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58
Advanced/Symmetric Groups/main.tex Executable file
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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}
\usetikzlibrary{calc}
\uptitlel{Advanced 2}
\uptitler{\smallurl{}}
\title{Symmetric Groups}
\subtitle{Prepared by Mark on \today{}}
\def\line#1#2{
\draw[line width = 0.3mm, ->, ocyan]
(#1)
-- ($(#1) + (0, -1)$)
-- ($(#2) + (0,1)$)
-- (#2);
}
\begin{document}
\maketitle
\input{parts/0 intro}
\input{parts/1 cycle}
\input{parts/2 groups}
\input{parts/3 subgroup}
\section{Bonus problems}
\problem{}
Show that $x \in \mathbb{Z}^+$ has a multiplicative inverse mod $n$ iff $\text{gcd}(x, n) = 1$
\vfill
\problem{}
Let $\sigma = (\sigma_1 \sigma_2 ... \sigma_k)$ be a $k$-cycle in $S_n$, and let $\tau$ be an arbitrary element of $S_n$. \par
Show that $\tau \sigma \tau^{-1}$ = $\bigl(\tau(\sigma_1), \tau(\sigma_2), ..., \tau(\sigma_k)\bigr)$ \par
\hint{As usual, $\tau$ is a permutation. Thus, $\tau(x)$ is the value at position $x$ after applying $\tau$.}
\vfill
\problem{}
Show that the set $\Bigl\{ (1, 2),~ (1,2,...,n) \Bigr\}$ generates $S_n$.
\vfill
% TODO: (a second day?)
% alternating group
% type and sign and conjugation
% isomorphisms & automorphisms
% automorphism groups
\end{document}

6
Advanced/Symmetric Groups/meta.toml Normal file
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[metadata]
title = "Symmetric Groups"
[publish]
handout = true
solutions = true

199
Advanced/Symmetric Groups/parts/0 intro.tex Normal file
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\section{Introduction}
\definition{}
Informally, a \textit{permutation} of a collection of $n$ objects is an ordering of these $n$ objects. \par
For example, a few permutations of $\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}$ are $\texttt{ABCD}$,
$\texttt{BCDA}$, and $\texttt{DACB}$. \par
\vspace{2mm}
This, however, isn't the definition we'll use today. Instead of defining permutations as \say{ordered lists,}
(as we do above), we'll define them as functions. Our first goal today is to make sense of this definition.
\definition{Permutations}
Let $\Omega$ be an arbitrary set of $n$ objects. \par
A \textit{permutation} on $\Omega$ is a map from $\Omega$ to itself that produces a \textit{unique} output for each input. \par
\note{In other words, if $a$ and $b$ are different, $f(a)$ and $f(b)$ must also be different.}
\footnotetext{The words \say{function} and \say{map} are equivalent.}
\vspace{2mm}
For example, consider $\{1, 2, 3\}$. \par
One permutation on this set can be defined as follows: \par
\begin{itemize}
\item $f(1) = 3$
\item $f(2) = 1$
\item $f(3) = 2$
\end{itemize}
If we take the array $123$ and apply
\problem{}
List all permutations on three objects. \par
How many permutations of $n$ objects are there?
\vfill
\problem{}
What map corresponds to the permutation $[321]$?
\vfill
\problem{}
What map corresponds to the \say{do-nothing} permutation? \par
Write it as a function and in square-bracket notation. \par
\note[Note]{We usually call this the \textit{trivial permutation}}
\vfill
\pagebreak
We can visualize permutations with a \textit{string diagram}, shown below. \par
The arrows in this 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};
\line{1a}{1b}
\line{2a}{2b}
\line{3a}{3b}
\line{4a}{4b}
\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};
\line{1a}{1b}
\line{2a}{2b}
\line{3a}{3b}
\line{4a}{4b}
\end{tikzpicture}
\hfill\null
\vspace{2mm}
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};
\line{1a}{1b}
\line{2a}{2b}
\line{3a}{3b}
\line{4a}{4b}
\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};
\line{1a}{1b}
\line{2a}{2b}
\line{3a}{3b}
\line{4a}{4b}
\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) {$\boxdot$};
\line{1a}{1b}
\line{2a}{2b}
\line{3a}{3b}
\line{4a}{4b}
\end{tikzpicture}
\hfill\null
\vspace{2mm}
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 names of the items in our set do not have any meaning. \par
$\Omega$ is just a set of $n$ arbitrary 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}
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 right.}
\problem{}
Draw string diagrams for $[4123]$ and $[2341]$.
\vfill
\pagebreak

536
Advanced/Symmetric Groups/parts/1 cycle.tex Executable file
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\section{Cycle Notation}
\definition{Order}
The \textit{order} of a permutation $f$ is the \textbf{smallest} positive $n$ so that $f^n(x) = x$ for all $x$. \par
If we repeatedly apply a permutation with order $n$, we will get back to where we started after $n$ steps. \par
\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}
Swapping the first two elements of a list twice changes nothing. \par
Thus, $[2134]$ has an order of two.
