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handouts/Advanced/Symmetric Group/parts/1 cycle.tex
Mark bd88b894a1
Added cycle notation
2023-12-18 10:55:08 -08:00

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\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
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