Started symmetric group handout

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Mark 2023-12-17 12:10:22 -08:00
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% Q/Z problems (mod generalization)
% isomorphism groups (which are iso to symmetric group)
% Another handout:
%
% symmetric group, number of permutations,
% cycle notation, type and sign,
% proofs about generators, alternating group
% alternating group generators, fun problems.
\end{document}

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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering
]{../../resources/ormc_handout}
\usepackage{../../resources/macros}
\uptitlel{Advanced 2}
\uptitler{Winter 2023}
\title{Symmetric Groups}
\subtitle{Prepared by \githref{Mark} on \today{}}
\begin{document}
\maketitle
\input{parts/0 intro}
% cycle notation
% decomposition into transpositions
% few more problems?
% inline functions
% symmetric group
% order & generators
% subgroups
% alternating group
% type and sign
% isomorphisms & automorphisms
% automorphism groups
\end{document}

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\section{Introduction}
\definition{Intuitive permutations}
Intuitively, a \textit{permutation} is an ordered arrangement of a set of objects. \par
For example, $123$, $312$, and $231$ are all permutations of 1, 2, and 3.
\problem{}
List all permutations on three objects. \par
How many permutations of $n$ objects are there?
\vfill
\definition{Formal permutations}<permadef>
Let $\Omega$ be an arbitrary set of $n$ objects. \par
A \textit{permutation} on $\Omega$ is a bijective map $f: \Omega \to \Omega$.
\vspace{2mm}
For example, consider the objects 1, 2, and 3. \par
The permutation $[312]$ is given by a map $f$ defined by the following table:
\begin{itemize}
\item $f(1) = 3$
\item $f(2) = 1$
\item $f(3) = 2$
\end{itemize}
Similarly, the \textit{trivial permutation} $[123]$ is given by the identity map $f(x) = x$.
\problem{}
What map corresponds to the permutation $[321]$?
\vfill
\problem{}
Why do we define permutations as a \textit{bijective} map?
\vfill
\pagebreak
We can visualize permutations with a diagram we'll call the \say{braid.}
The arrows in the diagram denote the image of $f$ for each possible input.
Two examples are below:
\vspace{2mm}
\hfill
\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 (1b) at (0, -2) {1};
\node (3b) at (1, -2) {3};
\node (4b) at (2, -2) {4};
\node (2b) at (3, -2) {2};
\draw[line width = 0.3mm, ->, ocyan] (1a.south) -- (1a.south |- 0, -0.5) -- (1b.north |- 0,-1) -- (1b.north);
\draw[line width = 0.3mm, ->, ocyan] (2a.south) -- (2a.south |- 0, -0.5) -- (2b.north |- 0,-1) -- (2b.north);
\draw[line width = 0.3mm, ->, ocyan] (3a.south) -- (3a.south |- 0, -0.5) -- (3b.north |- 0,-1) -- (3b.north);
\draw[line width = 0.3mm, ->, ocyan] (4a.south) -- (4a.south |- 0, -0.5) -- (4b.north |- 0,-1) -- (4b.north);
\end{tikzpicture}
\hfill
\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};
\draw[line width = 0.3mm, ->, ocyan] (1a.south) -- (1a.south |- 0, -0.5) -- (1b.north |- 0,-1) -- (1b.north);
\draw[line width = 0.3mm, ->, ocyan] (2a.south) -- (2a.south |- 0, -0.5) -- (2b.north |- 0,-1) -- (2b.north);
\draw[line width = 0.3mm, ->, ocyan] (3a.south) -- (3a.south |- 0, -0.5) -- (3b.north |- 0,-1) -- (3b.north);
\draw[line width = 0.3mm, ->, ocyan] (4a.south) -- (4a.south |- 0, -0.5) -- (4b.north |- 0,-1) -- (4b.north);
\end{tikzpicture}
\hfill\null
Note that in all our examples thus far, the objects in our set have an implicit order.
This is only for convenience. The elements of $\Omega$ are not ordered (it is a \textit{set}, after all),
