Started QG handout

Advanced/Quotient Groups/main.tex

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+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+\documentclass[
+	solutions,
+	singlenumbering,
+	shortwarning,
+	unfinished
+]{../../resources/ormc_handout}
+\usepackage{../../resources/macros}
+
+\usepackage{units}
+
+\uptitlel{Advanced 2}
+\uptitler{Fall 2023}
+\title{Quotient Groups}
+\subtitle{Prepared by \githref{Mark} on \today{}}
+
+
+\def\znz#1{\nicefrac{\mathbb{Z}}{#1\mathbb{Z}}}
+
+\begin{document}
+
+	\maketitle
+
+	\input{parts/0 mod}
+	\input{parts/1 groups}
+	\input{parts/2 subgroups}
+
+
+	% Rough outline:
+	%
+	% Part 1: (DONE)
+	% mod, eqrel, eqclass.
+	%
+	% Part 2: (IN PROGRESS)
+	% groups, Z/nZx, graphs, isomorphism.
+	%   generators, generating sets.
+	%
+	% Part 3: (IN PROGRESS)
+	% subgroups, isomorphic subgroups,
+
+	% TODO:
+	%
+	% cosets
+	% normal subgroups
+	% quotient groups
+	% Understand Z/nZ
+	% Functions as objects (groups of functions)
+	% 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}

Advanced/Quotient Groups/parts/0 mod.tex

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+\section{Modular Arithmetic}
+
+I'm sure you're all familiar with modular arithmetic.
+In this section, our goal is to meet \textit{equivalence relations},
+\textit{equivalence classes}, and use them to formally define arithmetic in mod $n$.
+
+
+\problem{}
+Compute the following:
+
+\begin{itemize}
+	\item $5 + 3 \pmod{4}$
+	\item $7 \times 4 \pmod{9}$
+	\item $-4 \pmod{5}$
+	\item $3^{-1} \pmod{7}$
+\end{itemize}
+
+\vfill
+
+
+\definition{}
+An \textit{equivalence relation} on a set $A$
+is a symbol that makes a statement about two elements of $A$.
+For example, $=$ is an equivalence relation on the set of integers.
+
+\vspace{2mm}
+
+An equivalence relation must satisfy the following properties:
+\begin{itemize}
+	\item Reflexivity: $x \sim x$ for all $x \in A$
+	\item Symmetry: if $x \sim y$, $y \sim x$ for any $x, y \in A$
+	\item Transitivity: if $x \sim y$ and $y \sim z$, then $x \sim z$
+\end{itemize}
+
+
+\problem{}<abseq>
+Which of the following are equivalence relations on $\mathbb{Z}$?
+\begin{itemize}
+	\item $>$
+	\item $\leq$
+	\item $\Bumpeq$, where $a \Bumpeq b$ if $|a| = |b|$
+	\item $\neq$
+\end{itemize}
+
+
+\vfill
+\pagebreak
+
+
+
+
+
+
+
+
+\problem{}
+Consider the relation $\equiv_n$ on $\mathbb{Z}$, where $a \equiv_n b$ holds iff $a \equiv b \pmod{n}$. \par
+Show that $\equiv_n$ is an equivalence relation.
+
+\vfill
+
+
+\definition{}
+Say we have an equivalence relation $\sim$ on a set $A$. \par