\problem{}
What is the order of $[2314]$? \par
How about $[4321]$? \par
\note[Note]{You shouldn't need to draw any strings to solve this problem.}
\vfill
\problem{}
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
\definition{Composition}<compdef>
The \textit{composition} of two permutations $f$ and $g$ is the permutation $h(x) = f(g(x))$. \par
We'll denote this as $fg$---that is, by simply writing the permutations we're composing next to each other.
\problem{}
Show that function composition is associative. \par
That is, show that $f(gh) = (fg)h$.
\vfill
\problem{}
What is $[1324][4321]$? \par
How about $[321][213][231]$? \par
\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{Cycles}
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 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}
Another name we'll often use for two-cycles is \textit{transposition}. \par
Any permutation that swaps two adjacent elements is called a transposition. \par
\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) is formed by 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{Cycle Notation}
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}
Any permutation $\sigma$ may be written as a product (i.e, composition) of disjoint cycles $\sigma_1\sigma_2...\sigma_k$. \par
Make sure you believe this fact. If you don't, ask an instructor. \par
Also, the identity $f(x) = x$ is written as $()$ in cycle notation.
\problem{}
Convince yourself that disjoint cycles commute. \par
That is, that $(1324)(56) = (56)(1324) = [431265]$ since $(1324)$ and $(56)$ do not overlap. \par
\problem{}<insquare>
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}
\note{
Note that $(12)$ refers the \say{swap first two} permutation on a set of \textit{any} size. \\
We can now use the same name for the same permutation on two different sets! \\
}
\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
\problem{}
Look at the last two permutations in \ref{insquare}, $(1234)$ and $(3412)$. \par
These are \textit{identical}---they are the same cycle written in two different ways. \par
List all other ways to write this cycle. \hint{There are two more.} \par
\note{Also, note that the last two permutations in \ref{insquare} are the same.}
\pagebreak
\problem{}
What is the inverse of $(12)$? \par
How about $(123)$? And $(4231)$? \par
\note{
Note that again, we don't need to know how big our set is. \\
The inverse of $(12)$ is the same in all sets.
}
\vfill
\problem{}
Say $\sigma$ is a permutation composed of disjoint cycles $\sigma_1\sigma_2...\sigma_k$. \par
Say we know the order of all $\sigma_i$. What is the order of $\sigma$?
\begin{solution}
$\text{lcm}\Bigl(\text{ord}(\sigma_1),~ \text{ord}(\sigma_2),~ ..., ~ \text{ord}(\sigma_k)\Bigr)$
\end{solution}
\vfill
\problem{}<cycletrans>
Show that any cycle $(123...n)$ is equal to the product $(12)(23)...(n-1, n)$.
\begin{solution}
TODO
\end{solution}
\vfill
\problem{}
Write $(7126453)$ as a product of transpositions. \par
\vfill
\pagebreak
\problem{}<simpletrans>
Show that any permutation is a product of transpositions.
\begin{solution}
Use \ref{cycletrans}.
\end{solution}
\vfill
\problem{}
Show that any permutation is a product of transpositions of the form $(1, k)$. \par
\begin{solution}
Use \ref{simpletrans} and rewrite each $(a, b)$ as $(1, a)(1, b)(1, a)$. \par
Showing that $(a, b) = (1, a)(1, b)(1, a)$ is fairly easy.
\end{solution}
\vfill
\pagebreak
\problem{}
Show that any transposition $(a, b)$ is equal to the product $(a, a+1)(a+1, b)(a, a+1)$.
\begin{solution}
This is the same as the $(1, a)(1, b)(1, a)$ case above, but we use $a + 1$
as a \say{working slot} instead of $1$.
\end{solution}
\vfill
\problem{}
Show that any permutation is a product of adjacent transpositions. \par
(An \textit{adjacent transposition} swaps two adjacent elements, and thus looks like $(n, n+1)$)
\begin{solution}
As before, we will use \ref{simpletrans} and rewrite the transpositions it produces in a form that fits the problem.
We thus need to show that every transposition $(a, b)$ is a product of adjacent transpositions.
\vspace{8mm}
In the proof below, assume that $a < b$ and perform induction on $b - a$. \par
\textbf{Base Case:}\par
If $b - a = 1$, we clearly see that $(a, b)$ is a product of adjacent. \par
In fact, it \textit{is} an adjacent transposition.
\vspace{4mm}
\textbf{Induction:}\par
Now, say $b - a = n + 1$. \par
Assume that all $(a, b)$ where $b - a \leq n$ are products of adjacent transpositions.\par
Note that $(a, b) = (a, a+1)(a+1, b)(a, a+1)$.
