Added initial parts of group theory handout

Advanced/Group Theory/main.tex

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+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+\documentclass[
+	solutions
+]{../../resources/ormc_handout}
+
+\usepackage{tikz}
+
+\begin{document}
+
+	\maketitle
+		<Advanced 2>
+		<Fall 2022>
+		{Group Theory}
+		{
+			Based on a lesson by Janet Chen \\
+			Prepared by Mark on \today
+		}
+
+
+	\input{parts/00 review}
+	\input{parts/01 groups}
+
+\end{document}

Advanced/Group Theory/parts/00 review.tex

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+\section{Review: Functions}
+
+\definition{}
+A \textit{function} or \textit{map} $f$ from a set $A$ (the \textit{domain}, $\mathcal{D}$) to a set $B$ (the \textit{range}, $\mathcal{R}$) is a rule that assigns each element of $A$ to an element of $B$. We write this as $f: A \to B$.
+
+\vspace{2mm}
+
+Consider a function $f: \mathbb{Z} \to \mathbb{Z}$. If $f(1) = 2$, we say that 2 is the \textit{image} of 1 and 1 is the \textit{preimage} of 2 under $f$.
+
+\vspace{2mm}
+
+An element in a function's domain must have exactly one image. However, an element in the range may have more than one preimage.
+
+\problem{}
+Consider the function $f: \mathbb{R} \to \mathbb{R}^+$ defined by $f(x) = x^2$
+\begin{itemize}
+	\item[-] What is the image of 2?
+	\item[-] What are the preimages of 9?
+\end{itemize}
+
+\vfill
+
+\definition{}
+We say a map is \textit{one-to-one} if $a \neq b \implies f(a) \neq f(b)$. In other words, this means that each element of the range has at most one preimage.
+
+\definition{}
+We say a map $f$ is \textit{onto} if, for every $y \in \mathcal{R}$, there exists an $x \in \mathcal{D}$ so that $f(x) = y$. In other words, this means that every $y$ in the range has a preimage in the domain.
+
+\problem{}
+Find a function that is...
+\begin{enumerate}
+	\item[-] not one-to-one, not onto
+	\item[-] one-to-one and not onto
+	\item[-] not one-to-one, but onto
+	\item[-] both one-to-one and onto
+\end{enumerate}
+We say a function that is both one-to-one and onto is \textit{bijective}.
+
+\vfill
+\pagebreak
+
+\definition{}
+Let $f: A \to B$ and $g: B \to C$. We can define a new function $(g \circ f): A \to C$, where $(g \circ f)(a) = g(f(a))$. This is called \textit{composition}.
+
+\problem{}
+Suppose $f: A \to B$ and $g: B \to C$ are both one-to-one. Must $(g \circ f)$ be one-to-one? Provide a proof or a counterexample.
+
+\vfill
+
+\problem{}
+Suppose $f: A \to B$ and $g: B \to C$ are both onto. Must $(g \circ f)$ be onto? Provide a proof or a counterexample.
+
+\vfill
+\pagebreak
+
+\section{Review: Modular Arithmetic}
+
+\definition{}
+$\mathbb{Z} / n$ is the set of integers mod $n$. For example, $Z/5 = \{0, 1, 2, 3, 4\}$. \\
+You should all be familiar with modular arithmetic.
+
+\definition{}
+The inverse of an element $a$ in $\mathbb{Z}/n$ is a $b$ so that $a \times b \equiv 1$. \\
+
+Not all elements of $\mathbb{Z}/n$ have an inverse. Those that do are called \textit{units}. \\
+
+\vspace{2mm}
+
+The set of all units in $\mathbb{Z}/n$ is written $(\mathbb{Z}/n)^\times$ \\
+Read this as \say{$\mathbb{Z}$ mod $n$ cross}
+
+\problem{}
+What are the elements of $(\mathbb{Z}/5)^\times$?
+
+\begin{solution}
+	$\{1, 2, 3, 4\}$
+\end{solution}
+
+\vfill
+
+\problem{}<modtables>
+Create an addition table for $\mathbb{Z}/4$ and a multiplication table for $(\mathbb{Z}/5)^\times$
+
+\begin{center}
+\begin{tabular}{c | c c c c}
+	+ & 0 & 1 & 2 & 3 \\
+	\hline
+	0 & ? & ? & ? & ? \\
+	1 & ? & ? & ? & ? \\
+	2 & ? & ? & ? & ? \\
+	3 & ? & ? & ? & ? \\
+\end{tabular}
+\end{center}
+
+\begin{solution}
+	\begin{center}
+	\begin{tabular}{c | c c c c}
+		+ & 0 & 1 & 2 & 3 \\
+		\hline
+		0 & 0 & 1 & 2 & 3 \\
+		1 & 1 & 2 & 3 & 0 \\
+		2 & 2 & 3 & 0 & 1 \\
+		3 & 3 & 0 & 1 & 2 \\
+	\end{tabular}
+	\hspace{1cm}
+	\begin{tabular}{c | c c c c}
+		\times & 1 & 2 & 3 & 4 \\
+		\hline
+		1 & 1 & 2 & 4 & 3 \\
+		2 & 2 & 4 & 3 & 1 \\
+		3 & 4 & 3 & 1 & 2 \\
+		4 & 3 & 1 & 2 & 4 \\
+	\end{tabular}
+	\end{center}
+\end{solution}
+
+
+
+\vfill
+\vfill
+\pagebreak

