cycle and also the order of the cycles, and thus there is no freedom left
for two different DCDs to fit the bill.
\end{proof}

\begin{remark}
\label{rmk.perm.dcd.1}Some authors omit $1$-element cycles from the DCD of a
permutation, since we have $\operatorname*{cyc}\nolimits_{i}%
=\operatorname*{id}$ for each $i\in X$. However, if you do this, then the
condition \textquotedblleft Each element of $X$ appears exactly once in the
composite list\textquotedblright\ in the definition of a DCD has to be
replaced by \textquotedblleft Each element of $X$ appears at most once in the
composite list\textquotedblright.
\end{remark}

\subsubsection{Composing with a transposition}

Before we move on with counting questions, we shall prove a lemma that we will
soon need. The lemma is about how a transposition $t_{i,j}$ affects the orbits
of a permutation $\sigma$ when it is multiplied onto $\sigma$ from the left --
i.e., about how the orbits of $t_{i,j}\circ\sigma$ differ from those of
$\sigma$.

Recall the notion of a transposition $t_{i,j}$, defined in Definition 4.2.1
(Lecture 26): If $i$ and $j$ are two distinct elements of a set $X$, then the
transposition $t_{i,j}$ is the permutation of $X$ that swaps $i$ with $j$,
while leaving all other elements of $X$ unchanged. We recall that this
permutation $t_{i,j}$ satisfies $t_{i,j}\circ t_{i,j}=\operatorname*{id}$
(since swapping $i$ with $j$ twice results in every element of $X$ returning
where it was).

\begin{lemma}
\label{lem.perm.dcd.transp}Let $X$ be a finite set. Let $i$ and $j$ be two
distinct elements of $X$. Let $\sigma\in S_{X}$ be a permutation, and let
$\tau=t_{i,j}\circ\sigma$. Then:

\textbf{(a)} If $i\overset{\sigma}{\sim}j$, then the permutation $\tau$ has
$1$ more orbit than $\sigma$.

\textbf{(b)} If $i\overset{\sigma}{\sim}j$, then we don't have $i\overset{\tau
}{\sim}j$.

\textbf{(c)} If we don't have $i\overset{\sigma}{\sim}j$, then the permutation
$\tau$ has $1$ fewer orbit than $\sigma$.

\textbf{(d)} If we don't have $i\overset{\sigma}{\sim}j$, then we have
$i\overset{\tau}{\sim}j$.
\end{lemma}

\begin{example}
\label{exa.perm.dcd.transp.1}Let $X$ be a $10$-element set $\left\{
a,b,c,\ldots,j\right\}  $, and let $\sigma\in S_{X}$ be the permutation whose
cycle digraph looks as follows:%
\[%
%TCIMACRO{\TeXButton{tikz sigma}{\begin{tikzpicture}%
%[->,shorten >=1pt,auto,node distance=3cm, thick,
%main node/.style={circle, fill=blue!20,draw},
%yellow node/.style={circle, fill=yellow!50, draw}]
%\node[main node] (a) at (0:2) {$a$};
%\node[main node] (b) at (360/7:2) {$b$};
%\node[yellow node] (j) at (2*360/7:2) {$j$};
%\node[main node] (c) at (3*360/7:2) {$c$};
%\node[main node] (d) at (4*360/7:2) {$d$};
%\node[main node] (e) at (5*360/7:2) {$e$};
%\node[yellow node] (i) at (6*360/7:2) {$i$};
%\node[main node] (f) at (4,1) {$f$};
%\node[main node] (g) at (6,1) {$g$};
%\node[main node] (h) at (4,-1) {$h$};
%\path[-{Stealth[length=4mm]}]
%(a) edge [bend right] (b)
%(b) edge [bend right, draw=blue] (j)
%(j) edge [bend right] (c)
%(c) edge [bend right] (d)
%(d) edge [bend right] (e)
%(e) edge [bend right, draw=blue] (i)
%(i) edge [bend right] (a)
%(f) edge [bend right] (g)
%(g) edge [bend right] (f)
%(h) edge [loop right] (h);
%\end{tikzpicture}}}%
%BeginExpansion
\begin{tikzpicture}[->,shorten >=1pt,auto,node distance=3cm, thick,
main node/.style={circle, fill=blue!20,draw},
yellow node/.style={circle, fill=yellow!50, draw}]
\node[main node] (a) at (0:2) {$a$};
\node[main node] (b) at (360/7:2) {$b$};
\node[yellow node] (j) at (2*360/7:2) {$j$};
\node[main node] (c) at (3*360/7:2) {$c$};
\node[main node] (d) at (4*360/7:2) {$d$};
\node[main node] (e) at (5*360/7:2) {$e$};
\node[yellow node] (i) at (6*360/7:2) {$i$};
\node[main node] (f) at (4,1) {$f$};
\node[main node] (g) at (6,1) {$g$};
\node[main node] (h) at (4,-1) {$h$};
\path[-{Stealth[length=4mm]}]
(a) edge [bend right] (b)
(b) edge [bend right, draw=blue] (j)
(j) edge [bend right] (c)
(c) edge [bend right] (d)
(d) edge [bend right] (e)
(e) edge [bend right, draw=blue] (i)
(i) edge [bend right] (a)
(f) edge [bend right] (g)
(g) edge [bend right] (f)
(h) edge [loop right] (h);
\end{tikzpicture}%
%EndExpansion
\ \ .
\]
Note that $i$ and $j$ visibly belong to the same orbit of $\sigma$, so that
$i\overset{\sigma}{\sim}j$ (and, specifically, $j=\sigma^{3}\left(  i\right)
$).

Let $\tau=t_{i,j}\circ\sigma$. This permutation $\tau$ has the following cycle
digraph:
\[%
%TCIMACRO{\TeXButton{tikz tau}{\begin{tikzpicture}%
%[->,shorten >=1pt,auto,node distance=3cm, thick,
%main node/.style={circle, fill=blue!20,draw},
%yellow node/.style={circle, fill=yellow!50, draw}]
%\node[main node] (a) at (0:2) {$a$};
%\node[main node] (b) at (360/7:2) {$b$};
%\node[yellow node] (j) at (2*360/7:2) {$j$};
%\node[main node] (c) at (3*360/7:2) {$c$};
%\node[main node] (d) at (4*360/7:2) {$d$};
%\node[main node] (e) at (5*360/7:2) {$e$};
%\node[yellow node] (i) at (6*360/7:2) {$i$};
%\node[main node] (f) at (4,1) {$f$};
%\node[main node] (g) at (6,1) {$g$};
%\node[main node] (h) at (4,-1) {$h$};
%\path[-{Stealth[length=4mm]}]
%(a) edge [bend right] (b)
%(b) edge [bend right, draw=red] (i)
%(j) edge [bend right] (c)
%(c) edge [bend right] (d)
%(d) edge [bend right] (e)
%(e) edge [bend right, draw=red] (j)
%(i) edge [bend right] (a)
%(f) edge [bend right] (g)
%(g) edge [bend right] (f)
%(h) edge [loop right] (h);
%\end{tikzpicture}}}%
%BeginExpansion
\begin{tikzpicture}[->,shorten >=1pt,auto,node distance=3cm, thick,
main node/.style={circle, fill=blue!20,draw},
yellow node/.style={circle, fill=yellow!50, draw}]
\node[main node] (a) at (0:2) {$a$};
\node[main node] (b) at (360/7:2) {$b$};
\node[yellow node] (j) at (2*360/7:2) {$j$};
\node[main node] (c) at (3*360/7:2) {$c$};
\node[main node] (d) at (4*360/7:2) {$d$};
\node[main node] (e) at (5*360/7:2) {$e$};
\node[yellow node] (i) at (6*360/7:2) {$i$};
\node[main node] (f) at (4,1) {$f$};
\node[main node] (g) at (6,1) {$g$};
\node[main node] (h) at (4,-1) {$h$};
\path[-{Stealth[length=4mm]}]
(a) edge [bend right] (b)
(b) edge [bend right, draw=red] (i)
(j) edge [bend right] (c)
(c) edge [bend right] (d)
(d) edge [bend right] (e)
(e) edge [bend right, draw=red] (j)
(i) edge [bend right] (a)
(f) edge [bend right] (g)
(g) edge [bend right] (f)
(h) edge [loop right] (h);
\end{tikzpicture}%
%EndExpansion
\ \ .
\]


Observe that the cycle digraphs of $\sigma$ and $\tau$ differ only in two of
their arcs: namely, the arcs that end in the nodes $i$ and $j$. (These are the
two arcs that we colored blue in the cycle digraph of $\sigma$ and red in the
cycle digraph of $\tau$.) The change from $\sigma$ to $\tau$ causes these two
arcs to swap their targets (= ending points). All other arcs of the cycle
digraph of $\sigma$ (colored black) appear equally in the cycle digraph of
$\tau$. By comparing the two cycle digraphs, we see that the cycle digraph of
$\tau$ contains $1$ more cycle than that of $\sigma$. In other words, the
permutation $\tau$ has $1$ more orbit than $\sigma$. This confirms Lemma
\ref{lem.perm.dcd.transp} \textbf{(a)} in our example. Furthermore, $i$ and
$j$ lie in different cycles of the cycle digraph of $\tau$, thus belong to
different orbits of $\tau$. In other words, we don't have $i\overset{\tau
}{\sim}j$. This confirms Lemma \ref{lem.perm.dcd.transp} \textbf{(b)} in our
example. Essentially, by composing $t_{i,j}$ with $\sigma$, we have
\textquotedblleft broken up\textquotedblright\ the cycle of $\sigma$ that
contained both $i$ and $j$ into two smaller cycles, one of which contains $i$
while the other contains $j$.