\vspace{2mm}
$(a, a+1)$ is an adjacent transposition, and $b - (a+1) = n$. \par
Thus, $(a, b)$ is a product of adjacent transpositions.
\end{solution}
\vfill
\pagebreak

165
Advanced/Symmetric Groups/parts/2 groups.tex Executable file
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@ -0,0 +1,165 @@
\section{Groups (review)}
\definition{}
Before we continue, we must introduce a bit of notation:
\begin{itemize}
\item $S_n$ is the set of permutations on $n$ objects.
\item $\mathbb{Z}_n$ is the set of integers mod $n$.
\item $\mathbb{Z}_n^\times$ is the set of integers mod $n$ with multiplicative inverses. \par
In other words, it is the set of integers smaller than $n$ and coprime to $n$.\footnotemark{} \par
For example, $\mathbb{Z}_{12}^\times = \{1, 5, 7, 11\}$.
\footnotetext{We proved this in another handout, but you may take it as fact here.}
\end{itemize}
\problem{}
What are the elements of $S_3$? \tab\hint{Use cycle notation}\par
How about $\mathbb{Z}_{17}^\times$?
\vfill
\definition{}
A \textit{group} $(G, \ast)$ consists of a set $G$ and an operator $\ast$. \par
Groups always have the following properties:
\begin{enumerate}
\item $G$ is closed under $\ast$. In other words, $a, b \in G \implies a \ast b \in G$.
\item $\ast$ is \textit{associative}: $(a \ast b) \ast c = a \ast (b \ast c)$ for all $a,b,c \in G$
\item There is an \textit{identity} $e \in G$, so that $a \ast e = a \ast e = a$ for all $a \in G$.
\item For any $a \in G$, there exists a $b \in G$ so that $a \ast b = b \ast a = e$. $b$ is called the \textit{inverse} of $a$. \par
This element is written as $-a$ if our operator is addition and $a^{-1}$ otherwise.
\end{enumerate}
Any pair $(G, \ast)$ that satisfies these properties is a group.
\problem{}
Is $(\mathbb{Z}_5, +)$ a group? \par
Is $(\mathbb{Z}_5, -)$ a group? \par
\note[Note]{$+$ and $-$ refer to the usual operations in modular arithmetic.}
\vfill
\problem{}
What is the group with the fewest elements?
\begin{solution}
Let $(G, \star)$ be our group, where $G = \{x\}$ and $\star$ is defined by $x \star x = x$
Verifying that the trivial group is a group is trivial.
\end{solution}
\vfill
\pagebreak
\problem{}
Show that function composition is associative
\vfill
\problem{}
Show that $S_n$ is a group under composition.
\vfill
\problem{}
Let $(G, \ast)$ be a group with finitely many elements, and let $a \in G$. \par
Show that $\exists n \in \mathbb{Z}^+$ so that $a^n = e$ \par
\hint{$a^n = a \ast a \ast ... \ast a$ repeated $n$ times.}
\vspace{2mm}
The smallest such $n$ defines the \textit{order} of $g$.
\begin{examplesolution}
We've already done a special case of this problem! \par
Find it in this handout, then rewrite your proof for an arbitrary (finite) group.
\end{examplesolution}
\vfill
\problem{}
What is the order of 5 in $(\mathbb{Z}_{25}, +)$? \par
What is the order of 2 in $(\mathbb{Z}_{17}^\times, \times)$? \par
\vfill
\pagebreak
\definition{}<gendef>
Let $G$ be a group, and let $g$ be an element of $G$. \par
We say $g$ is a \textit{generator} if every other element of $G$ may be written as a power of $g$. \par
\problem{}
Say the size of a group $G$ is $n$. \par
If $g$ is a generator, what is its order? \par
Provide a proof.
\vfill
\problem{}
Find the two generators in $(\mathbb{Z}, +)$ \par
Then, find all generators of $(\mathbb{Z}_5, +)$
\vfill
\problem{}
How many groups have only one generator?
\begin{solution}
Only one: the trivial group. The inverse of a generator is also a generator!
\end{solution}
\vfill
\definition{}
Let $S$ be a subset of the elements in $G$. \par
We say that $S$ \textit{generates} $G$ if every element of $G$ may be written as a product of elements in $S$. \par
\note{Note that this is an extension of \ref{gendef}.}
\problem{}
We've already found a few generating sets of $S_n$. What are they?
\begin{solution}
The following sets generate $S_n$:
\begin{itemize}
\item All transpositions
\item All transpositions of the form $(1, k)$
\item All adjacent transpositions
\end{itemize}
\vspace{2mm}
The smallest generating set of $S_n$ consists of the transposition $(12)$ and the $n$-cycle $(1,2,...,n)$. \par
The proof of this is a bonus problem later in the handout.
