\section{Cycle Notation}

\definition{}
The \textit{order} of a permutation $f$ is the smallest $n$ so that $f^n(x) = x$ for all $x$. \par
In other words, if we repeat this permutation $n$ times, we get back to where we started.

\vspace{2mm}

For example, consider $[2134]$. This permutation has order $2$, as we clearly see below:

\begin{center}
\begin{tikzpicture}[scale=0.5]
	\node (1a) at (0, 0.5) {1};
	\node (2a) at (1, 0.5) {2};
	\node (3a) at (2, 0.5) {3};
	\node (4a) at (3, 0.5) {4};

	\node (2b) at (0, -2) {2};
	\node (1b) at (1, -2) {1};
	\node (3b) at (2, -2) {3};
	\node (4b) at (3, -2) {4};

	\node (1c) at (0, -4.5) {1};
	\node (2c) at (1, -4.5) {2};
	\node (3c) at (2, -4.5) {3};
	\node (4c) at (3, -4.5) {4};

	\line{1a}{1b}
	\line{2a}{2b}
	\line{3a}{3b}
	\line{4a}{4b}
	\line{1b}{1c}
	\line{2b}{2c}
	\line{3b}{3c}
	\line{4b}{4c}
\end{tikzpicture}
\end{center}
Of course, swapping the first two elements twice results in the identity map. \par
$[2134]$ happens to be its own inverse. Also, the order of $[1234]$ is (of course) one.


\problem{}
What is the order of $[2314]$? \par
How about $[4321]$? \par
\note[Note]{You shouldn't need to draw any braids to solve this problem.}


\vfill

\problem{Bonus}
Show that all permutations (on a finite set) have a well-defined order. \par
In other words, show that there is always an integer $n$ so that $f^n(x) = x$.

\vfill
\pagebreak


As you may have noticed, the square-bracket notation we've been using thus far is a bit unwieldy.
Permutations are verbs---but we've been referring to them using a noun (namely, their output when
applied to an ordered sequence of numbers). Our notation fails to capture the meaning of the
underlying object.

\vspace{2mm}

Think about it: is the permutation $[1234]$ different than the permutation $[12345]$? \par
Indeed, these permutations operate on different sets---but they are both the identity! \par
What should we do if we want to talk about the identity on $\{1, 2, ..., 10\}$?

\vspace{2mm}

We need something better.




\definition{}
Any permutation is composed of a number of \textit{cycles}. \par

For example, consider the permutation $[2134]$, which consists of one two-cycle: $1 \to 2 \to 1$ \par
\note[Note]{$3 \to 3$ and $4 \to 4$ are also cycles, but we'll ignore them. One-cycles aren't aren't interesting.}

\begin{center}
\begin{tikzpicture}[scale=0.5]
	\node (1) at (0, 0) {1};
	\node (2) at (1, 0) {2};
	\node (3) at (2, 0) {3};
	\node (4) at (3, 0) {4};

	\draw[line width = 0.3mm, ->, ocyan]
		(1)
		-- ($(1) + (0,-1)$)
		-- ($(2) + (0,-1)$)
		-- (2);

	\draw[line width = 0.3mm, ->, ocyan]
		(2)
		-- ($(2) + (0, 1)$)
		-- ($(1) + (0, 1)$)
		-- (1);
\end{tikzpicture}
\end{center}


The permutation $[431265]$ is a bit more interesting---it contains of two cycles: \par
($1 \to 3 \to 2 \to 4 \to 1$ and $5 \to 6 \to 5$)


\begin{center}
\begin{tikzpicture}[scale=0.5]
	\node (1) at (0, 0) {1};
	\node (2) at (1, 0) {2};
	\node (3) at (2, 0) {3};
	\node (4) at (3, 0) {4};
	\node (5) at (4, 0) {5};
	\node (6) at (5, 0) {6};

	\draw[line width = 0.3mm, ->, ocyan]
		(3)
		-- ($(3) + (0,-1)$)
		-- ($(2) + (0,-1)$)
		-- (2);

