Removed group theory handout (crypto handout is better)
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% use [nosolutions] flag to hide solutions.
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% use [solutions] flag to show solutions.
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\documentclass[
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solutions
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]{../../resources/ormc_handout}
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\usepackage{tikz}
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\uptitlel{Advanced 2}
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\uptitler{Fall 2022}
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\title{Group Theory}
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\subtitle{Prepared by Mark on \today}
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\begin{document}
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\maketitle
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\input{parts/00 review}
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\input{parts/01 groups}
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\input{parts/02 isomorphism}
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\input{parts/03 bonus}
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\end{document}
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\section{A Review of Functions}
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\definition{}
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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 an element of $B$ to each element of $A$. We write this as $f: A \to B$.
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\vspace{2mm}
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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 a \textit{preimage} of 2 under $f$.
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\vspace{2mm}
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An element in a function's domain must have exactly one image. However, an element in the range may have more than one preimage.
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\problem{}
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Consider the function $f: \mathbb{R} \to \mathbb{R}^+ \cap \{0\}$ defined by $f(x) = x^2$
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\begin{itemize}
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\item[-] What is the image of 2?
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\item[-] What are the preimages of 9?
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\end{itemize}
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\vfill
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\definition{}
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We say a map is \textit{one-to-one} if $a = b \implies f(a) = f(b)$ for all $a, b$ in the domain. In other words, this means that each element of the range has at most one preimage.
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\definition{}
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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.
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\problem{}
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Find a function that is...
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\begin{enumerate}
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\item[-] neither one-to-one nor onto
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\item[-] one-to-one and not onto
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\item[-] not one-to-one, but onto
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\item[-] both one-to-one and onto
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\end{enumerate}
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We say a function that is both one-to-one and onto is \textit{bijective}.
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\vfill
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\pagebreak
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\definition{}
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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}.
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\problem{}
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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.
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\vfill
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\problem{}
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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.
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\vfill
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\pagebreak
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\section{A Review of Modular Arithmetic}
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\definition{}
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$\mathbb{Z}_n$ is the set of integers mod $n$. For example, $\mathbb{Z}_5 = \{0, 1, 2, 3, 4\}$. \\
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You should all be familiar with modular arithmetic.
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\definition{}
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The inverse of an element $a$ in $\mathbb{Z}_n$ is a $b$ so that $a \times b \equiv 1$. \\
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Not all elements of $\mathbb{Z}_n$ have an inverse. Those that do are called \textit{units}. \\
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\vspace{2mm}
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The set of all units in $\mathbb{Z}_n$ is written $\mathbb{Z}_n^\times$ \\
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Read this as \say{$\mathbb{Z}$ mod $n$ cross}
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\problem{}
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What are the elements of $\mathbb{Z}_5^\times$?
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\begin{solution}
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$\{1, 2, 3, 4\}$
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\end{solution}
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\vfill
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\problem{}<modtables>
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Create an addition table for $\mathbb{Z}_4$ and a multiplication table for $(\mathbb{Z}_5)^\times$
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\begin{center}
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\begin{tabular}{c | c c c c}
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+ & 0 & 1 & 2 & 3 \\
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\hline
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0 & ? & ? & ? & ? \\
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1 & ? & ? & ? & ? \\
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2 & ? & ? & ? & ? \\
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3 & ? & ? & ? & ? \\
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\end{tabular}
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\end{center}
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\begin{solution}
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\begin{center}
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\begin{tabular}{c | c c c c}
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+ & 0 & 1 & 2 & 3 \\
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\hline
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0 & 0 & 1 & 2 & 3 \\
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1 & 1 & 2 & 3 & 0 \\
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2 & 2 & 3 & 0 & 1 \\
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3 & 3 & 0 & 1 & 2 \\
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\end{tabular}
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\hspace{1cm}
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\begin{tabular}{c | c c c c}
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$\times$ & 1 & 2 & 3 & 4 \\
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\hline
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1 & 1 & 2 & 3 & 4 \\
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2 & 2 & 4 & 1 & 3 \\
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3 & 3 & 1 & 4 & 2 \\
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4 & 4 & 3 & 2 & 1 \\
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\end{tabular}
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\end{center}
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\end{solution}
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\vfill
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\vfill
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\pagebreak
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\section{Groups}
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Group theory gives us a set tools for understanding complex systems. We can use groups to solve the Rubik's cube, to solve problems in physics and chemistry, and to understand complex geometric symmetries. It's also worth noting that all modern cryptography relies heavily on group theory.