The same example can also be used to illustrate parts \textbf{(c)} and
\textbf{(d)} of Lemma \ref{lem.perm.dcd.transp}, once we swap $\sigma$ with
$\tau$. Indeed, from $\tau=t_{i,j}\circ\sigma$, we obtain $\sigma=t_{i,j}%
\circ\tau$ (because $t_{i,j}\circ\underbrace{\tau}_{=t_{i,j}\circ\sigma
}=\underbrace{t_{i,j}\circ t_{i,j}}_{=\operatorname*{id}}\circ\sigma
=\operatorname*{id}\circ\sigma=\sigma$), and therefore the relationship
between $\sigma$ and $\tau$ is mutual. As we know that we don't have
$i\overset{\tau}{\sim}j$, we thus see that the assumptions of Lemma
\ref{lem.perm.dcd.transp} \textbf{(c)} and \textbf{(d)} are satisfied if we
swap $\sigma$ with $\tau$. The conclusions, too, are satisfied, as we see from
the cycle digraphs. Essentially, when we compose $t_{i,j}$ with $\tau$, the
two cycles containing $i$ and $j$ get \textquotedblleft
merged\textquotedblright\ into a common cycle.
\end{example}

\begin{fineprint}
\begin{proof}
[Proof of Lemma \ref{lem.perm.dcd.transp} (sketched).]We generalize and
formalize what we already saw in Example \ref{exa.perm.dcd.transp.1}. If $a$
and $b$ are two elements of $X$, then the notation \textquotedblleft%
$a\overset{\sigma}{\mapsto}b$\textquotedblright\ shall mean \textquotedblleft%
$\sigma\left(  a\right)  =b$\textquotedblright, whereas the notation
\textquotedblleft$a\overset{\tau}{\mapsto}b$\textquotedblright\ shall mean
\textquotedblleft$\tau\left(  a\right)  =b$\textquotedblright. We note that if
$k$ elements $a_{1},a_{2},\ldots,a_{k}$ of $X$ (with $k\geq1$) form a circular
chain of the form%
\begin{equation}
a_{1}\overset{\tau}{\mapsto}a_{2}\overset{\tau}{\mapsto}\cdots\overset{\tau
}{\mapsto}a_{k}\overset{\tau}{\mapsto}a_{1},
\label{pf.lem.perm.dcd.transp.chain}%
\end{equation}
\medskip then the set $\left\{  a_{1},a_{2},\ldots,a_{k}\right\}  $ must be an
orbit of $\tau$ (since (\ref{pf.lem.perm.dcd.transp.chain}) shows that
$\left\{  a_{1},a_{2},\ldots,a_{k}\right\}  =\left\{  \tau^{0}\left(
a_{1}\right)  ,\ \tau^{1}\left(  a_{1}\right)  ,\ \tau^{2}\left(
a_{1}\right)  ,\ \ldots\right\}  $, but we know from
(\ref{pf.prop.perm.orbit-struc.all-N2}) (applied to $\tau$ and $a_{1}$ instead
of $\sigma$ and $a$) that the set $\left\{  \tau^{0}\left(  a_{1}\right)
,\ \tau^{1}\left(  a_{1}\right)  ,\ \tau^{2}\left(  a_{1}\right)
,\ \ldots\right\}  $ is an orbit of $\tau$). \medskip

\begin{noncompile}
We have $\tau=t_{i,j}\circ\sigma$. In other words, the permutation $\tau$ is
obtained from $\sigma$ by swapping the output values $i$ and $j$; that is, we
have%
\[
\tau\left(  k\right)  =%
\begin{cases}
\sigma\left(  k\right)  , & \text{if }\sigma\left(  k\right)  \neq i\text{ and
}\sigma\left(  k\right)  \neq j;\\
j, & \text{if }\sigma\left(  k\right)  =i;\\
i, & \text{if }\sigma\left(  k\right)  =j
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for each }k\in X.
\]
Hence, the cycle digraph of $\tau$ looks almost the same as the cycle digraph
of $\sigma$; more precisely, the former cycle digraph can be obtained from the
latter by swapping the targets of the two arcs that end in the nodes $i$ and
$j$. We need to explain why this swap results in the cycle ...
\end{noncompile}

\textbf{(a)} Assume that $i\overset{\sigma}{\sim}j$. Thus, $i$ and $j$ belong
to the $\overset{\sigma}{\sim}$-equivalence class, i.e., to the same orbit of
$\sigma$. Let $X_{1}$ be this orbit, and let $X_{2},X_{3},\ldots,X_{k}$ be all
the remaining orbits of $\sigma$ (listed without repetition). Thus, $\sigma$
has $k$ orbits. We shall now understand what the orbits of $\tau$ are.

Let $\sim$ be the relation $\overset{\sigma}{\sim}$. Thus, $i\sim j$ (since
$i\overset{\sigma}{\sim}j$). Proposition \ref{prop.perm.orbit-struc}
\textbf{(b)} (applied to $a=i$) yields that there exists some positive integer
$m$ such that%
\begin{align*}
\sigma^{m}\left(  i\right)   &  =i\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left[  i\right]  _{\sim}  &  =\left\{  \sigma^{0}\left(  i\right)
,\ \sigma^{1}\left(  i\right)  ,\ \ldots,\ \sigma^{m-1}\left(  i\right)
\right\}  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\left\vert \left[  i\right]  _{\sim}\right\vert  &  =m.
\end{align*}
Consider this $m$. From $i\sim j$, we obtain $j\in\left[  i\right]  _{\sim
}=\left\{  \sigma^{0}\left(  i\right)  ,\ \sigma^{1}\left(  i\right)
,\ \ldots,\ \sigma^{m-1}\left(  i\right)  \right\}  $. In other words,
$j=\sigma^{p}\left(  i\right)  $ for some $p\in\left\{  0,1,\ldots
,m-1\right\}  $. Consider this $p$. Thus, $\sigma^{p}\left(  i\right)  =j\neq
i=\sigma^{0}\left(  i\right)  $, so that $p\neq0$. Hence, $p\in\left[
m-1\right]  $ (since $p\in\left\{  0,1,\ldots,m-1\right\}  $ but $p\neq0$).

Recall that $X_{1}$ is the orbit of $\sigma$ that contains $i$. In other
words, $X_{1}=\left[  i\right]  _{\sim}$ (since $\left[  i\right]  _{\sim}$ is
the orbit of $\sigma$ that contains $i$). Therefore,%
\[
X_{1}=\left[  i\right]  _{\sim}=\left\{  \sigma^{0}\left(  i\right)
,\ \sigma^{1}\left(  i\right)  ,\ \ldots,\ \sigma^{m-1}\left(  i\right)
\right\}  =\left\{  i,\ \sigma^{1}\left(  i\right)  ,\ \sigma^{2}\left(
i\right)  ,\ \ldots,\ \sigma^{m-1}\left(  i\right)  \right\}
\]
(since $\sigma^{0}\left(  i\right)  =i$). The action of the map $\sigma$ on
the elements of $X_{1}$ looks as follows:%
\[
i\overset{\sigma}{\mapsto}\sigma^{1}\left(  i\right)  \overset{\sigma
}{\mapsto}\sigma^{2}\left(  i\right)  \overset{\sigma}{\mapsto}\cdots
\overset{\sigma}{\mapsto}\sigma^{m-1}\left(  i\right)  \overset{\sigma
}{\mapsto}i
\]
(since $\sigma\left(  \sigma^{m-1}\left(  i\right)  \right)  =\sigma
^{m}\left(  i\right)  =i$). Since $j=\sigma^{p}\left(  i\right)  $, we can
rewrite the element $\sigma^{p}\left(  i\right)  $ on this circular chain as
$j$, so that the chain becomes%
\begin{align}
i  &  \overset{\sigma}{\mapsto}\sigma^{1}\left(  i\right)  \overset{\sigma
}{\mapsto}\sigma^{2}\left(  i\right)  \overset{\sigma}{\mapsto}\cdots
\overset{\sigma}{\mapsto}\sigma^{p-1}\left(  i\right) \nonumber\\
\overset{\sigma}{\mapsto}j  &  \overset{\sigma}{\mapsto}\sigma^{p+1}\left(
i\right)  \overset{\sigma}{\mapsto}\sigma^{p+2}\left(  i\right)
\overset{\sigma}{\mapsto}\cdots\overset{\sigma}{\mapsto}\sigma^{m-1}\left(
i\right)  \overset{\sigma}{\mapsto}i. \label{pf.lem.perm.dcd.transp.a.chain}%
\end{align}


However, the transposition $t_{i,j}$ swaps $i$ with $j$ while leaving all
other elements unchanged. Thus, the permutation $\tau=t_{i,j}\circ\sigma$
differs from $\sigma$ only in two things:

\begin{itemize}
\item The element of $X$ that is sent to $i$ by $\sigma$ is instead sent to
$j$ by $\tau$ (because $t_{i,j}$ sends $i$ to $j$).