+The \textit{equivalence class} of $x$ is the set of all elements that are $\sim$ to $x$. \par
+Here are a few examples: \par
+\begin{itemize}[itemsep=2mm]
+	\item
+	The equivalence class of $2$ in $\mathbb{Z}$ under the relation $=$ is $\{2\}$, \par
+	since the only $x$ that satisfies $x = 2$ is $2$.
+
+	\item
+	The equivalence class of $9$ in $\mathbb{Z}$ under the relation $\Bumpeq$
+	from \ref{abseq} is $\{-9, 9\}$.
+\end{itemize}
+
+
+\problem{}
+What is the equivalence class of $3$ in $\mathbb{Z}$ under $\equiv_5$? \par
+\hint{Remember that $\mathbb{Z}$ contains both positive and negative numbers.}
+
+\begin{solution}
+	$\{..., -7, -2, 3, 8, 12, ... \}$
+\end{solution}
+
+
+\vfill
+
+\problem{}
+Let $A$ be a set and $\sim$ an equivalence relation. \par
+Show that every element of $A$ is in \textit{exactly one} equivalence class\footnotemark{}\hspace{-1ex}. \par
+\hint{What properties does an equivalence relation satisfy?}
+
+\footnotetext{
+	We could also say \say{$A$ is partitioned by $[A ~/ \sim]$}
+	or \say{$A$ is the disjoint union of $[A ~/ \sim]$,} \par
+	where $[A ~/ \sim]$ is the set of equivalence classes of $\sim$.
+}
+
+\vfill
+
+
+
+We now have a proper definition of \say{mod $n$:} \par
+it is the equivalence relation $a \equiv_n b$, which is usually written as $a \equiv b \pmod{n}$. \par
+We will use this definition thoughout this handout.
+
+\note[Note]{
+	This is different than the \say{mod} operator $a ~\%~ b $,
+	which is defined as the remainder of $a \div b$.
+}
+
+
+\pagebreak
+
+
+\definition{}
+Given any $x \in \mathbb{Z}$, $[x]_n$ is the equivalence class of $x$ under $\equiv_n$.
+
+
+\problem{}
+Compute the following:
+\begin{itemize}[itemsep = 1mm]
+	\item $[5]_3 + [4]_3$
+	\item $[-2]_7 + [9]_7$
+\end{itemize}
+
+\vfill
+
+\problem{}
+Does $[4]_3 + [7]_5$ make sense?
+
+\vfill
+
+\problem{}
+Find all $n$ that satisfy
+$[5]_n \times [17]_n = [3]_n + [2]_n$ \par
+\hint{$[a]_n = [b]_n$ iff $n$ divides $a - b$, by definition of mod.}
+
+\begin{solution}
+	$[85] = [12] ~\implies~ n ~|~ 85 - 12 ~\implies~ n ~|~ 73 ~\implies~ n \in \{1, 73\}$
+\end{solution}
+
+
+\vfill
+
+\definition{}
+$\znz{n}$ (pronounced \say{$\mathbb{Z}$ mod $n \mathbb{Z}$}) is the set of equivalence classes of $\equiv_n$ on $\mathbb{Z}$. \par
+For example, $\znz{5} = \{~ [0]_5,~ [1]_5,~ [2]_5,~ [3]_5,~ [4]_5 ~\}$. \par
+
+\vspace{2mm}
+
+This notation may seem a bit odd, but don't let it confuse you. \par
+One of our goals today is to understand what exactly $\znz{n}$ means.
+
+\problem{}
+What is $\znz{6}$?
+
+\vfill
+
+
+\pagebreak