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\definition{}
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A \textit{group} $(G, \ast)$ consists of a set $G$ and an operator $\ast$. \\
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A group must have the following properties: \\
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\begin{enumerate}
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\item $G$ is closed under $\ast$. In other words, $a, b \in G \implies a \ast b \in G$.
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\item $\ast$ is associative: $(a \ast b) \ast c = a \ast (b \ast c)$ for all $a,b,c \in G$
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\item There is an \textit{identity} $e \in G$, so that $a \ast e = a \ast e = a$ for all $a \in G$.
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\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 $a$. \\
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This element is written as $-a$ if our operator is addition and $a^{-1}$ otherwise.
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\end{enumerate}
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Any pair $(G, \ast)$ that satisfies these properties is a group.
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\problem{}
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Is $(\mathbb{Z}_5, +)$ a group? \\
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Is $(\mathbb{Z}_5, -)$ a group? \\
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\hint{$+$ and $-$ refer to our usual definition of modular arithmetic.}
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\vfill
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\problem{}
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Show that $(\mathbb{R}, \times)$ is not a group, then make it one by modifying $\mathbb{R}$. \\
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\begin{solution}
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$(\mathbb{R}, \times)$ is not a group because $0$ has no inverse. \\
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The solution is simple: remove the problem.
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\vspace{3mm}
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$(\mathbb{R} - \{0\}, \times)$ is a group.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Show that a group has exactly one identity element.
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\vfill
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\problem{}
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Show that each element in a group has exactly one inverse.
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\vfill
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\problem{}
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Show that $(\mathbb{Z}_n^\times, \times)$ is a group for any $n \in \mathbb{Z}^+$.
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\vfill
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\problem{}
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Let $(G, \ast)$ be a group and $a, b, c \in G$. Show that...
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\begin{itemize}
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\item $a \ast b = a \ast c \implies b = c$
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\item $b \ast a = c \ast a \implies b = c$
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\end{itemize}
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This means that we can \say{cancel} operations in groups, much like we do in algebra.
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\vfill
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\pagebreak
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\problem{}
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What is the smallest group we can create?
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\begin{solution}
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Let $(G, \circledcirc)$ be our group, where $G = \{\star\}$ and $\circledcirc$ is defined by the identity $\star \circledcirc \star = \star$
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Verifying that the trivial group is a group is trivial.
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\end{solution}
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\vfill
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\problem{}
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Let $G$ be the set of all bijections $A \to A$. \\
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Let $\circ$ be the usual composition operator. \\
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Is $(G, \circ)$ a group?
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\vfill
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\definition{}
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Note that our definition of a group does \textbf{not} state that $a \ast b = b \ast a$. \\
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Many interesting groups do not have this property.
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Those that do are called \textit{abelian} groups. \\
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\vspace{2mm}
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One example of a non-abelian group is the set of invertible 2x2 matrices under matrix multiplication. In this handout, all groups are abelian.\\
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\problem{}
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Show that if $G$ has four elements, $(G, \ast)$ is abelian.
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\vfill
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\pagebreak
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\problem{}
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Let $(G, \ast)$ be a finite group (i.e, $G$ has finitely many elements), and let $g \in G$. \\
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Show that $\exists~n \in \mathbb{Z}^+$ so that $g^n = e$ \\
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\hint{$g^n = g \ast g \ast ... \ast g$ $n$ times.}
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\vspace{2mm}
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The smallest such $n$ defines the \textit{order} of $g$.
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\vfill
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\problem{}
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What is the order of 5 in $(\mathbb{Z}_{25}, +)$? \\
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What is the order of 2 in $(\mathbb{Z}_{17}^\times, \times)$? \\
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\vfill
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\pagebreak
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% \problem{}
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% Let $e, a, b, c$ be counterclockwise rotations of a square by $0, \frac{\pi}{2}, \pi,$ and $\frac{3\pi}{2}$. \\
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% Create a multiplication table for this group.
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% \vfill
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%
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% \problem{}
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% Let $d, f, g, h$ correspond to reflections of the square along the following axis. \\
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% Create a multiplication table for this group.
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%
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% \begin{center}
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% \begin{tikzpicture}[scale=2]
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% \draw (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
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%
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% \draw[gray] (1.25,1.25) -- (-0.25,-0.25) node[below left]{$d$};
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% \draw[gray] (1.25,-0.25) -- (-0.25,1.25) node[above left]{$f$};
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% \draw[gray] (0.5,-0.25) -- (0.5,1.25) node[above]{$g$};
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% \draw[gray] (-0.25, 0.5) -- (1.25,0.5) node[right]{$h$};
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%
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% \end{tikzpicture}
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% \end{center}
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% \vfill
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%
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% \problem{}
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% Create a multiplication table for all symmetries of a square.
