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PMATH336_note.html
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<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>README</title>
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font-weight: bold;
font-style: italic;
padding: 0;
margin: 15px 0 5px; }
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dl dt > :first-child {
margin-top: 0; }
dl dt > :last-child {
margin-bottom: 0; }
dl dd {
margin: 0 0 15px;
padding: 0 15px; }
dl dd > :first-child {
margin-top: 0; }
dl dd > :last-child {
margin-bottom: 0; }
blockquote {
border-left: 4px solid #dddddd;
padding: 0 15px;
color: #777777; }
blockquote > :first-child {
margin-top: 0; }
blockquote > :last-child {
margin-bottom: 0; }
table {
padding: 0;border-collapse: collapse; }
table tr {
border-top: 1px solid #cccccc;
background-color: white;
margin: 0;
padding: 0; }
table tr:nth-child(2n) {
background-color: #f8f8f8; }
table tr th {
font-weight: bold;
border: 1px solid #cccccc;
margin: 0;
padding: 6px 13px; }
table tr td {
border: 1px solid #cccccc;
margin: 0;
padding: 6px 13px; }
table tr th :first-child, table tr td :first-child {
margin-top: 0; }
table tr th :last-child, table tr td :last-child {
margin-bottom: 0; }
img {
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span.frame {
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overflow: hidden; }
span.frame > span {
border: 1px solid #dddddd;
display: block;
float: left;
overflow: hidden;
margin: 13px 0 0;
padding: 7px;
width: auto; }
span.frame span img {
display: block;
float: left; }
span.frame span span {
clear: both;
color: #333333;
display: block;
padding: 5px 0 0; }
span.align-center {
display: block;
overflow: hidden;
clear: both; }
span.align-center > span {
display: block;
overflow: hidden;
margin: 13px auto 0;
text-align: center; }
span.align-center span img {
margin: 0 auto;
text-align: center; }
span.align-right {
display: block;
overflow: hidden;
clear: both; }
span.align-right > span {
display: block;
overflow: hidden;
margin: 13px 0 0;
text-align: right; }
span.align-right span img {
margin: 0;
text-align: right; }
span.float-left {
display: block;
margin-right: 13px;
overflow: hidden;
float: left; }
span.float-left span {
margin: 13px 0 0; }
span.float-right {
display: block;
margin-left: 13px;
overflow: hidden;
float: right; }
span.float-right > span {
display: block;
overflow: hidden;
margin: 13px auto 0;
text-align: right; }
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overflow: auto;
padding: 6px 10px;
border-radius: 3px; }
pre {
background-color: #f8f8f8;
border: 1px solid #cccccc;
font-size: 13px;
line-height: 19px;
overflow: auto;
padding: 6px 10px;
border-radius: 3px; }
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sup {
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* {
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@media print {
table, pre {
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pre {
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</style>
<style type="text/css">
/**
* prism.js default theme for JavaScript, CSS and HTML
* Based on dabblet (http://dabblet.com)
* @author Lea Verou
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code[class*="language-"],
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white-space: pre;
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word-break: normal;
word-wrap: normal;
line-height: 1.5;
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pre[class*="language-"]::-moz-selection, pre[class*="language-"] ::-moz-selection,
code[class*="language-"]::-moz-selection, code[class*="language-"] ::-moz-selection {
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pre[class*="language-"]::selection, pre[class*="language-"] ::selection,
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background: #b3d4fc;
}
@media print {
code[class*="language-"],
pre[class*="language-"] {
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/* Code blocks */
pre[class*="language-"] {
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margin: .5em 0;
overflow: auto;
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:not(pre) > code[class*="language-"],
pre[class*="language-"] {
background: #f5f2f0;
}
/* Inline code */
:not(pre) > code[class*="language-"] {
padding: .1em;
border-radius: .3em;
white-space: normal;
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.token.comment,
.token.prolog,
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color: slategray;