\item The element of $X$ that is sent to $j$ by $\sigma$ is instead sent to
$i$ by $\tau$ (because $t_{i,j}$ sends $j$ to $i$).

\item All other elements of $X$ are sent by $\tau$ to the same value that they
are sent to by $\sigma$.
\end{itemize}

Hence, when we replace $\sigma$ by $\tau$, then the two arrows $\sigma
^{p-1}\left(  i\right)  \overset{\sigma}{\mapsto}j$ and $\sigma^{m-1}\left(
i\right)  \overset{\sigma}{\mapsto}i$ in the circular chain
(\ref{pf.lem.perm.dcd.transp.a.chain}) turn into $\sigma^{p-1}\left(
i\right)  \overset{\tau}{\mapsto}i$ and $\sigma^{m-1}\left(  i\right)
\overset{\tau}{\mapsto}j$, whereas all the other arrows remain unchanged.
Hence, the circular chain (\ref{pf.lem.perm.dcd.transp.a.chain}) turns into
two separate circular chains%
\begin{align*}
&  i\overset{\tau}{\mapsto}\sigma^{1}\left(  i\right)  \overset{\tau}{\mapsto
}\sigma^{2}\left(  i\right)  \overset{\tau}{\mapsto}\cdots\overset{\tau
}{\mapsto}\sigma^{p-1}\left(  i\right)  \overset{\tau}{\mapsto}%
i\ \ \ \ \ \ \ \ \ \ \text{and}\\
&  j\overset{\tau}{\mapsto}\sigma^{p+1}\left(  i\right)  \overset{\tau
}{\mapsto}\sigma^{p+2}\left(  i\right)  \overset{\tau}{\mapsto}\cdots
\overset{\tau}{\mapsto}\sigma^{m-1}\left(  i\right)  \overset{\tau}{\mapsto}j.
\end{align*}
Hence, in place of the orbit $X_{1}$ of $\sigma$, the permutation $\tau$ has
two disjoint orbits%
\begin{align*}
X_{1}^{\prime}  &  :=\left\{  i,\ \sigma^{1}\left(  i\right)  ,\ \sigma
^{2}\left(  i\right)  ,\ \ldots,\ \sigma^{p-1}\left(  i\right)  \right\}
\ \ \ \ \ \ \ \ \ \ \text{and}\\
X_{1}^{\prime\prime}  &  :=\left\{  j,\ \sigma^{p+1}\left(  i\right)
,\ \sigma^{p+2}\left(  i\right)  ,\ \ldots,\ \sigma^{m-1}\left(  i\right)
\right\}  .
\end{align*}
All the remaining orbits $X_{2},X_{3},\ldots,X_{k}$ of $\sigma$ remain orbits
of $\tau$ (because the map $\tau$ sends them to the same values that the map
$\sigma$ sends them to\footnote{\textit{Proof.} Let $x$ be an element of one
of the orbits $X_{2},X_{3},\ldots,X_{k}$. We must prove that $\tau\left(
x\right)  =\sigma\left(  x\right)  $.
\par
The orbits $X_{1},X_{2},\ldots,X_{k}$ of $\sigma$ are disjoint. Hence, $x$
does not belong to $X_{1}$ (since $x$ belongs to one of the orbits
$X_{2},X_{3},\ldots,X_{k}$).
\par
We have $\sigma\left(  x\right)  \overset{\sigma}{\sim}x$ (since
$\sigma\left(  x\right)  =\sigma^{k}\left(  x\right)  $ for $k=1$). Thus, the
elements $\sigma\left(  x\right)  $ and $x$ belong to the same
$\overset{\sigma}{\sim}$-equivalence class, i.e., to the same orbit of
$\sigma$. This shows that the element $\sigma\left(  x\right)  $ does not
belong to the orbit $X_{1}$ (since $x$ does not belong to $X_{1}$). Hence,
$\sigma\left(  x\right)  $ equals neither $i$ nor $j$ (since both $i$ and $j$
belong to $X_{1}$). Therefore, $t_{i,j}\left(  \sigma\left(  x\right)
\right)  =\sigma\left(  x\right)  $ (since the transposition $t_{i,j}$ leaves
every element other than $i$ and $j$ fixed).
\par
Now, $\tau=t_{i,j}\circ\sigma$, so that $\tau\left(  x\right)  =\left(
t_{i,j}\circ\sigma\right)  \left(  x\right)  =t_{i,j}\left(  \sigma\left(
x\right)  \right)  =\sigma\left(  x\right)  $, qed.}).

To summarize: we have shown that the orbit $X_{1}$ of $\sigma$ splits into two
orbits $X_{1}^{\prime}$ and $X_{1}^{\prime\prime}$ of $\tau$ (in the sense
that $X_{1}^{\prime}$ and $X_{1}^{\prime\prime}$ are two disjoint orbits of
$\tau$, and their union is $X_{1}$), whereas all the remaining orbits
$X_{2},X_{3},\ldots,X_{k}$ of $\sigma$ remain orbits of $\tau$. Therefore,
$\tau$ has $1$ more orbit than $\sigma$. This proves Lemma
\ref{lem.perm.dcd.transp} \textbf{(a)}. \medskip

\textbf{(b)} We continue with the notations from our above proof of Lemma
\ref{lem.perm.dcd.transp} \textbf{(a)}. The two orbits $X_{1}^{\prime}$ and
$X_{1}^{\prime\prime}$ of $\tau$ are disjoint, thus different, and furthermore
they contain $i$ and $j$, respectively. Hence, $i$ and $j$ belong to different
orbits of $\tau$. In other words, $i$ and $j$ belong to different
$\overset{\tau}{\sim}$-equivalence classes. Hence, we don't have
$i\overset{\tau}{\sim}j$. This proves Lemma \ref{lem.perm.dcd.transp}
\textbf{(b)}. \medskip

\textbf{(c)} Assume that we don't have $i\overset{\sigma}{\sim}j$. Thus, $i$
and $j$ belong to different $\overset{\sigma}{\sim}$-equivalence classes,
i.e., to different orbits of $\sigma$. Let $X_{1}$ and $X_{2}$ be these two
orbits, so that $i\in X_{1}$ and $j\in X_{2}$. Let $X_{3},X_{4},\ldots,X_{k}$
be all the remaining orbits of $\sigma$ (listed without repetition). Thus,
$\sigma$ has $k$ orbits. We shall now understand what the orbits of $\tau$ are.

Let $\sim$ be the relation $\overset{\sigma}{\sim}$. Proposition
\ref{prop.perm.orbit-struc} \textbf{(b)} (applied to $a=i$) yields that there
exists some positive integer $m$ such that%
\begin{align*}
\sigma^{m}\left(  i\right)   &  =i\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left[  i\right]  _{\sim}  &  =\left\{  \sigma^{0}\left(  i\right)
,\ \sigma^{1}\left(  i\right)  ,\ \ldots,\ \sigma^{m-1}\left(  i\right)
\right\}  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\left\vert \left[  i\right]  _{\sim}\right\vert  &  =m.
\end{align*}
Consider this $m$. Proposition \ref{prop.perm.orbit-struc} \textbf{(b)}
(applied to $a=j$) yields that there exists some positive integer $n$ (we
cannot call it $m$ now) such that%
\begin{align*}
\sigma^{n}\left(  j\right)   &  =j\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left[  j\right]  _{\sim}  &  =\left\{  \sigma^{0}\left(  j\right)
,\ \sigma^{1}\left(  j\right)  ,\ \ldots,\ \sigma^{n-1}\left(  j\right)
\right\}  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\left\vert \left[  j\right]  _{\sim}\right\vert  &  =n.
\end{align*}
Consider this $n$.

