Removed quotient group handout
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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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singlenumbering,
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shortwarning,
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unfinished
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]{../../resources/ormc_handout}
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\usepackage{../../resources/macros}
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\usepackage{units}
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\uptitlel{Advanced 2}
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\uptitler{Fall 2023}
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\title{Quotient Groups}
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\subtitle{Prepared by \githref{Mark} on \today{}}
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\def\znz#1{\nicefrac{\mathbb{Z}}{#1\mathbb{Z}}}
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\begin{document}
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\maketitle
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\input{parts/0 mod}
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\input{parts/1 groups}
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\input{parts/2 subgroups}
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% Rough outline:
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%
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% Part 1: (DONE)
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% mod, eqrel, eqclass.
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%
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% Part 2: (IN PROGRESS)
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% groups, Z/nZx, graphs, isomorphism.
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% generators, generating sets.
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%
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% Part 3: (IN PROGRESS)
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% subgroups, isomorphic subgroups,
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% TODO:
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%
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% cosets
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% normal subgroups
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% quotient groups
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% Understand Z/nZ
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% Functions as objects (groups of functions)
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% Q/Z problems (mod generalization)
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% isomorphism groups (which are iso to symmetric group)
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\end{document}
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@ -1,157 +0,0 @@
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\section{Modular Arithmetic}
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I'm sure you're all familiar with modular arithmetic.
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In this section, our goal is to define \textit{equivalence relations},
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\textit{equivalence classes}, and use them to formally define arithmetic in mod $n$.
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\problem{}
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Compute the following:
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\begin{itemize}
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\item $5 + 3 \pmod{4}$
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\item $7 \times 4 \pmod{9}$
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\item $-4 \pmod{5}$
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\item $3^{-1} \pmod{7}$
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\end{itemize}
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\vfill
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\definition{}
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An \textit{equivalence relation} on a set $A$
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is a symbol that makes a statement about two elements of $A$.
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For example, $=$ is an equivalence relation on the set of integers.
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\vspace{2mm}
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An equivalence relation must satisfy the following properties:
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\begin{itemize}
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\item Reflexivity: $x \sim x$ for all $x \in A$
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\item Symmetry: if $x \sim y$, $y \sim x$ for any $x, y \in A$
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\item Transitivity: if $x \sim y$ and $y \sim z$, then $x \sim z$
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\end{itemize}
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\problem{}<abseq>
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Which of the following are equivalence relations on $\mathbb{Z}$?
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\begin{itemize}
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\item $>$
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\item $\leq$
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\item $\Bumpeq$, where $a \Bumpeq b$ if $|a| = |b|$
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\item $\neq$
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\end{itemize}
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\vfill
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\pagebreak
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\problem{}
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Consider the relation $\equiv_n$ on $\mathbb{Z}$, where $a \equiv_n b$ holds iff $a \equiv b \pmod{n}$. \par
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Show that $\equiv_n$ is an equivalence relation.
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\vfill
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\definition{}
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Say we have an equivalence relation $\sim$ on a set $A$. \par
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The \textit{equivalence class} of $x$ is the set of all elements that are $\sim$ to $x$. \par
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Here are a few examples: \par
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\begin{itemize}[itemsep=2mm]
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\item
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The equivalence class of $2$ in $\mathbb{Z}$ under the relation $=$ is $\{2\}$, \par
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since the only $x$ that satisfies $x = 2$ is $2$.
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\item
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The equivalence class of $9$ in $\mathbb{Z}$ under the relation $\Bumpeq$
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from \ref{abseq} is $\{-9, 9\}$.