Advanced/Quotient Groups/parts/1 groups.tex

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+\section{Groups}
+
+\definition{}
+A \textit{group} $G = (S, \ast)$ consists of a set $S$ and a binary operator $\ast$. \par
+By definition, a group always has 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 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 Any $a \in G$ has an \textit{inverse} $a^{-1} \in G$ that satisfies $a \ast a^{-1} = a^{-1} \ast a = e$. \par
+\end{enumerate}
+
+\note[Note]{
+	Commutativity is \textit{not} a required property of a group! \\
+	In most cases, $a \ast b \neq b \ast a$.
+}
+
+
+
+\problem{}
+Is $(\znz{5}, +)$ a group? \par
+How about $(\znz{5}, -)$? \par
+\hint{In this problem, $+$ and $-$ work just as you'd expect.}
+
+\vfill
+
+\problem{}
+What is the smallest possible group?
+
+\begin{solution}
+	Let $(G, \ast)$ be our group, where $G = \{e\}$ and $\ast$ is defined by the identity $e \ast e = e$
+
+	Verifying that the trivial group is a group is trivial.
+\end{solution}
+
+\vfill
+
+\problem{}
+How many distinct groups have two elements? \par
+\hint{
+	Two groups are \say{the same} if the elements of one can be renamed to get the other. \\
+	A group is fully defined by its multiplication table.
+}
+
+
+\vfill
+\pagebreak
+
+
+
+%\problem{}<firstcross>
+%Is $(\znz{17}, \times)$ a group? \par
+%How should we modify $\znz{17}$ to make it one?
+
+%\problem{}<secondcross>
+%Is $(\znz{6}, \times)$ a group? \par
+%How should we modify $\znz{6}$ to make it one? \par
+%\hint{
+%	Be careful, this isn't as easy as \ref{firstcross}. \\
+%	Which elements aren't invertible?
+%}
+
+
+%\definition{}
+%Building on problems \ref{num:firstcross} and \ref{num:secondcross}, we'll define $(\znz{n})^\times$ as the multiplicative
+%group of integers mod $n$. \par
+
+%Specifically, $(\znz{n})^\times$ is the set of all integers coprime to $n$. \par
+
+%\vspace{2mm}
+
+%For example, $(\znz{6})^\times = \{1, 5\}$ \par
+%and $(\znz{15})^\times = \{1, 2, 4, 7, 8, 11, 13, 14\}$ \par
+
+%\vspace{2mm}
+
+%Note that $0$ is the identity in $\znz{n}$ and $1$ is the identity in $(\znz{n})^\times$\hspace{-1.5ex}. \par
+%\note[Note]{
+%	Also, notice that we've omitted the operations $+$ and $\times$ in the two groups above. \\
+%	These operations are implicitly \say{attached} to $\znz{n}$ and $(\znz{n})^\times$\hspace{-1.5ex}, \\
+%	and we rarely write them for the sake of cleaner notation.
+%}
+
+
+\vfill
+
+
+\definition{}
+Let $G$ be a group, $a$ an element of $G$, and $n \in \mathbb{Z}^+$. \par
+$a^n$ is the defined as $a \ast a \ast ... \ast a$, repeated $n$ times.
+
+\vspace{1mm}
+
+Note that this is \textbf{not} \say{normal} exponentiation! \par
+If our group's operator is $+$ (for example, $\znz{5}$), $a^n = a + ... + a$, \par
+which you'll recognize as multipication.
+
+\vspace{1mm}
+
+Beware of this odd notation. By convention, we use \say{multiplicative} notation
+when working with groups---so, $a \ast b$ may also be written as $ab$,
+and $a \ast a \ast a$ may be written as $a^3$.
+
+\vspace{1mm}
+
+Again, remember that $a^n$ simply means \say{$\ast$ $a$ with itself $n$ times,} \par
+regardless of the specific operator our group uses.
+
+\problem{}
+Let $a$ be an element of a finite group. \par
+Show that there is a positive integer $n$ so that $a^n = e$. \par
+
+\vspace{2mm}
+
+The smallest such $n$ defines the \textit{order} of $g$.
+
+\vfill
+
+\problem{}
+Find the order of 5 in $(\znz{25}, +)$. \par
+%Find the order of 2 in $((\znz{17})^\times, \times)$. \par
+Find the order of 2 in $(\znz{7}, +)$. \par
+
+\vfill
+\pagebreak
+
+
+\definition{}
+Let $G$ be a group. \par
+We say a $g \in G$ is a \textit{generator} of $G$
+if every element in $G$ can be written as some power of $g$.
+
+\vspace{2mm}
+
+If $G$ has a generator, we say $G$ is \textit{cyclic.}
+
+\problem{}
+Find a generator of $\znz{7}$. Then, find a generator of $(\znz{7})^\times$
+
+\vfill
+
+\definition{}
+Let $G$ be a group. \par
+The \textit{order} of $G$ is the number of elements in $G$. \par
+We'll write this as $|G|$, using the same notation we use with sets. \par
+\note[Note]{
+	Don't confuse the order of an \textbf{element}
+	with the order of a \textbf{group}!
+}
+
+\problem{}
+Let $G$ be a cyclic group, and let $g$ be any generator in $G$. \par
+Show that $\text{ord}(g) = |G|$. \par
+\hint{Contradiction.}
+
+\vfill
+\pagebreak

Advanced/Quotient Groups/parts/2 subgroups.tex

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+\section{Subgroups}
+
+\definition{}
+Let $G$ be a group, and let $H$ be a subset of $G$. \par
+We say $H$ is a \textit{subgroup} of $G$ if $H$ is also a group
+(with the operation $\ast$).
+
+\definition{}
+Let $S$ be a subset of $G$. \par
+The \textit{group generated by $S$} consists of all elements of $G$ \par
+that may be written as a combination of elements in $S$
+
+\vspace{2mm}
+
+We will denote this group as $\langle S \rangle$. \par
+Convince yourself that $\langle g \rangle = G$ if $g$ generates $G$.
+
+\problem{}
+What is the subgroup generated by $\{7, 8\}$ in $(\znz{15})^\times$? \par
+Is this the whole group?
+
+\problem{}
+Show that the group generated by $S$ is indeed a group.