Recall that $X_{1}$ is the orbit of $\sigma$ that contains $i$. In other
words, $X_{1}=\left[  i\right]  _{\sim}$ (since $\left[  i\right]  _{\sim}$ is
the orbit of $\sigma$ that contains $i$). Therefore,%
\[
X_{1}=\left[  i\right]  _{\sim}=\left\{  \sigma^{0}\left(  i\right)
,\ \sigma^{1}\left(  i\right)  ,\ \ldots,\ \sigma^{m-1}\left(  i\right)
\right\}  =\left\{  i,\ \sigma^{1}\left(  i\right)  ,\ \sigma^{2}\left(
i\right)  ,\ \ldots,\ \sigma^{m-1}\left(  i\right)  \right\}
\]
(since $\sigma^{0}\left(  i\right)  =i$). The action of the map $\sigma$ on
the elements of $X_{1}$ looks as follows:%
\begin{equation}
i\overset{\sigma}{\mapsto}\sigma^{1}\left(  i\right)  \overset{\sigma
}{\mapsto}\sigma^{2}\left(  i\right)  \overset{\sigma}{\mapsto}\cdots
\overset{\sigma}{\mapsto}\sigma^{m-1}\left(  i\right)  \overset{\sigma
}{\mapsto}i \label{pf.lem.perm.dcd.transp.c.chain1}%
\end{equation}
(since $\sigma\left(  \sigma^{m-1}\left(  i\right)  \right)  =\sigma
^{m}\left(  i\right)  =i$). Similarly, the action of $\sigma$ on the elements
of $X_{2}$ looks as follows:%
\begin{equation}
j\overset{\sigma}{\mapsto}\sigma^{1}\left(  j\right)  \overset{\sigma
}{\mapsto}\sigma^{2}\left(  j\right)  \overset{\sigma}{\mapsto}\cdots
\overset{\sigma}{\mapsto}\sigma^{n-1}\left(  j\right)  \overset{\sigma
}{\mapsto}j. \label{pf.lem.perm.dcd.transp.c.chain2}%
\end{equation}


However, the transposition $t_{i,j}$ swaps $i$ with $j$ while leaving all
other elements unchanged. Thus, the permutation $\tau=t_{i,j}\circ\sigma$
differs from $\sigma$ only in two things:

\begin{itemize}
\item The element of $X$ that is sent to $i$ by $\sigma$ is instead sent to
$j$ by $\tau$ (because $t_{i,j}$ sends $i$ to $j$).

\item The element of $X$ that is sent to $j$ by $\sigma$ is instead sent to
$i$ by $\tau$ (because $t_{i,j}$ sends $j$ to $i$).

\item All other elements of $X$ are sent by $\tau$ to the same value that they
are sent to by $\sigma$.
\end{itemize}

Hence, when we replace $\sigma$ by $\tau$, then the two arrows $\sigma
^{m-1}\left(  i\right)  \overset{\sigma}{\mapsto}i$ and $\sigma^{n-1}\left(
j\right)  \overset{\sigma}{\mapsto}j$ in the circular chains
(\ref{pf.lem.perm.dcd.transp.c.chain1}) and
(\ref{pf.lem.perm.dcd.transp.c.chain2}) turn into $\sigma^{m-1}\left(
i\right)  \overset{\tau}{\mapsto}j$ and $\sigma^{n-1}\left(  j\right)
\overset{\tau}{\mapsto}i$, whereas all the other arrows remain unchanged.
Hence, these two circular chains get merged into one single circular chain%
\begin{align*}
i  &  \overset{\tau}{\mapsto}\sigma^{1}\left(  i\right)  \overset{\tau
}{\mapsto}\sigma^{2}\left(  i\right)  \overset{\tau}{\mapsto}\cdots
\overset{\tau}{\mapsto}\sigma^{m-1}\left(  i\right) \\
\overset{\tau}{\mapsto}j  &  \overset{\tau}{\mapsto}\sigma^{1}\left(
j\right)  \overset{\tau}{\mapsto}\sigma^{2}\left(  j\right)  \overset{\tau
}{\mapsto}\cdots\overset{\tau}{\mapsto}\sigma^{n-1}\left(  j\right)
\overset{\tau}{\mapsto}i.
\end{align*}
Hence, in place of the orbits $X_{1}$ and $X_{2}$ of $\sigma$, the permutation
$\tau$ has a single orbit%
\[
X_{1}\cup X_{2}=\left\{  i,\ \sigma^{1}\left(  i\right)  ,\ \sigma^{2}\left(
i\right)  ,\ \ldots,\ \sigma^{m-1}\left(  i\right)  ,\ j,\ \sigma^{1}\left(
j\right)  ,\ \ldots,\ \sigma^{n-1}\left(  j\right)  \right\}  .
\]
All the remaining orbits $X_{3},X_{4},\ldots,X_{k}$ of $\sigma$ remain orbits
of $\tau$ (because the map $\tau$ sends them to the same values that the map
$\sigma$ sends them to\footnote{This is proved similarly to the analogous
claim in our above proof of Lemma \ref{lem.perm.dcd.transp} \textbf{(c)}.}).

To summarize: we have shown that the orbits $X_{1}$ and $X_{2}$ of $\sigma$
are merged into one single orbit $X_{1}\cup X_{2}$ of $\tau$ (this is just
saying that $X_{1}\cup X_{2}$ is an orbit of $\tau$), whereas all the
remaining orbits $X_{3},X_{4},\ldots,X_{k}$ of $\sigma$ remain orbits of
$\tau$. Therefore, $\tau$ has $1$ fewer orbit than $\sigma$. This proves Lemma
\ref{lem.perm.dcd.transp} \textbf{(c)}. \medskip

\textbf{(d)} We continue with the notations from our above proof of Lemma
\ref{lem.perm.dcd.transp} \textbf{(c)}. Both $i$ and $j$ are contained in the
orbit $X_{1}\cup X_{2}$ of $\tau$. Hence, $i$ and $j$ belong to the same orbit
of $\tau$. In other words, $i$ and $j$ belong to the same $\overset{\tau
}{\sim}$-equivalence class. Hence, $i\overset{\tau}{\sim}j$. This proves Lemma
\ref{lem.perm.dcd.transp} \textbf{(d)}.
\end{proof}
\end{fineprint}

\subsubsection{Stirling numbers of the first kind}

Recall that $S_{n}$ denotes the symmetric group $S_{\left[  n\right]  }$ (for
any $n\in\mathbb{N}$).

Let us now count permutations with a given \# of orbits (or, equivalently,
cycles in their DCD):

\begin{definition}
\label{def.perm.stir1}Let $n\in\mathbb{N}$ and $k\in\mathbb{Z}$. Then, the
\textbf{unsigned Stirling number of the first kind} $%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
$ is defined to be the \# of all permutations $\sigma\in S_{n}$ that have
exactly $k$ orbits (i.e., that have exactly $k$ cycles in their DCD, provided
that we \textbf{don't} omit the $1$-element cycles).

The \textbf{signed Stirling number of the first kind} $s\left(  n,k\right)  $
is defined to be $\left(  -1\right)  ^{n-k}%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
$.
\end{definition}

\begin{example}
We have $%
%TCIMACRO{\QDATOPD{[}{]}{4}{2}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{4}{2}%
%EndExpansion
=11$, because there are exactly $11$ permutations $\sigma\in S_{4}$ that have
exactly $2$ orbits. The OLNs of these $11$ permutations are (written without
parentheses and without commas)%
\begin{align*}
&  1342,\ \ \ \ 1423,\ \ \ \ 2143,\ \ \ \ 2314,\ \ \ \ 2431,\ \ \ \ 3124,\\
&  3241,\ \ \ \ 3412,\ \ \ \ 4132,\ \ \ \ 4213,\ \ \ \ 4321.
\end{align*}
Here are DCDs of two of them:%
\[
1342=\operatorname*{cyc}\nolimits_{2,3,4}\circ\operatorname*{cyc}%
\nolimits_{1};\ \ \ \ \ \ \ \ \ \ 2143=\operatorname*{cyc}\nolimits_{1,2}%
\circ\operatorname*{cyc}\nolimits_{3,4}.
\]
Eight of the $11$ permutations we listed have a DCD of the form
$\operatorname*{cyc}\nolimits_{a,b,c}\circ\operatorname*{cyc}\nolimits_{d}$
(and thus have an orbit of size $3$ and an orbit of size $1$); the other three
have a DCD of the form $\operatorname*{cyc}\nolimits_{a,b}\circ
\operatorname*{cyc}\nolimits_{c,d}$ (and thus have two orbits of size $2$).
\end{example}

\begin{convention}
We will abbreviate \textquotedblleft Stirling numbers of the first
kind\textquotedblright\ as \textquotedblleft\textbf{Stirling-1 numbers}%
\textquotedblright, and \textquotedblleft Stirling numbers of the second
kind\textquotedblright\ as \textquotedblleft\textbf{Stirling-2 numbers}%
\textquotedblright.
\end{convention}

Let us compute some \textquotedblleft easy\textquotedblright\ values of $%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
$:

\begin{proposition}
\label{prop.perm.stir1.easy}\textbf{(a)} We have $%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
=0$ for any $n,k\in\mathbb{N}$ satisfying $k>n$.