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\end{itemize}
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\problem{}
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What is the equivalence class of $3$ in $\mathbb{Z}$ under $\equiv_5$? \par
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\hint{Remember that $\mathbb{Z}$ contains both positive and negative numbers.}
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\begin{solution}
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$\{..., -7, -2, 3, 8, 12, ... \}$
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\end{solution}
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\vfill
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\problem{}
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Let $A$ be a set and $\sim$ an equivalence relation. \par
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Show that every element of $A$ is in \textit{exactly one} equivalence class. \par
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\hint{What properties does an equivalence relation satisfy?}
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\vfill
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We now have a proper definition of \say{mod $n$:} \par
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it is the equivalence relation $a \equiv_n b$, which is usually written as $a \equiv b \pmod{n}$. \par
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We will use this definition throughout this handout.
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\note[Note]{
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This is different than the \say{mod} operator $a ~\%~ b $,
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which is defined as the remainder of $a \div b$.
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}
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\pagebreak
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\definition{}
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Given any $x \in \mathbb{Z}$, $[x]_n$ is the equivalence class of $x$ under $\equiv_n$.
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\problem{}
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Compute the following:
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\begin{itemize}[itemsep = 1mm]
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\item $[5]_3 + [4]_3$
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\item $[-2]_7 + [9]_7$
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\end{itemize}
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\vfill
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\problem{}
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Does $[4]_3 + [7]_5$ make sense?
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\vfill
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\problem{}
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Find all $n$ that satisfy
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$[5]_n \times [17]_n = [3]_n + [2]_n$ \par
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\hint{$[a]_n = [b]_n$ iff $n$ divides $a - b$, by definition of mod.}
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\begin{solution}
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$[85] = [12] ~\implies~ n ~|~ 85 - 12 ~\implies~ n ~|~ 73 ~\implies~ n \in \{1, 73\}$
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\end{solution}
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\vfill
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\definition{}
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$\znz{n}$ (pronounced \say{$\mathbb{Z}$ mod $n \mathbb{Z}$}) is the set of equivalence classes of $\equiv_n$ on $\mathbb{Z}$. \par
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For example, $\znz{5} = \{~ [0]_5,~ [1]_5,~ [2]_5,~ [3]_5,~ [4]_5 ~\}$. \par
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\vspace{2mm}
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This notation may seem a bit odd, but don't let it confuse you. \par
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One of our goals today is to understand what exactly $\znz{n}$ means.
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\problem{}
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What is $\znz{6}$?
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\vfill
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\pagebreak
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\section{Groups}
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\definition{}
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A \textit{group} $G = (S, \ast)$ consists of a set $S$ and a binary operator $\ast$. \par
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By definition, a group always has 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 Any $a \in G$ has an \textit{inverse} $a^{-1} \in G$ that satisfies $a \ast a^{-1} = a^{-1} \ast a = e$. \par
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\end{enumerate}
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\note[Note]{
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Commutativity is \textit{not} a required property of a group! \\
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In most cases, $a \ast b \neq b \ast a$.
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}
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\problem{}
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Is $(\znz{5}, +)$ a group? \par
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How about $(\znz{5}, -)$? \par
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\hint{In this problem, $+$ and $-$ work just as you'd expect.}
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\vfill
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\problem{}
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What is the smallest possible group?
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\begin{solution}
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Let $(G, \ast)$ be our group, where $G = \{e\}$ and $\ast$ is defined by the identity $e \ast e = e$
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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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How many distinct groups have two elements? \par
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\hint{
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Two groups are \say{the same} if the elements of one can be renamed to get the other. \\
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A group is fully defined by its multiplication table.
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}
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\vfill
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\pagebreak
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%\problem{}<firstcross>
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%Is $(\znz{17}, \times)$ a group? \par
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%How should we modify $\znz{17}$ to make it one?
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%\problem{}<secondcross>
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%Is $(\znz{6}, \times)$ a group? \par
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%How should we modify $\znz{6}$ to make it one? \par
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%\hint{
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% Be careful, this isn't as easy as \ref{firstcross}. \\
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% Which elements aren't invertible?