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color: #999;
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opacity: .7;
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<p><strong>Important</strong>: since github markdown doesn't support LaTeX, view the complete note in the following link instead: <a href="https://rawgit.com/oldfatcrab/PMATH336/master/PMATH336_note.html">https://rawgit.com/oldfatcrab/PMATH336/master/PMATH336_note.html</a></p>
<hr>
<h1 id="toc_0">Table of Content</h1>
<ul>
<li><a href="#toc_1">Chapter 0: Course Administration</a> (May 02)</li>
<li><a href="#toc_2">Chapter 1: Groups</a> (May 02)
<ul>
<li><a href="#toc_3">1.1 Definitions and examples</a> (May 02)</li>
<li><a href="#toc_4">1.2 Subgroups</a> (May 06)</li>
<li><a href="#toc_5">1.3 Finite groups</a> (May 13)</li>
<li><a href="#toc_6">1.4 Cyclic groups</a> (May 18)</li>
</ul></li>
<li><a href="#toc_7">Chapter 2: Group homomorphisms</a> (May 30)
<ul>
<li><a href="#toc_8">2.1 Homomorphism and Isomorphism</a> (May 30)</li>
<li><a href="#toc_9">2.2 Permutation Groups</a> (Jun 10)</li>
<li><a href="#toc_10">2.3 Automorphisms</a> (Jun 22)</li>
</ul></li>
<li><a href="#toc_11">Chapter 3: Lagrange Theorem and Group Actions</a> (Jun 27)
<ul>
<li><a href="#toc_12">3.1 Langrange Theorem and Cosets</a> (Jun 27)</li>
<li><a href="#toc_13">3.2 Group Actions</a> (Jul 4)</li>
<li><a href="#toc_14">3.3 Stabilizers and orbits of an action</a> (Jul 6)</li>
</ul></li>
</ul>
<hr>
<p><em>2016/05/02</em></p>
<h1 id="toc_1">Chapter 0: Course Administration</h1>
<ul>
<li>Course structure
<ul>
<li>Group Theory Basics
<ul>
<li>Examples</li>
<li>Axioms</li>
<li>Subgroups and Lagrange's Theorem</li>
<li>Cyclic groups</li>
<li>Permutation groups</li>
<li>Normal subgroups</li>
<li>quotients</li>
<li>homomorphisms</li>
<li>Isomorphism theorems</li>
<li>automorphism groups and conjugation</li>
</ul></li>
<li>Group Actions
<ul>
<li>The orbit-statiliser theorem</li>
<li>Cauchy's theorem</li>
<li>Burnside's lemma</li>
<li>Pòlya enumeration</li>
<li>the class equation</li>
</ul></li>
<li>Other Topics
<ul>
<li>Platonic solids</li>
<li>classification of finite abelian groups</li>
<li>cryptography</li>
<li>application to physics (if time permitting)</li>
</ul></li>
</ul></li>
<li>Grading
<ul>
<li>6 assignments: 30%</li>
<li>Midterm: 20%</li>
<li>Final: 50%</li>
</ul></li>
<li>Office
<ul>
<li>MC 5427</li>
<li>Monday 14:00-16:00, Thursday 15:00-17:00, or by appointment</li>
</ul></li>
</ul>
<hr>
<h1 id="toc_2">Chapter 1: Groups</h1>
<h2 id="toc_3">1.1 Definitions and examples:</h2>
<ul>
<li>Let \(G\) be a non-empty set.
<ul>
<li>Def'n: A binary operation on \(G\) is a map that
<ul>
<li>\(G \times G \implies G\)</li>
<li>\((a,b) \mapsto a\star b\) or \(a\cdot b\) or \(ab\) (notation)</li>
</ul></li>
<li>A binary operation is often called a <strong>product</strong></li>
</ul></li>
<li>Ex 1)
<ul>
<li>\(G=\mathbb{Z}\) and</li>
<li>and \(+: \mathbb{Z}\times \mathbb{Z} \implies \mathbb{Z}\), \((m,n) \mapsto m+n\)</li>
<li>or \(\cdot: \mathbb{Z}\times \mathbb{Z} \implies \mathbb{Z}\), \((a,b) \mapsto ab\)</li>
</ul></li>
<li>Ex 2)
<ul>
<li>\(G=\mathbb{R}^3\)</li>
<li>and \( +: \mathbb{R}^3 \times \mathbb{R}^3 \implies \mathbb{R}^3, ((x_1, y_1, z_1), (x_2, y_2, z_2)) \mapsto (x_1+x_2, y_1+y_2, z_1+z_2)\)</li>
<li>and \(\times: \mathbb{R}^3 \times \mathbb{R}^3 \implies \mathbb{R}^3, ((x_1, y_1, z_1), (x_2, y_2, z_2)) \mapsto (x_1, y_1, z_1) \cdot (x_2, y_2, z_2)\)</li>
<li>but \(\dot: \mathbb{R}^3 \times \mathbb{R}^3 \implies \mathbb{R}, ((x_1, y_1, z_1), (x_2, y_2, z_2)) \mapsto (x_1x_2, y_1y_2, z_1z_2)\) is <strong>not</strong> a binary operation because the target space is not \(G=\mathbb{R}^3\)</li>
</ul></li>
<li>Def'n (<strong>Groups</strong>): Let \(G\) be a non-empty set with a binary operation, \(G\times G\implies G\), then \(G\) is a <strong>group</strong> if the binary operation has the following properties:
<ul>
<li>(i) (<strong>Associativity</strong>): \(a(bc) = (ab)c\), \(\forall a,b,c \in G\)</li>
<li>(ii) (<strong>Identity</strong>): \(\exists e \in G\) such that: \(ea = ae =a, \forall a\) in \(G\)</li>
<li>(iii) (<strong>Inverse</strong>): \(\forall a \in G\), \(\exists a^{-1} \in G\) such that: \(aa^{-1} = a^{-1} a = e\)</li>
</ul></li>
<li>Ex 1)
<ul>
<li>The integers: \((\mathbb{Z},+) \leadsto \) this is a group</li>
<li><strong>HERE</strong>: \(G = \mathbb{Z}\) and \(+: \mathbb{Z}\times\mathbb{Z} \implies \mathbb{Z}, (m,n) \mapsto m+n\)</li>
<li>Does \(+\) satisfy properties (i) to (iii)?