\textbf{(b)} We have $%
%TCIMACRO{\QDATOPD{[}{]}{n}{n}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{n}%
%EndExpansion
=1$ for any $n\in\mathbb{N}$.

\textbf{(c)} We have $%
%TCIMACRO{\QDATOPD{[}{]}{n}{1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{1}%
%EndExpansion
=\left(  n-1\right)  !$ for any $n>0$.

\textbf{(d)} We have $%
%TCIMACRO{\QDATOPD{[}{]}{n}{0}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{0}%
%EndExpansion
=\left[  n=0\right]  $ for any $n\in\mathbb{N}$.

\textbf{(e)} We have $%
%TCIMACRO{\QDATOPD{[}{]}{0}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{0}{k}%
%EndExpansion
=\left[  k=0\right]  $ for any $k\in\mathbb{Z}$.

\textbf{(f)} We have $%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
=0$ for any $n\in\mathbb{N}$ and any negative $k\in\mathbb{Z}$.
\end{proposition}

\begin{proof}
[Proof (sketched).]Part \textbf{(e)} is obvious (there is only one permutation
$\sigma\in S_{0}$, namely the identity map $\operatorname*{id}%
\nolimits_{\left[  0\right]  }$, and it has $0$ orbits). So is part
\textbf{(f)} (since a permutation cannot have a negative \# of orbits). Thus,
it remains to prove parts \textbf{(a)}, \textbf{(b)}, \textbf{(c)} and
\textbf{(d)}. To that purpose, we fix $n\in\mathbb{N}$. \medskip

\textbf{(a)} If $\sigma\in S_{n}$ is any permutation, then the orbits of
$\sigma$ are disjoint nonempty subsets of $\left[  n\right]  $ (by Proposition
\ref{cor.perm.orbits-sp}), and thus there cannot be more than $n$ of them
(since $\left[  n\right]  $ has only $n$ elements, and each of the orbits of
$\sigma$ must contain at least one of these $n$ elements). In other words, a
permutation $\sigma\in S_{n}$ cannot have $k$ orbits for $k>n$. In other
words, $%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
=0$ for any $k\in\mathbb{N}$ satisfying $k>n$. This proves Proposition
\ref{prop.perm.stir1.easy} \textbf{(a)}. \medskip

\textbf{(b)} We start as in part \textbf{(a)}: If $\sigma\in S_{n}$ is any
permutation, then the orbits of $\sigma$ are disjoint nonempty subsets of
$\left[  n\right]  $. Hence, if there are $n$ of these orbits, then each of
them must be a $1$-element set (otherwise, their sizes would sum up to a
number larger than $n$, which is impossible for disjoint subsets of $\left[
n\right]  $). However, this means that each element of $\left[  n\right]  $ is
a fixed point of $\sigma$ (by Remark \ref{rmk.perm.orbit-1elt}), and therefore
$\sigma$ is the identity permutation $\operatorname*{id}\nolimits_{\left[
n\right]  }$.

Thus, we have shown that the only permutation $\sigma\in S_{n}$ that has $n$
orbits is $\operatorname*{id}\nolimits_{\left[  n\right]  }$. Hence, $%
%TCIMACRO{\QDATOPD{[}{]}{n}{n}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{n}%
%EndExpansion
=1$. This proves Proposition \ref{prop.perm.stir1.easy} \textbf{(b)}. \medskip

\textbf{(c)} Proposition \ref{prop.perm.orbit-struc-2} shows that if a
permutation $\sigma\in S_{n}$ has only $1$ orbit, then it is an $n$-cycle
(because its one orbit must necessarily be the whole set $\left[  n\right]
$). Conversely, a permutation $\sigma\in S_{n}$ that is an $n$-cycle must have
only $1$ orbit (namely, the set $\left[  n\right]  $). Thus, we conclude that
the permutations $\sigma\in S_{n}$ that have only $1$ orbit are precisely the
$n$-cycles in $S_{n}$. Hence, their \# is the \# of $n$-cycles in $S_{n}$. In
other words,%
\[%
%TCIMACRO{\QDATOPD{[}{]}{n}{1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{1}%
%EndExpansion
=\left(  \text{\# of }n\text{-cycles in }S_{n}\right)  .
\]
However, Exercise 1 in Lecture 26 (applied to $X=\left[  n\right]  $ and
$k=n$) yields that%
\begin{align*}
\left(  \text{\# of }n\text{-cycles in }S_{n}\right)   &  =%
\begin{cases}
1, & \text{if }n=1;\\
\left(  n-1\right)  !\cdot\dbinom{n}{n}, & \text{if }n>1
\end{cases}
\\
&  =%
\begin{cases}
1, & \text{if }n=1;\\
\left(  n-1\right)  !, & \text{if }n>1
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\dbinom{n}{n}=1\right) \\
&  =\left(  n-1\right)  !\ \ \ \ \ \ \ \ \ \ \left(  \text{since }1=\left(
n-1\right)  !\text{ when }n=1\right)  .
\end{align*}
Thus,
\[%
%TCIMACRO{\QDATOPD{[}{]}{n}{1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{1}%
%EndExpansion
=\left(  \text{\# of }n\text{-cycles in }S_{n}\right)  =\left(  n-1\right)
!.
\]
This proves Proposition \ref{prop.perm.stir1.easy} \textbf{(c)}. \medskip

\textbf{(d)} If $\sigma\in S_{n}$ is any permutation, then the orbits of
$\sigma$ must cover the entire set $\left[  n\right]  $ (by Proposition
\ref{cor.perm.orbits-sp}). Hence, $\sigma$ cannot have $0$ orbits, unless the
set $\left[  n\right]  $ is empty, i.e., unless $n=0$. Thus, $%
%TCIMACRO{\QDATOPD{[}{]}{n}{0}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{0}%
%EndExpansion
=0$ for any $n>0$. Combining this with the obvious fact that $%
%TCIMACRO{\QDATOPD{[}{]}{0}{0}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{0}{0}%
%EndExpansion
=1$, we conclude that Proposition \ref{prop.perm.stir1.easy} \textbf{(d)} holds.
\end{proof}

Let us now further explore the Stirling-1 numbers. Unlike the Stirling-2
numbers, they cannot be computed by a simple explicit formula, even allowing
for a summation sign. But they satisfy the following recursion:

\begin{proposition}
[recurrence relation for the Stirling-1 numbers]\label{prop.perm.stir1.rec}For
any positive integer $n$ and any integer $k$, we have%
\[%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
=\left(  n-1\right)
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k}%
%EndExpansion
+%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k-1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k-1}%
%EndExpansion
.
\]

\end{proposition}

\begin{fineprint}
\begin{proof}
[Proof sketch.]Let $n$ be a positive integer. Let $k$ be any integer. We have%
\begin{align*}%
%TCIMACRO{\QDATOPD{[}{]}{n}{k} }%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}
%EndExpansion
&  =\left(  \text{\# of permutations }\sigma\in S_{n}\text{ with }k\text{
orbits}\right) \\
&  =\sum_{i=1}^{n}\left(  \text{\# of permutations }\sigma\in S_{n}\text{ with
}k\text{ orbits and with }\sigma\left(  n\right)  =i\right)
\end{align*}
(by the sum rule). Now, we claim the following:

\begin{statement}
\textit{Claim 1:} We have%
\[
\left(  \text{\# of permutations }\sigma\in S_{n}\text{ with }k\text{ orbits
and with }\sigma\left(  n\right)  =n\right)  =%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k-1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k-1}%
%EndExpansion
.
\]

\end{statement}

\begin{statement}
\textit{Claim 2:} For any $i\in\left[  n-1\right]  $, we have%
\[
\left(  \text{\# of permutations }\sigma\in S_{n}\text{ with }k\text{ orbits
and with }\sigma\left(  n\right)  =i\right)  =%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k}%
%EndExpansion
.
\]

\end{statement}

Once Claim 1 and Claim 2 are proved, we will conclude that%
\begin{align*}%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
&  =\sum_{i=1}^{n}\left(  \text{\# of permutations }\sigma\in S_{n}\text{ with
}k\text{ orbits and with }\sigma\left(  n\right)  =i\right) \\
&  =\sum_{i=1}^{n-1}\underbrace{\left(  \text{\# of permutations }\sigma\in
S_{n}\text{ with }k\text{ orbits and with }\sigma\left(  n\right)  =i\right)
}_{\substack{=%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k}%
%EndExpansion
\\\text{(by Claim 2)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{\# of permutations }%
\sigma\in S_{n}\text{ with }k\text{ orbits and with }\sigma\left(  n\right)
=n\right)  }_{\substack{=%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k-1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k-1}%
%EndExpansion
\\\text{(by Claim 1)}}}\\
&  =\underbrace{\sum_{i=1}^{n-1}%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k}%
%EndExpansion
}_{=\left(  n-1\right)
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k}%
%EndExpansion
}+%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k-1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k-1}%
%EndExpansion
=\left(  n-1\right)
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k}%
%EndExpansion
+%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k-1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k-1}%
%EndExpansion
,
\end{align*}
and thus Proposition \ref{prop.perm.stir1.rec} will be proved. So it remains
to verify Claim 1 and Claim 2. \medskip

