387 lines
8.1 KiB
TeX
387 lines
8.1 KiB
TeX
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\section{Cycle Notation}
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\definition{}
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The \textit{order} of a permutation $f$ is the smallest $n$ so that $f^n(x) = x$ for all $x$. \par
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In other words, if we repeat this permutation $n$ times, we get back to where we started.
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\vspace{2mm}
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For example, consider $[2134]$. This permutation has order $2$, as we clearly see below:
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\node (1a) at (0, 0.5) {1};
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\node (2a) at (1, 0.5) {2};
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\node (3a) at (2, 0.5) {3};
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\node (4a) at (3, 0.5) {4};
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\node (2b) at (0, -2) {2};
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\node (1b) at (1, -2) {1};
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\node (3b) at (2, -2) {3};
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\node (4b) at (3, -2) {4};
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\node (1c) at (0, -4.5) {1};
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\node (2c) at (1, -4.5) {2};
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\node (3c) at (2, -4.5) {3};
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\node (4c) at (3, -4.5) {4};
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\line{1a}{1b}
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\line{2a}{2b}
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\line{3a}{3b}
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\line{4a}{4b}
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\line{1b}{1c}
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\line{2b}{2c}
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\line{3b}{3c}
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\line{4b}{4c}
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\end{tikzpicture}
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\end{center}
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Of course, swapping the first two elements twice results in the identity map. \par
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$[2134]$ happens to be its own inverse. Also, the order of $[1234]$ is (of course) one.
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\problem{}
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What is the order of $[2314]$? \par
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How about $[4321]$? \par
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\note[Note]{You shouldn't need to draw any braids to solve this problem.}
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\vfill
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\problem{Bonus}
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Show that all permutations (on a finite set) have a well-defined order. \par
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In other words, show that there is always an integer $n$ so that $f^n(x) = x$.
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\vfill
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\pagebreak
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As you may have noticed, the square-bracket notation we've been using thus far is a bit unwieldy.
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Permutations are verbs---but we've been referring to them using a noun (namely, their output when
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applied to an ordered sequence of numbers). Our notation fails to capture the meaning of the
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underlying object.
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\vspace{2mm}
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Think about it: is the permutation $[1234]$ different than the permutation $[12345]$? \par
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Indeed, these permutations operate on different sets---but they are both the identity! \par
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What should we do if we want to talk about the identity on $\{1, 2, ..., 10\}$?
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\vspace{2mm}
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We need something better.
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\definition{}
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Any permutation is composed of a number of \textit{cycles}. \par
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For example, consider the permutation $[2134]$, which consists of one two-cycle: $1 \to 2 \to 1$ \par
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\note[Note]{$3 \to 3$ and $4 \to 4$ are also cycles, but we'll ignore them. One-cycles aren't aren't interesting.}
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\node (1) at (0, 0) {1};
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\node (2) at (1, 0) {2};
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\node (3) at (2, 0) {3};
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\node (4) at (3, 0) {4};
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\draw[line width = 0.3mm, ->, ocyan]
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(1)
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-- ($(1) + (0,-1)$)
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-- ($(2) + (0,-1)$)
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-- (2);
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\draw[line width = 0.3mm, ->, ocyan]
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(2)
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-- ($(2) + (0, 1)$)
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-- ($(1) + (0, 1)$)
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-- (1);
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\end{tikzpicture}
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\end{center}
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The permutation $[431265]$ is a bit more interesting---it contains of two cycles: \par
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($1 \to 3 \to 2 \to 4 \to 1$ and $5 \to 6 \to 5$)
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\node (1) at (0, 0) {1};
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\node (2) at (1, 0) {2};
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\node (3) at (2, 0) {3};
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\node (4) at (3, 0) {4};
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\node (5) at (4, 0) {5};
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\node (6) at (5, 0) {6};
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\draw[line width = 0.3mm, ->, ocyan]
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(3)
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-- ($(3) + (0,-1)$)
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-- ($(2) + (0,-1)$)
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-- (2);
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\draw[line width = 0.3mm, ->, ocyan]
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(2)
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-- ($(2) + (0,1.5)$)
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-- ($(4) + (0,1.5)$)
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-- (4);
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\draw[line width = 0.3mm, ->, ocyan]
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(4)
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-- ($(4) + (0,-1.5)$)
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-- ($(1) + (0,-1.5)$)
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-- (1);
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\draw[line width = 0.3mm, ->, ocyan]
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(1)
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-- ($(1) + (0,1)$)
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-- ($(3) + (0,1)$)
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-- (3);
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\draw[line width = 0.3mm, ->, ogreen]
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(5)
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-- ($(5) + (0,-1)$)
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-- ($(6) + (0,-1)$)
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-- (6);
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\draw[line width = 0.3mm, ->, ogreen]
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(6)
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-- ($(6) + (0,1)$)
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-- ($(5) + (0,1)$)
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-- (5);
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\end{tikzpicture}
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\end{center}
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\problem{}
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Find all cycles in $[5342761]$.