and we may present them however we wish.
\vspace{1cm}
For example, consider the diagrams below. \par
On the left, 1234 are ordered as usual. In the middle, they are ordered alphabetically. \par
The rightmost diagram uses arbitrary, meaningless labels.
\vspace{2mm}
\hfill
\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};
\draw[line width = 0.3mm, ->, ocyan] (1a.south) -- (1a.south |- 0, -0.5) -- (1b.north |- 0,-1) -- (1b.north);
\draw[line width = 0.3mm, ->, ocyan] (2a.south) -- (2a.south |- 0, -0.5) -- (2b.north |- 0,-1) -- (2b.north);
\draw[line width = 0.3mm, ->, ocyan] (3a.south) -- (3a.south |- 0, -0.5) -- (3b.north |- 0,-1) -- (3b.north);
\draw[line width = 0.3mm, ->, ocyan] (4a.south) -- (4a.south |- 0, -0.5) -- (4b.north |- 0,-1) -- (4b.north);
\end{tikzpicture}
\hfill
\begin{tikzpicture}[scale=0.5]
\node (4a) at (0, 0.5) {4};
\node (1a) at (1, 0.5) {1};
\node (3a) at (2, 0.5) {3};
\node (2a) at (3, 0.5) {2};
\node (1b) at (0, -2) {1};
\node (4b) at (1, -2) {4};
\node (3b) at (2, -2) {3};
\node (2b) at (3, -2) {2};
\draw[line width = 0.3mm, ->, ocyan] (1a.south) -- (1a.south |- 0, -0.5) -- (1b.north |- 0,-1) -- (1b.north);
\draw[line width = 0.3mm, ->, ocyan] (2a.south) -- (2a.south |- 0, -0.5) -- (2b.north |- 0,-1) -- (2b.north);
\draw[line width = 0.3mm, ->, ocyan] (3a.south) -- (3a.south |- 0, -0.5) -- (3b.north |- 0,-1) -- (3b.north);
\draw[line width = 0.3mm, ->, ocyan] (4a.south) -- (4a.south |- 0, -0.5) -- (4b.north |- 0,-1) -- (4b.north);
\end{tikzpicture}
\hfill
\begin{tikzpicture}[scale=0.5]
\node (1a) at (0, 0.5) {$\triangle$};
\node (2a) at (1, 0.5) {$\divideontimes$};
\node (3a) at (2, 0.5) {$\circledcirc$};
\node (4a) at (3, 0.5) {$\boxdot$};
\node (2b) at (0, -2) {$\divideontimes$};
\node (1b) at (1, -2) {$\triangle$};
\node (3b) at (2, -2) {$\circledcirc$};
\node (4b) at (3, -2) {$\boxdot$};
\draw[line width = 0.3mm, ->, ocyan] (1a.south) -- (1a.south |- 0, -0.5) -- (1b.north |- 0,-1) -- (1b.north);
\draw[line width = 0.3mm, ->, ocyan] (2a.south) -- (2a.south |- 0, -0.5) -- (2b.north |- 0,-1) -- (2b.north);
\draw[line width = 0.3mm, ->, ocyan] (3a.south) -- (3a.south |- 0, -0.5) -- (3b.north |- 0,-1) -- (3b.north);
\draw[line width = 0.3mm, ->, ocyan] (4a.south) -- (4a.south |- 0, -0.5) -- (4b.north |- 0,-1) -- (4b.north);
\end{tikzpicture}
\hfill\null
It shouldn't be hard to see that despite the different \say{output} order (2134 and 1432), \par
the same permutation is depicted in all three diagrams. This example demonstrates two things:
\begin{itemize}[itemsep=2mm]
\item First, the items of our set do not have any meaning. \par
$\Omega$ is just a set of arbitrary \textit{things}, which we may label however we like.
\item Second, permutations are verbs. We do not care about the \say{output} of a certain permutation,
we care about what it \textit{does}. We could, for example, describe the permutation above as
\say{swap the first two of four elements.}
\end{itemize}
\vspace{2mm}
\vspace{1cm}
Why, then, do we order our elements when we talk about permutations? As noted before, this is for convenience.
If we assign a natural order to the elements of $\Omega$ (say, 1234), we can identify permutations by simply listing
their output:
Clearly, $[1234]$ represents the trivial permutation, $[2134]$ represents \say{swap first two,}
and $[4123]$ represents \say{cycle left.}
\problem{}
Draw braids for $[4123]$ and $[2341]$.
\vfill
Finally, note that permutations (as defined in \ref{permadef}) are \textit{not} \say{orderings of a certain set.} \par
They are defined as \textit{bijective maps}, which can be \textit{thought of} as orderings. \par
Remember: permutations are verbs!
\pagebreak