\textit{Proof of Claim 1:} If a permutation $\sigma\in S_{n}$ has $k$ orbits
and satisfies $\sigma\left(  n\right)  =n$, then $n$ is a fixed point of
$\sigma$, and thus the $1$-element set $\left\{  n\right\}  $ is itself an
orbit of $\sigma$ (by Remark \ref{rmk.perm.orbit-1elt}). Hence, by restricting
such a permutation $\sigma$ to the subset $\left[  n-1\right]  $, we obtain a
permutation $\sigma\mid_{\left[  n-1\right]  }$ of $\left[  n-1\right]  $ that
has $k-1$ orbits (note that this is well-defined, because $\sigma\left(
n\right)  =n$ ensures that $\sigma\left(  i\right)  \in\left[  n-1\right]  $
for each $i\in\left[  n-1\right]  $). This yields a bijection%
\begin{align*}
&  \text{from }\left\{  \text{permutations }\sigma\in S_{n}\text{ with
}k\text{ orbits and with }\sigma\left(  n\right)  =n\right\} \\
&  \text{to }\left\{  \text{permutations }\tau\in S_{n-1}\text{ with
}k-1\text{ orbits}\right\}  ,
\end{align*}
which sends each permutation $\sigma$ to its restriction $\sigma\mid_{\left[
n-1\right]  }:\left[  n-1\right]  \rightarrow\left[  n-1\right]  $. (The
inverse map of this bijection simply extends a permutation $\tau\in S_{n-1}$
to $\left[  n\right]  $ by setting $\tau\left(  n\right)  :=n$. This extension
creates a new orbit $\left\{  n\right\}  $, thus raising the \# of orbits from
$k-1$ to $k$.)

Thus, the bijection principle yields%
\begin{align*}
&  \left(  \text{\# of permutations }\sigma\in S_{n}\text{ with }k\text{
orbits and with }\sigma\left(  n\right)  =n\right) \\
&  =\left(  \text{\# of permutations }\tau\in S_{n-1}\text{ with }k-1\text{
orbits}\right)  =%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k-1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k-1}%
%EndExpansion
\end{align*}
(by the definition of $%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k-1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k-1}%
%EndExpansion
$). This proves Claim 1. \medskip

\textit{Proof of Claim 2:} Let $i\in\left[  n-1\right]  $. Thus, $i\neq n$.
Consider the transposition $t_{i,n}$, which swaps $i$ with $n$. It satisfies
$t_{i,n}\circ t_{i,n}=\operatorname*{id}$ (since swapping $i$ with $n$ twice
puts all elements back where they started). Now, if $\sigma\in S_{n}$ is a
permutation with $k$ orbits and with $\sigma\left(  n\right)  =i$, then the
permutation $t_{i,n}\circ\sigma$ has $k+1$ orbits (this follows easily from
Lemma \ref{lem.perm.dcd.transp} \textbf{(a)}\footnote{Here is the proof in
more detail: Let $\sigma\in S_{n}$ be a permutation with $k$ orbits and with
$\sigma\left(  n\right)  =i$. Set $\tau:=t_{i,n}\circ\sigma$.
\par
We have $i=\sigma\left(  n\right)  =\sigma^{1}\left(  n\right)  $, so that
$i\overset{\sigma}{\sim}n$. Hence, Lemma \ref{lem.perm.dcd.transp}
\textbf{(a)} (applied to $j=n$) yields that the permutation $\tau$ has $1$
more orbit than $\sigma$. Thus, $\tau$ has $k+1$ orbits (since $\sigma$ has
$k$ orbits). In other words, $t_{i,n}\circ\sigma$ has $k+1$ orbits (since
$\tau=t_{i,n}\circ\sigma$). Qed.}) and satisfies $\left(  t_{i,n}\circ
\sigma\right)  \left(  n\right)  =n$\ \ \ \ \footnote{\textit{Proof.} We have
$\left(  t_{i,n}\circ\sigma\right)  \left(  n\right)  =t_{i,n}\left(
\underbrace{\sigma\left(  n\right)  }_{=i}\right)  =t_{i,n}\left(  i\right)
=n$ (by the definition of $t_{i,n}$).}. Hence, we obtain a map%
\begin{align*}
&  \text{from }\left\{  \text{permutations }\sigma\in S_{n}\text{ with
}k\text{ orbits and with }\sigma\left(  n\right)  =i\right\} \\
&  \text{to }\left\{  \text{permutations }\sigma\in S_{n}\text{ with
}k+1\text{ orbits and with }\sigma\left(  n\right)  =n\right\}  ,
\end{align*}
which sends each $\sigma$ to $t_{i,n}\circ\sigma$. Conversely, if $\sigma\in
S_{n}$ is a permutation with $k+1$ orbits and with $\sigma\left(  n\right)
=n$, then the permutation $t_{i,n}\circ\sigma$ has $k$ orbits (this follows
easily from Lemma \ref{lem.perm.dcd.transp} \textbf{(c)}\footnote{Here is the
proof in more detail: Let $\sigma\in S_{n}$ be a permutation with $k+1$ orbits
and with $\sigma\left(  n\right)  =n$. Set $\tau:=t_{i,n}\circ\sigma$.
\par
We have $\sigma\left(  n\right)  =n$, so that $n$ is a fixed point of $\sigma
$. Hence, Remark \ref{rmk.perm.orbit-1elt} (applied to $X=\left[  n\right]  $
and $a=n$) shows that the orbit of $\sigma$ that contains $n$ is a $1$-element
set. This orbit must therefore be $\left\{  n\right\}  $ (since $n$ belongs to
this orbit), and thus does not contain $i$ (since $i\neq n$). Therefore, we
don't have $i\overset{\sigma}{\sim}n$. Hence, Lemma \ref{lem.perm.dcd.transp}
\textbf{(c)} (applied to $j=n$) yields that the permutation $\tau$ has $1$
fewer orbit than $\sigma$. Thus, $\tau$ has $k$ orbits (since $\sigma$ has
$k+1$ orbits). In other words, $t_{i,n}\circ\sigma$ has $k$ orbits (since
$\tau=t_{i,n}\circ\sigma$). Qed.}) and satisfies $\left(  t_{i,n}\circ
\sigma\right)  \left(  n\right)  =i$\ \ \ \ \footnote{\textit{Proof.} We have
$\left(  t_{i,n}\circ\sigma\right)  \left(  n\right)  =t_{i,n}\left(
\underbrace{\sigma\left(  n\right)  }_{=n}\right)  =t_{i,n}\left(  n\right)
=i$ (by the definition of $t_{i,n}$).}. Hence, we obtain a map%
\begin{align*}
&  \text{from }\left\{  \text{permutations }\sigma\in S_{n}\text{ with
}k+1\text{ orbits and with }\sigma\left(  n\right)  =n\right\} \\
&  \text{to }\left\{  \text{permutations }\sigma\in S_{n}\text{ with }k\text{
orbits and with }\sigma\left(  n\right)  =i\right\}  ,
\end{align*}
which sends each $\sigma$ to $t_{i,n}\circ\sigma$.

Thus, we have found two maps between the two sets%
\begin{align*}
&  \left\{  \text{permutations }\sigma\in S_{n}\text{ with }k\text{ orbits and
with }\sigma\left(  n\right)  =i\right\} \\
\text{and }  &  \left\{  \text{permutations }\sigma\in S_{n}\text{ with
}k+1\text{ orbits and with }\sigma\left(  n\right)  =n\right\}
\end{align*}
(one map in either direction). These two maps are mutually inverse, since any
permutation $\sigma\in S_{n}$ satisfies $t_{i,n}\circ\left(  t_{i,n}%
\circ\sigma\right)  =\underbrace{t_{i,n}\circ t_{i,n}}_{=\operatorname*{id}%
}\circ\,\sigma=\operatorname*{id}\circ\,\sigma=\sigma$. Hence, they are
bijections. The bijection principle thus yields%
\begin{align*}
&  \left(  \text{\# of permutations }\sigma\in S_{n}\text{ with }k\text{
orbits and with }\sigma\left(  n\right)  =i\right) \\
&  =\left(  \text{\# of permutations }\sigma\in S_{n}\text{ with }k+1\text{
orbits and with }\sigma\left(  n\right)  =n\right) \\
&  =%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{\left(  k+1\right)  -1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{\left(  k+1\right)  -1}%
%EndExpansion
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Claim 1, applied to }k+1\text{ instead of
}k\right) \\
&  =%
%TCIMACRO{\QDATOPD{[}{]}{n-1}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n-1}{k}%
%EndExpansion
.
\end{align*}
This proves Claim 2. \medskip