	\draw[line width = 0.3mm, ->, ocyan]
		(2)
		-- ($(2) + (0,1.5)$)
		-- ($(4) + (0,1.5)$)
		-- (4);

	\draw[line width = 0.3mm, ->, ocyan]
		(4)
		-- ($(4) + (0,-1.5)$)
		-- ($(1) + (0,-1.5)$)
		-- (1);

	\draw[line width = 0.3mm, ->, ocyan]
		(1)
		-- ($(1) + (0,1)$)
		-- ($(3) + (0,1)$)
		-- (3);

	\draw[line width = 0.3mm, ->, ogreen]
		(5)
		-- ($(5) + (0,-1)$)
		-- ($(6) + (0,-1)$)
		-- (6);

	\draw[line width = 0.3mm, ->, ogreen]
		(6)
		-- ($(6) + (0,1)$)
		-- ($(5) + (0,1)$)
		-- (5);

\end{tikzpicture}
\end{center}



\problem{}
Find all cycles in $[5342761]$.

\begin{solution}
	\begin{center}
	\begin{tikzpicture}[scale=0.5]
		\node (1) at (0, 0) {1};
		\node (2) at (1, 0) {2};
		\node (3) at (2, 0) {3};
		\node (4) at (3, 0) {4};
		\node (5) at (4, 0) {5};
		\node (6) at (5, 0) {6};
		\node (7) at (6, 0) {7};

		\draw[line width = 0.3mm, ->, ocyan]
			(1)
			-- ($(1) + (0,2)$)
			-- ($(7) + (0,2)$)
			-- (7);

		\draw[line width = 0.3mm, ->, ocyan]
			(7)
			-- ($(7) + (0,-1.5)$)
			-- ($(5) + (0,-1.5)$)
			-- (5);

		\draw[line width = 0.3mm, ->, ocyan]
			(5)
			-- ($(5) + (0,1.5)$)
			-- ($(1) + (0.5,1.5)$)
			-- ($(1) + (0.5,-1)$)
			-- ($(1) + (0,-1)$)
			-- (1);

		\draw[line width = 0.3mm, ->, ogreen]
			(2)
			-- ($(2) + (0,-1.5)$)
			-- ($(4) + (0,-1.5)$)
			-- (4);

		\draw[line width = 0.3mm, ->, ogreen]
			(4)
			-- ($(4) + (0,1)$)
			-- ($(3) + (0,1)$)
			-- (3);

		\draw[line width = 0.3mm, ->, ogreen]
			(3)
			-- ($(3) + (0,-1)$)
			-- ($(2) + (0.5,-1)$)
			-- ($(2) + (0.5,1)$)
			-- ($(2) + (0,1)$)
			-- (2);
	\end{tikzpicture}
	\end{center}
\end{solution}

\vfill


\problem{}
What permutation (on five objects) consists of the cycles $3 \to 5 \to 3$ and $1 \to 2 \to 4 \to 1$?


\begin{solution}
	\begin{center}
	\begin{tikzpicture}[scale=0.5]
		\node (1) at (0, 0) {1};
		\node (2) at (1, 0) {2};
		\node (3) at (2, 0) {3};
		\node (4) at (3, 0) {4};
		\node (5) at (4, 0) {5};

		\draw[line width = 0.3mm, ->, ocyan]
			(3)
			-- ($(3) + (0,1)$)
			-- ($(5) + (0,1)$)
			-- (5);

		\draw[line width = 0.3mm, ->, ocyan]
			(5)
			-- ($(5) + (0,-1)$)
			-- ($(3) + (0,-1)$)
			-- (3);

		\draw[line width = 0.3mm, ->, ogreen]
			(1)
			-- ($(1) + (0,-1)$)
			-- ($(2) + (0,-1)$)
			-- (2);

		\draw[line width = 0.3mm, ->, ogreen]
			(2)
			-- ($(2) + (0,1.5)$)
			-- ($(4) + (0,1.5)$)
			-- (4);

		\draw[line width = 0.3mm, ->, ogreen]
			(4)
			-- ($(4) + (0,-1.5)$)
			-- ($(1) + (0.5,-1.5)$)
			-- ($(1) + (0.5,1)$)
			-- ($(1) + (0,1)$)
			-- (1);
	\end{tikzpicture}

	This is $[41523]$
	\end{center}
\end{solution}



\vfill
\pagebreak

\definition{}
We now have a solution to our problem of notation.
Instead of referring to permutations using their output, we will refer to them using their \textit{cycles}.

