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Removed quotient group handout

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52
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)
\end{document}

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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 define \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. \par
\hint{What properties does an equivalence relation satisfy?}
\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 throughout 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

158
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

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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.