Having proved both Claim 1 and Claim 2, we have now completed our proof of
Proposition \ref{prop.perm.stir1.rec}.
\end{proof}
\end{fineprint}

Using Proposition \ref{prop.perm.stir1.rec}, it is easy to construct a table
of Stirling-1 numbers:%
\[%
\begin{tabular}
[c]{|c||c|c|c|c|c|c|c|c|}\hline
$%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
$ & $n=0$ & $n=1$ & $n=2$ & $n=3$ & $n=4$ & $n=5$ & $n=6$ & $n=7$%
\\\hline\hline
$k=0$ & $1$ & $0$ & $0$ & $0$ & $0$ & $0$ & $0$ & $0$\\\hline
$k=1$ & $0$ & $1$ & $1$ & $2$ & $6$ & $24$ & $120$ & $720$\\\hline
$k=2$ & $0$ & $0$ & $1$ & $3$ & $11$ & $50$ & $274$ & $1764$\\\hline
$k=3$ & $0$ & $0$ & $0$ & $1$ & $6$ & $35$ & $225$ & $1624$\\\hline
$k=4$ & $0$ & $0$ & $0$ & $0$ & $1$ & $10$ & $85$ & $735$\\\hline
$k=5$ & $0$ & $0$ & $0$ & $0$ & $0$ & $1$ & $15$ & $175$\\\hline
$k=6$ & $0$ & $0$ & $0$ & $0$ & $0$ & $0$ & $1$ & $21$\\\hline
$k=7$ & $0$ & $0$ & $0$ & $0$ & $0$ & $0$ & $0$ & $1$\\\hline
\end{tabular}
\ \
\]
(see
\href{https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind}{the
Wikipedia page} for more).

\bigskip

It is time to answer the question that you will surely have asked by now,
namely: What do the Stirling-1 numbers have to do with the Stirling-2 numbers?
Both types of numbers have been invented by James Stirling in 1730, but a mere
coincidence like this does not usually lead to \textquotedblleft first
kind\textquotedblright/\textquotedblleft second kind\textquotedblright%
\ nomenclature. There is a deeper connection between the two -- a type of duality.

To best understand this duality, we need a little bit of linear algebra. We
consider polynomials in one indeterminate $X$ (with rational coefficients).
For each $k\in\mathbb{N}$, we consider the $k$\textbf{-th falling factorial}%
\[
X^{\underline{k}}=X\left(  X-1\right)  \left(  X-2\right)  \cdots\left(
X-k+1\right)  ;
\]
this is a degree-$k$ polynomial in $X$. Evaluating this polynomial at a number
$n$ will, of course, yield the falling factorial $n^{\underline{k}}$ we
introduced in Definition 2.4.2 (Lecture 16).

It is obvious that each polynomial in $X$ can be written as a linear
combination of the powers $X^{0},X^{1},X^{2},\ldots$. It is a bit less obvious
that each polynomial in $X$ can be written as a linear combination of the
falling factorials $X^{\underline{0}},X^{\underline{1}},X^{\underline{2}%
},\ldots$. The easiest way to see this is by writing each power $X^{n}$ as
such a combination. This can be done explicitly, and the coefficients are the
Stirling-2 numbers:

\begin{theorem}
\label{thm.perm.stir2-as-coeffs}Let $n\in\mathbb{N}$. Then,%
\begin{equation}
X^{n}=\sum_{k=0}^{n}%
%TCIMACRO{\QDATOPD{\{}{\}}{n}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{n}{k}%
%EndExpansion
X^{\underline{k}}. \label{eq.thm.perm.stir2-as-coeffs.eq}%
\end{equation}

\end{theorem}

\begin{fineprint}
\begin{proof}
[Proof sketch.]By the \textquotedblleft polynomial identity
trick\textquotedblright\ (Corollary 2.2.5 in Lecture 14), it suffices to show
that%
\[
x^{n}=\sum_{k=0}^{n}%
%TCIMACRO{\QDATOPD{\{}{\}}{n}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{n}{k}%
%EndExpansion
x^{\underline{k}}\ \ \ \ \ \ \ \ \ \ \text{for each }x\in\mathbb{N}.
\]
So let us do this. Fix $x\in\mathbb{N}$. Then, Theorem 2.5.1 in Lecture 17
(applied to $k=x$ and $m=n$) yields%
\begin{align*}
x^{n}  &  =\sum_{i=0}^{n}\operatorname*{sur}\left(  n,i\right)  \cdot
\dbinom{x}{i}=\sum_{k=0}^{n}\operatorname*{sur}\left(  n,k\right)
\cdot\underbrace{\dbinom{x}{k}}_{\substack{=\dfrac{x\left(  x-1\right)
\left(  x-2\right)  \cdots\left(  x-k+1\right)  }{k!}\\=\dfrac
{x^{\underline{k}}}{k!}\\\text{(since }x\left(  x-1\right)  \left(
x-2\right)  \cdots\left(  x-k+1\right)  =x^{\underline{k}}\text{)}}}\\
&  =\sum_{k=0}^{n}\underbrace{\operatorname*{sur}\left(  n,k\right)
\cdot\dfrac{x^{\underline{k}}}{k!}}_{=\dfrac{\operatorname*{sur}\left(
n,k\right)  }{k!}\cdot x^{\underline{k}}}=\sum_{k=0}^{n}\underbrace{\dfrac
{\operatorname*{sur}\left(  n,k\right)  }{k!}}_{\substack{=%
%TCIMACRO{\QDATOPD{\{}{\}}{n}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{n}{k}%
%EndExpansion
\\\text{(by the definition of}\\\text{the Stirling-2 numbers)}}}\cdot
x^{\underline{k}}=\sum_{k=0}^{n}%
%TCIMACRO{\QDATOPD{\{}{\}}{n}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{n}{k}%
%EndExpansion
x^{\underline{k}},
\end{align*}
qed.
\end{proof}
\end{fineprint}

So Theorem \ref{thm.perm.stir2-as-coeffs} expands any power $X^{n}$ as a
linear combination of falling factorials $X^{\underline{k}}$. The coefficients
are the Stirling-2 numbers $%
%TCIMACRO{\QDATOPD{\{}{\}}{n}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{n}{k}%
%EndExpansion
$.

What about the converse direction -- i.e., how can we expand a falling
factorial $X^{\underline{n}}$ as a linear combination of the powers $X^{k}$ ?
In other words, what are the coefficients of $X^{\underline{n}}=X\left(
X-1\right)  \left(  X-2\right)  \cdots\left(  X-n+1\right)  $ ?

Here, as it turns out, the coefficients will be the signed Stirling-1 numbers:

\begin{theorem}
\label{thm.perm.stir1-as-coeffs}Let $n\in\mathbb{N}$. Then,
\begin{align}
X^{\underline{n}}  &  =X\left(  X-1\right)  \left(  X-2\right)  \cdots\left(
X-n+1\right) \nonumber\\
&  =\sum_{k=0}^{n}s\left(  n,k\right)  X^{k}%
\label{eq.thm.perm.stir1-as-coeffs.eq}\\
&  =\sum_{k=0}^{n}\left(  -1\right)  ^{n-k}%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
X^{k}.\nonumber
\end{align}
Equivalently (by substituting $-X$ for $X$), this becomes%
\[
X\left(  X+1\right)  \left(  X+2\right)  \cdots\left(  X+n-1\right)
=\sum_{k=0}^{n}%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
X^{k}.
\]

\end{theorem}

\begin{fineprint}
\begin{proof}
[Proof sketch.]There is a nice combinatorial proof (see, e.g., \cite[proof of
(34)]{Galvin}). Three other proofs can be found in \cite[Proposition
1.3.7]{Stanley-EC1} (in which $%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
$ is denoted by $c\left(  n,k\right)  $).

The easiest proof at this point is just a straightforward induction on $n$,
similar to our proof of the binomial formula (Theorem 1.3.19 in Lecture 7).
Use Proposition \ref{prop.perm.stir1.rec} in the induction step.
\end{proof}
\end{fineprint}

A curious consequence of Theorems \ref{thm.perm.stir1-as-coeffs} and
\ref{thm.perm.stir2-as-coeffs} is the following relation between the two kinds
of Stirling numbers:

\begin{theorem}
\label{thm.perm.stir2.inv-matrices}For any $i,j\in\mathbb{N}$, we have%
\[
\sum_{k=0}^{i}s\left(  i,k\right)
%TCIMACRO{\QDATOPD{\{}{\}}{k}{j}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{k}{j}%
%EndExpansion
=\sum_{k=0}^{i}%
%TCIMACRO{\QDATOPD{\{}{\}}{i}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{i}{k}%
%EndExpansion
s\left(  k,j\right)  =\left[  i=j\right]  .
\]

\end{theorem}

\begin{fineprint}
\begin{proof}
[Proof sketch.](Some familiarity with vector spaces is required here. See
\cite[Theorem 2.64]{Loehr-2ed} for the details of this argument. See also
\url{https://math.stackexchange.com/questions/42018/stirling-numbers-and-inverse-matrices}
for a generalization.)