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% \vfill
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% \pagebreak
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%
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% \problem{}
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% Create a multiplication table for all symmetries of a rhombus.
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% \vfill
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% \pagebreak
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%
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% \problem{}
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% Find the order of each element in...
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% \begin{itemize}
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% \item The group of symmetries of a square
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% \item The group of symmetries of a rhombus
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% \end{itemize}
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%
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%
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% \vfill
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% \pagebreak
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\section{Isomorphisms}
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\definition{}
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We say two groups are \textit{isomorphic} if we can create a bijective mapping between them while preserving multiplication structure. This mapping is called an \textit{isomorphism}.\\
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\vspace{2mm}
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This means that if groups $A$ and $B$ are isomorphic under $f$, \\
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$a_1 \ast a_2 = a_3$ in A implies that $f(a_1) \ast f(a_2) = f(a_3)$ in B.
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\problem{}
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Recall your tables from \ref{modtables}: \\
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\begin{center}
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\begin{tabular}{c | c c c c}
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+ & 0 & 1 & 2 & 3 \\
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\hline
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0 & 0 & 1 & 2 & 3 \\
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1 & 1 & 2 & 3 & 0 \\
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2 & 2 & 3 & 0 & 1 \\
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3 & 3 & 0 & 1 & 2 \\
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\end{tabular}
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\hspace{1cm}
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\begin{tabular}{c | c c c c}
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$\times$ & 1 & 2 & 3 & 4 \\
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\hline
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1 & 1 & 2 & 3 & 4 \\
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2 & 2 & 4 & 1 & 3 \\
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3 & 3 & 1 & 4 & 2 \\
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4 & 4 & 3 & 2 & 1 \\
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\end{tabular}
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\end{center}
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Are $(\mathbb{Z}_4, +)$ and $(\mathbb{Z}_5^\times, \times)$ isomorphic? If they are, find a bijection that maps one to the other.
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\vfill
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\problem{}
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Let groups $A$ and $B$ be isomorphic under $f$. Show that $f(e_A) = e_B$, where $e_A$ and $e_B$ are the identities of $A$ and $B$.
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\vfill
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\problem{}
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Let groups $A$ and $B$ be isomorphic under $f$. \\
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Show that $f(a^{-1}) = f(a)^{-1}$ for all $a \in A$.
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\vfill
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\problem{}
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Let groups $A$ and $B$ be isomorphic under $f$. Show that $f(a)$ and $a$ have the same order.
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\vfill
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\pagebreak
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\problem{}<howmanygroups>
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Find all distinct groups of two elements. \\
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Find all distinct groups of three elements. \\
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Groups that are isomorphic are not distinct.
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\begin{solution}
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There is only one nonisomorphic two-element group. \\
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The same is true of a three-element group. \\
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See \texttt{https://oeis.org/A000001}, titled \say{Number of groups of order n}
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\end{solution}
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\vfill
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\problem{}
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Show that the groups $(\mathbb{R}, +)$ and $(\mathbb{R}^+, \times)$ are isomorphic.
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\vfill
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\pagebreak
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\section{Bonus}
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\problem{}
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Find the inverse of 19 in $\mathbb{Z}_{23}$ \\
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\hint{Recall the Euclidean Algorithm}
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\begin{solution}
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17
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\end{solution}
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\vfill
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\problem{}
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Prove Fermat's little theorem:
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$$
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a^p = a \text{ (mod p)}
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$$
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For positive integers $a, p$
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\vfill
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\problem{}
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Let $a$ and $m$ be integers so that $a < m$. \\
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Show that $a$ has an inverse mod $m$ iff $\gcd(a, m) = 1$ \\
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\begin{solution}
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Assume $a^\star$ is the inverse of $a \pmod{m}$. \\
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Then $a^\star \times a \equiv 1 \pmod{m}$ \\
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Therefore, $aa^\star - 1 = km$, and $aa^\star - km = 1$ \\
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We know that $\gcd(a, m)$ divides $a$ and $m$, therefore $\gcd(a, m)$ must divide $1$. \\
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$\gcd(a, m) = 1$ \\
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Now, assume $\gcd(a, m) = 1$. \\
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By the Extended Euclidean Algorithm, we can find $(u, v)$ that satisfy $au+mv=1$ \\
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So, $au-1 = mv$. \\
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$m$ divides $au-1$, so $au \equiv 1 \pmod{m}$ \\
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$u$ is $a^\star$.
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\end{solution}
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\vfill
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\problem{}
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Show that for any integers $a, b, c$, \\
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$\gcd(ac + b, a) = \gcd(a, b)$\\
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\vfill
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