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\begin{solution}
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\node (1) at (0, 0) {1};
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\node (2) at (1, 0) {2};
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\node (3) at (2, 0) {3};
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\node (4) at (3, 0) {4};
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\node (5) at (4, 0) {5};
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\node (6) at (5, 0) {6};
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\node (7) at (6, 0) {7};
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\draw[line width = 0.3mm, ->, ocyan]
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(1)
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-- ($(1) + (0,2)$)
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-- ($(7) + (0,2)$)
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-- (7);
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\draw[line width = 0.3mm, ->, ocyan]
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(7)
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-- ($(7) + (0,-1.5)$)
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-- ($(5) + (0,-1.5)$)
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-- (5);
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\draw[line width = 0.3mm, ->, ocyan]
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(5)
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-- ($(5) + (0,1.5)$)
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-- ($(1) + (0.5,1.5)$)
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-- ($(1) + (0.5,-1)$)
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-- ($(1) + (0,-1)$)
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-- (1);
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\draw[line width = 0.3mm, ->, ogreen]
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(2)
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-- ($(2) + (0,-1.5)$)
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-- ($(4) + (0,-1.5)$)
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-- (4);
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\draw[line width = 0.3mm, ->, ogreen]
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(4)
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-- ($(4) + (0,1)$)
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-- ($(3) + (0,1)$)
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-- (3);
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\draw[line width = 0.3mm, ->, ogreen]
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(3)
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-- ($(3) + (0,-1)$)
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-- ($(2) + (0.5,-1)$)
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-- ($(2) + (0.5,1)$)
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-- ($(2) + (0,1)$)
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-- (2);
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\end{tikzpicture}
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\end{center}
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\end{solution}
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\vfill
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\problem{}
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What permutation (on five objects) consists of the cycles $3 \to 5 \to 3$ and $1 \to 2 \to 4 \to 1$?
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\begin{solution}
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\node (1) at (0, 0) {1};
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\node (2) at (1, 0) {2};
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\node (3) at (2, 0) {3};
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\node (4) at (3, 0) {4};
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\node (5) at (4, 0) {5};
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\draw[line width = 0.3mm, ->, ocyan]
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(3)
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-- ($(3) + (0,1)$)
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-- ($(5) + (0,1)$)
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-- (5);
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\draw[line width = 0.3mm, ->, ocyan]
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(5)
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-- ($(5) + (0,-1)$)
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-- ($(3) + (0,-1)$)
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-- (3);
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\draw[line width = 0.3mm, ->, ogreen]
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(1)
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-- ($(1) + (0,-1)$)
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-- ($(2) + (0,-1)$)
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-- (2);
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\draw[line width = 0.3mm, ->, ogreen]
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(2)
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-- ($(2) + (0,1.5)$)
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-- ($(4) + (0,1.5)$)
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-- (4);
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\draw[line width = 0.3mm, ->, ogreen]
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(4)
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-- ($(4) + (0,-1.5)$)
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-- ($(1) + (0.5,-1.5)$)
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-- ($(1) + (0.5,1)$)
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-- ($(1) + (0,1)$)
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-- (1);
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\end{tikzpicture}
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This is $[41523]$
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\end{center}
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\end{solution}
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\vfill
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\pagebreak
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\definition{}
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We now have a solution to our problem of notation.
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Instead of referring to permutations using their output, we will refer to them using their \textit{cycles}.
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\vspace{2mm}
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For example, we'll write $[2134]$ as $(12)$, which denotes the cycle $1 \to 2 \to 1$:
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\node (1) at (0, 0) {1};
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\node (2) at (1, 0) {2};
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\node (3) at (2, 0) {3};
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\node (4) at (3, 0) {4};
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\draw[line width = 0.3mm, ->, ocyan]
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(1)
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-- ($(1) + (0,-1)$)
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-- ($(2) + (0,-1)$)
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-- (2);
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\draw[line width = 0.3mm, ->, ocyan]
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(2)
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-- ($(2) + (0, 1)$)
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-- ($(1) + (0, 1)$)
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-- (1);
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\end{tikzpicture}
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\end{center}
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As another example, $[431265]$ is $(1324)(56)$ in cycle notation. \par
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Note that we write $[431265]$ as a \textit{composition} of two cycles: \par
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applying the permutation $[431265]$ is the same as applying $(1324)$, then applying $(56)$.
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\node (1) at (0, 0) {1};
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\node (2) at (1, 0) {2};
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\node (3) at (2, 0) {3};
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\node (4) at (3, 0) {4};
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\node (5) at (4, 0) {5};
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\node (6) at (5, 0) {6};
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\draw[line width = 0.3mm, ->, ocyan]
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(3)
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-- ($(3) + (0,-1)$)
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-- ($(2) + (0,-1)$)
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-- (2);
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\draw[line width = 0.3mm, ->, ocyan]
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(2)
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-- ($(2) + (0,1.5)$)
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-- ($(4) + (0,1.5)$)
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-- (4);
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\draw[line width = 0.3mm, ->, ocyan]
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(4)
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-- ($(4) + (0,-1.5)$)
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-- ($(1) + (0,-1.5)$)
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-- (1);
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\draw[line width = 0.3mm, ->, ocyan]
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(1)
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-- ($(1) + (0,1)$)
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-- ($(3) + (0,1)$)
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-- (3);
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\draw[line width = 0.3mm, ->, ogreen]
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(5)
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-- ($(5) + (0,-1)$)
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-- ($(6) + (0,-1)$)
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-- (6);
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\draw[line width = 0.3mm, ->, ogreen]
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(6)
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-- ($(6) + (0,1)$)
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-- ($(5) + (0,1)$)
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-- (5);
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\end{tikzpicture}
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\end{center}
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\problem{}
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Convince yourself that disjoint cycles commute. \par
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That is, $(1324)(56) = (56)(1324)$ since $(1324)$ and $(56)$ do not overlap.
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\problem{}
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Write the following in square-bracket notation.
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\begin{itemize}
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\item $(12)$ \tab~\tab on a set of 2 elements
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\item $(12)(435)$ \tab on a set of 5 elements
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\vspace{2mm}
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\item $(321)$ \tab~\tab on a set of 3 elements
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\item $(321)$ \tab~\tab on a set of 6 elements
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\vspace{2mm}
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\item $(1234)$ \tab on a set of 4 elements
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\item $(3412)$ \tab on a set of 4 elements
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\end{itemize}
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|
|
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|
|
\vfill
|
||
|
|
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|
|
\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
|