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%}
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%\definition{}
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%Building on problems \ref{num:firstcross} and \ref{num:secondcross}, we'll define $(\znz{n})^\times$ as the multiplicative
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%group of integers mod $n$. \par
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%Specifically, $(\znz{n})^\times$ is the set of all integers coprime to $n$. \par
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%\vspace{2mm}
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%For example, $(\znz{6})^\times = \{1, 5\}$ \par
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%and $(\znz{15})^\times = \{1, 2, 4, 7, 8, 11, 13, 14\}$ \par
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%\vspace{2mm}
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%Note that $0$ is the identity in $\znz{n}$ and $1$ is the identity in $(\znz{n})^\times$\hspace{-1.5ex}. \par
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%\note[Note]{
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% Also, notice that we've omitted the operations $+$ and $\times$ in the two groups above. \\
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% These operations are implicitly \say{attached} to $\znz{n}$ and $(\znz{n})^\times$\hspace{-1.5ex}, \\
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% and we rarely write them for the sake of cleaner notation.
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%}
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\vfill
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\definition{}
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Let $G$ be a group, $a$ an element of $G$, and $n \in \mathbb{Z}^+$. \par
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$a^n$ is the defined as $a \ast a \ast ... \ast a$, repeated $n$ times.
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\vspace{1mm}
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Note that this is \textbf{not} \say{normal} exponentiation! \par
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If our group's operator is $+$ (for example, $\znz{5}$), $a^n = a + ... + a$, \par
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which you'll recognize as multipication.
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\vspace{1mm}
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Beware of this odd notation. By convention, we use \say{multiplicative} notation
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when working with groups---so, $a \ast b$ may also be written as $ab$,
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and $a \ast a \ast a$ may be written as $a^3$.
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\vspace{1mm}
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Again, remember that $a^n$ simply means \say{$\ast$ $a$ with itself $n$ times,} \par
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regardless of the specific operator our group uses.
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\problem{}
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Let $a$ be an element of a finite group. \par
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Show that there is a positive integer $n$ so that $a^n = e$. \par
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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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Find the order of 5 in $(\znz{25}, +)$. \par
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%Find the order of 2 in $((\znz{17})^\times, \times)$. \par
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Find the order of 2 in $(\znz{7}, +)$. \par
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\vfill
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\pagebreak
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\definition{}
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Let $G$ be a group. \par
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We say a $g \in G$ is a \textit{generator} of $G$
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if every element in $G$ can be written as some power of $g$.
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\vspace{2mm}
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If $G$ has a generator, we say $G$ is \textit{cyclic.}
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\problem{}
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Find a generator of $\znz{7}$. Then, find a generator of $(\znz{7})^\times$
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\vfill
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\definition{}
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Let $G$ be a group. \par
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The \textit{order} of $G$ is the number of elements in $G$. \par
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We'll write this as $|G|$, using the same notation we use with sets. \par
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\note[Note]{
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Don't confuse the order of an \textbf{element}
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with the order of a \textbf{group}!
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}
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\problem{}
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Let $G$ be a cyclic group, and let $g$ be any generator in $G$. \par
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Show that $\text{ord}(g) = |G|$. \par
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\hint{Contradiction.}
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\vfill
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\pagebreak
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@ -1,23 +0,0 @@
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\section{Subgroups}
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\definition{}
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Let $G$ be a group, and let $H$ be a subset of $G$. \par
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We say $H$ is a \textit{subgroup} of $G$ if $H$ is also a group
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(with the operation $\ast$).
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\definition{}
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Let $S$ be a subset of $G$. \par
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The \textit{group generated by $S$} consists of all elements of $G$ \par
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that may be written as a combination of elements in $S$
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\vspace{2mm}
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We will denote this group as $\langle S \rangle$. \par
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Convince yourself that $\langle g \rangle = G$ if $g$ generates $G$.
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\problem{}
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What is the subgroup generated by $\{7, 8\}$ in $(\znz{15})^\times$? \par
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Is this the whole group?
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\problem{}
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Show that the group generated by $S$ is indeed a group.
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