Advanced/Group Theory/parts/01 groups.tex

@@ -0,0 +1,95 @@
+\section{Groups}
+
+\definition{}
+A \textit{group} $(G, \ast)$ consists of a set $G$ and an operator $\ast$. \\
+A group must 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 associative: $a \ast b = b \ast a$
+	\item There is an \textit{identity} $\overline{0} \in G$, so that $a \ast \overline{0} = a$ for all $a \in G$.
+	\item For any $a \in G$, there exists a $b \in G$ so that $a \ast b = \overline{0}$. $b$ is called the \textit{inverse} of $a$. \\
+	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.
+
+\definition{}
+Note that our definition of a group does \textbf{not} state that $a \ast b = b \ast a$. \\
+Many interesting groups do not have this property. \\
+Those that do are called \textit{abelian} groups.
+
+\problem{}
+Is $(\mathbb{Z}/5, +)$ a group? \\
+Is $(\mathbb{Z}/5, -)$ a group? \\
+\hint{$+$ and $-$ refer to our usual definition of modular arithmetic.}
+
+\vfill
+
+\problem{}
+$(\mathbb{R}, \times)$ is not a group. \\
+Make it one by modifying $\mathbb{R}$. \\
+
+\begin{solution}
+	$(\mathbb{R}, \times)$ is not a group because $0$ has no inverse. \\
+	The solution is simple: remove the problem.
+
+	\vspace{3mm}
+
+	$(\mathbb{R} - \{0\}, \times)$ is a group.
+\end{solution}
+
+\vfill
+
+\problem{}
+Can you construct a group that contains a single element?
+
+\begin{solution}
+	Let $(G, \circledcirc)$ be our group, where $G = \{\star\}$ and $\circledcirc$ is defined by the identity $\star \circledcirc \star = \star$
+
+	Verifying that the trivial group is a group is trivial.
+\end{solution}
+
+\vfill
+\pagebreak
+
+\definition{}
+Recall your tables from \ref{modtables}: \\
+
+\begin{center}
+\begin{tabular}{c | c c c c}
+	+ & 0 & 1 & 2 & 3 \\
+	\hline
+	0 & 0 & 1 & 2 & 3 \\
+	1 & 1 & 2 & 3 & 0 \\
+	2 & 2 & 3 & 0 & 1 \\
+	3 & 3 & 0 & 1 & 2 \\
+\end{tabular}
+\hspace{1cm}
+\begin{tabular}{c | c c c c}
+	\times & 1 & 2 & 3 & 4 \\
+	\hline
+	1 & 1 & 2 & 4 & 3 \\
+	2 & 2 & 4 & 3 & 1 \\
+	3 & 4 & 3 & 1 & 2 \\
+	4 & 3 & 1 & 2 & 4 \\
+\end{tabular}
+\end{center}
+
+Convince yourself that $(\mathbb{Z}/4, +)$ and $( (\mathbb{Z}/5)^\times, \times )$ are the same group. \\
+We say that two such groups are \textit{isomorphic}.
+
+\vspace{2mm}
+
+Intuitively, this means that $(\mathbb{Z}/4, +)$ and $( (\mathbb{Z}/5)^\times, \times )$ have the same algebraic structure. We can translate statements about addition in $\mathbb{Z}/4$ into statements about multiplication in $(\mathbb{Z}/5)^\times$ \\
+
+\problem{}
+Show that a group has exactly one identity element.
+\vfill
+
+\problem{}
+Show that each element in a group has exactly one inverse.
+
+\vfill
+\pagebreak
+