\end{solution}
\vfill
\pagebreak

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\section{Subgroups}
\problem{}<s2s3share>
What elements do $S_2$ and $S_3$ share?
\vspace{2cm}
Consider the sets $\{1, 2\}$ and $\{1,2,3\}$. Clearly, $\{1, 2\} \subset \{1, 2, 3\}$. \par
Can we say something similar about $S_2$ and $S_3$?
\vspace{2mm}
Looking at \ref{s2s3share}, we may want to say that $S_2 \subset S_3$ since every element of $S_2$ is in $S_3$. \par
This however, isn't as interesting as it could be. Remember that $S_2$ and $S_3$ are \textit{groups}, not \textit{sets}: \par
their elements come with structure, which the \say{subset} relation does not capture.
\vspace{2mm}
To account for this, we'll define a similar relation: subgroups.
\definition{}
Let $G$ and $G'$ be groups. We say $G'$ is a \textit{subgroup} of $G$ (and write $G' \subset G$) if the following are true:\par
(Note that $x, y$ are elements of $G$, and $xy$ is multiplication in $G$)
\begin{itemize}
\item the set of elements in $G'$ is a subset of the set of elements in $G$.
\item the identity of $G$ is in $G'$
\item $x,y \in G' \implies xy \in G'$
\item $x \in G' \implies x^{-1} \in G'$
\end{itemize}
The above definition may look faily scary, but the idea behind a subgroup is simple. \par
Consider $S_3$ and $S_4$, the groups of permutations of $3$ and $4$ elements. \par
\vspace{2mm}
Say we have a set of four elements and only look at the first three. \par
$S_3$ fully describes all the ways we can arrange those three elements:
\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 (3b) at (1, -2) {3};
\node (1b) at (2, -2) {1};
\node (4b) at (3, -2) {4};
\draw[line width = 0.3mm, ->, ogreen]
(4a)
-- ($(4a) + (0, -1)$)
-- ($(4b) + (0,1)$)
-- (4b);
\line{1a}{1b}
\line{2a}{2b}
\line{3a}{3b}
\node[fill=white,draw=oblue,line width=0.3mm] at (1, -0.75) {$S_3$};
\end{tikzpicture}
\end{center}
\problem{}
Show that $S_3$ is a subgroup of $S_4$.
\vfill
\pagebreak
\definition{}
Let $G$ and $H$ be groups. We say that $G$ and $H$ are \textit{isomorphic} (and write $A \simeq B$) \par
if there is a bijection $f: G \to H$ with the following properties:
\begin{itemize}
\item $f(e_G) = e_H$, where $e_G$ is the identity in $G$
\item $f(x^{-1}) = f(x)^{-1}$ for all $x$ in $G$
\item $f(xy) = f(x)f(y)$ for all $x, y$ in $G$
\end{itemize}
Intuitively, you can think of isomorphism as a form of equivalence. \par
If two groups are isomorphic, they only differ by the names of their elements. \par
The function $f$ above tells us how to map one set of labels to the other.
\problem{}
Show that $\mathbb{Z}_7^\times$ and $\mathbb{Z}_9^\times$ are isomorphic.
\hint{
Build a bijection with the above properties. \\
Remember that a group is fully defined by its multiplication table.
}
\vfill
\problem{}
Show that $\mathbb{Z}_{10}^\times$ and $\mathbb{Z}_5^\times$, and $\mathbb{Z}_4$ are isomorphic.
\hint{
Build a bijection with the above properties. \\
Remember that a group is fully defined by its multiplication table.
}
\vfill
\problem{}
Show that isomorphism is transitive. \par
That is, if $A \simeq B$ and $B \simeq C$, then $A \simeq C$.
\vfill
\pagebreak
\problem{}<firstindex>
How many subgroups of $S_4$ are isomorphic to $S_3$? \par
\vfill
\problem{}
What are the orders of $S_3$ and $S_4$? \par
How is this related to \ref{firstindex}?
\begin{solution}
$|S_4| = |S_3| \times [S_4 : S_3]$
\vspace{2mm}
This solution is written using index notation, \par
but the class doesn't need to know what it means yet.
\end{solution}
\vfill
\problem{}
$S_4$ also has $S_2$ and the trivial group as subgroups. \par
How many instances of each does $S_4$ contain?
\vfill
\problem{}
$(\mathbb{Z}_4, +)$ is also a subgroup of $S_4$. Find it! \par
How many subgroups of $\mathbb{Z}_4$ are isomorphic to $S_4$?.
\begin{solution}
A good hint is \say{look at generators.}
\vspace{4mm}
There are four instances of $\mathbb{Z}_4$ in $S_4$, each of which is generated by a 4-cycle of $S_n$. \par
(i.e, the group generated by $(1234)$ is isomorphic to $\mathbb{Z}_4$)
\end{solution}
\vfill
\pagebreak