Fix $n\in\mathbb{N}$. Consider the $\mathbb{Q}$-vector space $P_{n}$ of all
polynomials of degree $\leq n-1$ (including the zero polynomial) in one
indeterminate $X$ (with rational coefficients). This vector space $P_{n}$ has
two important bases:

\begin{itemize}
\item the \textbf{power basis} $\left(  X^{0},X^{1},\ldots,X^{n-1}\right)  $,
which consists of the first $n$ powers of $X$;

\item the \textbf{falling factorial basis} $\left(  X^{\underline{0}%
},X^{\underline{1}},\ldots,X^{\underline{n-1}}\right)  $.
\end{itemize}

(The power basis is clearly a basis. The falling factorial basis can easily be
seen to be a basis, since it expands triangularily in the power basis -- i.e.,
each falling factorial $X^{\underline{k}}$ can be written as $X^{k}$ plus a
linear combination of lower powers $X^{k-1},X^{k-2},\ldots,X^{0}$.)

For each $i\in\left\{  0,1,\ldots,n-1\right\}  $, we have%
\begin{align*}
X^{i} &  =\sum_{k=0}^{i}%
%TCIMACRO{\QDATOPD{\{}{\}}{i}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{i}{k}%
%EndExpansion
X^{\underline{k}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.thm.perm.stir2-as-coeffs.eq}), applied to }i\text{ instead of
}n\right)  \\
&  =\sum_{k=0}^{n-1}%
%TCIMACRO{\QDATOPD{\{}{\}}{i}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{i}{k}%
%EndExpansion
X^{\underline{k}}%
\end{align*}
(here, we have extended the range of the summation from $k\in\left\{
0,1,\ldots,i\right\}  $ to $k\in\left\{  0,1,\ldots,n-1\right\}  $, which does
not change the value because $%
%TCIMACRO{\QDATOPD{\{}{\}}{i}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{i}{k}%
%EndExpansion
=0$ for all $k>i$). This shows that the $n\times n$-matrix\footnote{This is an
$n\times n$-matrix, but we number its rows and its columns by $0,1,\ldots,n-1$
rather than by $1,2,\ldots,n$.}%
\[
S_{2}:=\left(
%TCIMACRO{\QDATOPD{\{}{\}}{i}{j}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{i}{j}%
%EndExpansion
\right)  _{0\leq i,j\leq n-1}%
\]
is the transition matrix (= change-of-basis matrix) from the falling factorial
basis to the power basis. Likewise, using the formula
(\ref{eq.thm.perm.stir1-as-coeffs.eq}), we can see that the $n\times n$-matrix%
\[
S_{1}:=\left(  s\left(  i,j\right)  \right)  _{0\leq i,j\leq n-1}%
\]
is the transition matrix from the power basis to the falling factorial basis.
However, it is well-known that the transition matrices between two bases of a
vector space are always mutually inverse (see, e.g., \cite[\S 10.2, $RS=I$%
]{LaNaSc16}). Thus, we conclude that the two matrices%
\[
S_{1}=\left(  s\left(  i,j\right)  \right)  _{0\leq i,j\leq n-1}%
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ S_{2}=\left(
%TCIMACRO{\QDATOPD{\{}{\}}{i}{j}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{i}{j}%
%EndExpansion
\right)  _{0\leq i,j\leq n-1}%
\]
are mutually inverse. In other words, they satisfy $S_{1}S_{2}=I_{n}$ and
$S_{2}S_{1}=I_{n}$ (where $I_{n}$ denotes the $n\times n$-identity matrix).
Explicitly, this means that for any $i,j\in\left\{  0,1,\ldots,n-1\right\}  $,
the $\left(  i,j\right)  $-th entries of both products $S_{1}S_{2}$ and
$S_{2}S_{1}$ equal the $\left(  i,j\right)  $-th entry of $I_{n}$, which of
course is $\left[  i=j\right]  $. In other words, for any $i,j\in\left\{
0,1,\ldots,n-1\right\}  $, we have the equality
\[
\sum_{k=0}^{n-1}s\left(  i,k\right)
%TCIMACRO{\QDATOPD{\{}{\}}{k}{j}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{k}{j}%
%EndExpansion
=\sum_{k=0}^{n-1}%
%TCIMACRO{\QDATOPD{\{}{\}}{i}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{i}{k}%
%EndExpansion
s\left(  k,j\right)  =\left[  i=j\right]  .
\]
Since both numbers $s\left(  i,k\right)  $ and $%
%TCIMACRO{\QDATOPD{\{}{\}}{i}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{i}{k}%
%EndExpansion
$ are $0$ for $k>i$, we can replace the upper limits of both summation signs
by $i$, thus rewriting this equality as follows:
\[
\sum_{k=0}^{i}s\left(  i,k\right)
%TCIMACRO{\QDATOPD{\{}{\}}{k}{j}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{k}{j}%
%EndExpansion
=\sum_{k=0}^{i}%
%TCIMACRO{\QDATOPD{\{}{\}}{i}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{i}{k}%
%EndExpansion
s\left(  k,j\right)  =\left[  i=j\right]  .
\]
Since we can choose $n$ arbitrarily high, we thus conclude that this equality
holds for all $i,j\in\mathbb{N}$. This proves Theorem
\ref{thm.perm.stir2.inv-matrices}.

Alternatively, Theorem \ref{thm.perm.stir2.inv-matrices} can also be proved by
induction. See
\url{https://proofwiki.org/wiki/First_Inversion_Formula_for_Stirling_Numbers}
and
\url{https://proofwiki.org/wiki/Second_Inversion_Formula_for_Stirling_Numbers}
for such a proof. See also \cite[Identity 194]{BenQui03} for a combinatorial
proof of $\sum_{k=0}^{i}s\left(  i,k\right)
%TCIMACRO{\QDATOPD{\{}{\}}{k}{j}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{k}{j}%
%EndExpansion
=\left[  i=j\right]  $.
\end{proof}
\end{fineprint}

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[17f-hw7s]{17f-hw7s}Darij Grinberg, \textit{UMN Fall 2017 Math 4990
homework set \#7 with solutions}, \url{http://www.cip.ifi.lmu.de/~grinberg/t/17f/hw7os.pdf}

\bibitem[21s]{21s}Darij Grinberg, \textit{Algebraic Combinatorics (Drexel
Spring 2021 Math 701 lecture notes)}, 4 May 2021.\newline\url{https://www.cip.ifi.lmu.de/~grinberg/t/21s/lecs.pdf}

\bibitem[BenQui03]{BenQui03}Arthur T. Benjamin and Jennifer J. Quinn,
\textit{Proofs that Really Count: The Art of Combinatorial Proof}, Dolciani
Mathematical Expositions \textbf{27}, The Mathematical Association of America, 2003.

\bibitem[Bourba74]{Bourba74}%
\href{http://libgen.rs/book/index.php?md5=3270565F6D0052635A1550883588204C}{Nicolas
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\bibitem[Galvin17]{Galvin}David Galvin, \textit{Basic discrete mathematics},
13 December 2017.\newline%
\url{http://www-users.math.umn.edu/~dgrinber/comb/60610lectures2017-Galvin.pdf}
\newline(The URL might change, and the text may get updated. In order to
reliably obtain the version of 13 December 2017, you can use the archive.org
Wayback Machine:
\url{https://web.archive.org/web/20180205122609/http://www-users.math.umn.edu/~dgrinber/comb/60610lectures2017-Galvin.pdf}
.)

\bibitem[Goodma15]{Goodman}Frederick M. Goodman, \textit{Algebra: Abstract and
Concrete}, edition 2.6, 1 May 2015.\newline%
\url{http://homepage.math.uiowa.edu/~goodman/algebrabook.dir/book.2.6.pdf} .

\bibitem[LaNaSc16]{LaNaSc16}Isaiah Lankham, Bruno Nachtergaele, Anne
Schilling, \textit{Linear Algebra As an Introduction to Abstract Mathematics},
2016.\newline\url{https://www.math.ucdavis.edu/~anne/linear_algebra/mat67_course_notes.pdf}

\bibitem[Loehr17]{Loehr-2ed}%
\href{https://www.routledge.com/Combinatorics/Loehr/p/book/9781498780254}{Nicholas
A. Loehr, \textit{Combinatorics}, 2nd edition, CRC Press 2017.}

\bibitem[Stanle11]{Stanley-EC1}Richard P. Stanley, \textit{Enumerative
Combinatorics, volume 1}, Second edition, version of 15 July 2011. Available
at \url{http://math.mit.edu/~rstan/ec/} .\newline See
\url{http://math.mit.edu/~rstan/ec/} for errata.
\end{thebibliography}


\end{document}