diff --git a/Advanced/Group Theory/main.tex b/Advanced/Group Theory/main.tex new file mode 100755 index 0000000..566da71 --- /dev/null +++ b/Advanced/Group Theory/main.tex @@ -0,0 +1,24 @@ +% use [nosolutions] flag to hide solutions. +% use [solutions] flag to show solutions. +\documentclass[ + solutions +]{../../resources/ormc_handout} + +\usepackage{tikz} + +\begin{document} + + \maketitle + + + {Group Theory} + { + Based on a lesson by Janet Chen \\ + Prepared by Mark on \today + } + + + \input{parts/00 review} + \input{parts/01 groups} + +\end{document} \ No newline at end of file diff --git a/Advanced/Group Theory/parts/00 review.tex b/Advanced/Group Theory/parts/00 review.tex new file mode 100755 index 0000000..706f63c --- /dev/null +++ b/Advanced/Group Theory/parts/00 review.tex @@ -0,0 +1,121 @@ +\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{} +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 \ No newline at end of file diff --git a/Advanced/Group Theory/parts/01 groups.tex b/Advanced/Group Theory/parts/01 groups.tex new file mode 100755 index 0000000..a58bdfb --- /dev/null +++ b/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 +