From e3baa4f86ed5f7875e31063957b1d881999d84d7 Mon Sep 17 00:00:00 2001 From: mark Date: Sat, 18 Nov 2023 19:17:23 -0800 Subject: [PATCH] Started QG handout --- Advanced/Quotient Groups/main.tex | 60 +++++++ Advanced/Quotient Groups/parts/0 mod.tex | 162 ++++++++++++++++++ Advanced/Quotient Groups/parts/1 groups.tex | 158 +++++++++++++++++ .../Quotient Groups/parts/2 subgroups.tex | 23 +++ 4 files changed, 403 insertions(+) create mode 100755 Advanced/Quotient Groups/main.tex create mode 100644 Advanced/Quotient Groups/parts/0 mod.tex create mode 100644 Advanced/Quotient Groups/parts/1 groups.tex create mode 100644 Advanced/Quotient Groups/parts/2 subgroups.tex diff --git a/Advanced/Quotient Groups/main.tex b/Advanced/Quotient Groups/main.tex new file mode 100755 index 0000000..7cebf99 --- /dev/null +++ b/Advanced/Quotient Groups/main.tex @@ -0,0 +1,60 @@ +% 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) + + + % Another handout: + % + % symmetric group, number of permutations, + % cycle notation, type and sign, + % proofs about generators, alternating group + % alternating group generators, fun problems. + +\end{document} diff --git a/Advanced/Quotient Groups/parts/0 mod.tex b/Advanced/Quotient Groups/parts/0 mod.tex new file mode 100644 index 0000000..6eb2d08 --- /dev/null +++ b/Advanced/Quotient Groups/parts/0 mod.tex @@ -0,0 +1,162 @@ +\section{Modular Arithmetic} + +I'm sure you're all familiar with modular arithmetic. +In this section, our goal is to meet \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{} +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\footnotemark{}\hspace{-1ex}. \par +\hint{What properties does an equivalence relation satisfy?} + +\footnotetext{ + We could also say \say{$A$ is partitioned by $[A ~/ \sim]$} + or \say{$A$ is the disjoint union of $[A ~/ \sim]$,} \par + where $[A ~/ \sim]$ is the set of equivalence classes of $\sim$. +} + +\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 thoughout 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 \ No newline at end of file diff --git a/Advanced/Quotient Groups/parts/1 groups.tex b/Advanced/Quotient Groups/parts/1 groups.tex new file mode 100644 index 0000000..3df0aaa --- /dev/null +++ b/Advanced/Quotient Groups/parts/1 groups.tex @@ -0,0 +1,158 @@ +\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{} +%Is $(\znz{17}, \times)$ a group? \par +%How should we modify $\znz{17}$ to make it one? + +%\problem{} +%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 \ No newline at end of file diff --git a/Advanced/Quotient Groups/parts/2 subgroups.tex b/Advanced/Quotient Groups/parts/2 subgroups.tex new file mode 100644 index 0000000..ff6d5e2 --- /dev/null +++ b/Advanced/Quotient Groups/parts/2 subgroups.tex @@ -0,0 +1,23 @@ +\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. \ No newline at end of file