\vspace{2mm}

For example, we'll write $[2134]$ as $(12)$, which denotes the cycle $1 \to 2 \to 1$:


\begin{center}
\begin{tikzpicture}[scale=0.5]
	\node (1) at (0, 0) {1};
	\node (2) at (1, 0) {2};
	\node (3) at (2, 0) {3};
	\node (4) at (3, 0) {4};

	\draw[line width = 0.3mm, ->, ocyan]
		(1)
		-- ($(1) + (0,-1)$)
		-- ($(2) + (0,-1)$)
		-- (2);

	\draw[line width = 0.3mm, ->, ocyan]
		(2)
		-- ($(2) + (0, 1)$)
		-- ($(1) + (0, 1)$)
		-- (1);
\end{tikzpicture}
\end{center}



As another example, $[431265]$ is $(1324)(56)$ in cycle notation. \par
Note that we write $[431265]$ as a \textit{composition} of two cycles: \par
applying the permutation $[431265]$ is the same as applying $(1324)$, then applying $(56)$.

\begin{center}
\begin{tikzpicture}[scale=0.5]
	\node (1) at (0, 0) {1};
	\node (2) at (1, 0) {2};
	\node (3) at (2, 0) {3};
	\node (4) at (3, 0) {4};
	\node (5) at (4, 0) {5};
	\node (6) at (5, 0) {6};

	\draw[line width = 0.3mm, ->, ocyan]
		(3)
		-- ($(3) + (0,-1)$)
		-- ($(2) + (0,-1)$)
		-- (2);

	\draw[line width = 0.3mm, ->, ocyan]
		(2)
		-- ($(2) + (0,1.5)$)
		-- ($(4) + (0,1.5)$)
		-- (4);

	\draw[line width = 0.3mm, ->, ocyan]
		(4)
		-- ($(4) + (0,-1.5)$)
		-- ($(1) + (0,-1.5)$)
		-- (1);

	\draw[line width = 0.3mm, ->, ocyan]
		(1)
		-- ($(1) + (0,1)$)
		-- ($(3) + (0,1)$)
		-- (3);

	\draw[line width = 0.3mm, ->, ogreen]
		(5)
		-- ($(5) + (0,-1)$)
		-- ($(6) + (0,-1)$)
		-- (6);

	\draw[line width = 0.3mm, ->, ogreen]
		(6)
		-- ($(6) + (0,1)$)
		-- ($(5) + (0,1)$)
		-- (5);

\end{tikzpicture}
\end{center}


\problem{}
Convince yourself that disjoint cycles commute. \par
That is, $(1324)(56) = (56)(1324)$ since $(1324)$ and $(56)$ do not overlap.

\problem{}
Write the following in square-bracket notation.
\begin{itemize}
	\item $(12)$       \tab~\tab on a set of 2 elements
	\item $(12)(435)$  \tab on a set of 5 elements
	\vspace{2mm}
	\item $(321)$      \tab~\tab on a set of 3 elements
	\item $(321)$      \tab~\tab on a set of 6 elements
	\vspace{2mm}
	\item $(1234)$      \tab on a set of 4 elements
	\item $(3412)$      \tab on a set of 4 elements
\end{itemize}

\vfill

\problem{}
Write the following in square-bracket notation.
Be careful.
\begin{itemize}
	\item $(13)(243)$ \tab on a set of 4 elements
	\item $(243)(13)$ \tab on a set of 4 elements
\end{itemize}

\vfill
\pagebreak