<ul>
<li>(i) Let \(m, n, r \in \mathbb{Z}\), then:
<ul>
<li>\(m+(n+r) = (m+n)+r\) ? YES!</li>
</ul></li>
<li>(ii) Identity of \(+\) is \(0 \in \mathbb{Z}\)
<ul>
<li>because \(0 + m = m + 0 = m, \forall m \in \mathbb{Z}\)</li>
</ul></li>
<li>(iii) \(\forall m \in \mathbb{Z}\), we have that
<ul>
<li>\(m + (-m) = (-m) + m = 0\)</li>
<li>\(\implies -m \in \mathbb{Z}\) is the inverse of \(m \in \mathbb{Z}\)</li>
</ul></li>
<li>Thus, \((\mathbb{Z},+)\) is a group.</li>
</ul></li>
</ul></li>
<li>Ex 2)
<ul>
<li>The integers: \((\mathbb{Z},\cdot)\leadsto\) this is a group</li>
<li><strong>HERE</strong>: \(G = \mathbb{Z}\) and \(\cdot: \mathbb{Z}\times\mathbb{Z} \implies \mathbb{Z}, (m,n) \mapsto m\cdot n\)</li>
<li>Does \(\cdot\) satisfy properties (i) to (iii)?
<ul>
<li>(i) Let \(m, n, r \in \mathbb{Z}\), then:
<ul>
<li>\(m\cdot(n\cdot r) = (m\cdot n)\cdot r\) ? YES!</li>
</ul></li>
<li>(ii) Identity of \(\cdot\) is \(1 \in \mathbb{Z}\)
<ul>
<li>because \(1 \cdot m = m \cdot 1 = m\), \(\forall m \in \mathbb{Z}\)</li>
</ul></li>
<li>(iii) \(\forall m \in \mathbb{Z}\), the inverse of m with respect to multiplication is \(\frac{1}{m}\) (if \(m \neq 0\)), but \(\frac{1}{m} \notin \mathbb{Z}\) if \(m\neq\pm1\). So property (iii) fails!</li>
<li>Thus, \((\mathbb{Z},\cdot)\) is not a group.</li>
</ul></li>
</ul></li>
<li>Ex 3)
<ul>
<li>\(G = \{1, -1\}\) set of 2 elements with the binary operation given by:</li>
</ul></li>
</ul>
<table>
<thead>
<tr>
<th>\(\cdot\)</th>
<th>\(1\)</th>
<th>\(-1\)</th>
</tr>
</thead>
<tbody>
<tr>
<td>\(1\)</td>
<td>\(1\)</td>
<td>\(-1\)</td>
</tr>
<tr>
<td>\(-1\)</td>
<td>\(-1\)</td>
<td>\(1\)</td>
</tr>
</tbody>
</table>
<ul>
<li>(continue)
<ul>
<li>Note that \(\cdot\) is just the usual product in \(\mathbb{R}\) restricted to element in \(G\). Then \(G\) is a group because:
<ul>
<li>(i) \(\cdot\) is associative because multiplication in \(\mathbb{R}\) is associative</li>
<li>(ii) \(1\) is the identity.</li>
<li>(iii) From the table, we see that \(1\) is the inverse of \(1\) and \(-1\) is the inverse of \(-1\leadsto\) every element in \(G\) has an inverse.</li>
</ul></li>
</ul></li>
<li>Ex 4)
<ul>
<li>\((\mathbb{Q}, +)\) is a group (exercise)</li>
</ul></li>
<li>Ex 5)
<ul>
<li>\( (\mathbb{Q}^*, \cdot)\) is a group where \(\mathbb{Q}^* = \mathbb{Q}\setminus\{0\}\)</li>
<li>Indeed:
<ul>
<li>(i) Multiplication in \(\mathbb{Q}^*\) is associative as in \(\mathbb{Z}\) and \(\mathbb{R}\)</li>
<li>(ii) \(1 \in \mathbb{Q}^*\) is the identity with respect to \dot.</li>
<li>(iii) \(\forall \frac{a}{b} \in \mathbb{Q}^*\), we have that \(\frac{b}{a} \in \mathbb{Q}^*\) and \(\frac{a}{b} \cdot \frac{b}{a} = \frac{b}{a}\cdot \frac{a}{b} = 1 \implies \frac{a}{b}^{-1} = \frac{b}{a} \implies\) every element in \(\mathbb{Q}^*\) has an inverse</li>
</ul></li>
<li>Note:
<ul>
<li>\((\mathbb{Q}, \cdot)\) is <strong>not</strong> a group because even though (i) and (ii) hold, property (iii) fails for \(0 \in \mathbb{Q}\)</li>
</ul></li>
</ul></li>
<li>Ex 6)
<ul>
<li>\((\mathbb{R}, +)\) and \((\mathbb{C}, +)\) are groups</li>
</ul></li>
<li>Ex 7)
<ul>
<li>\((\mathbb{R}^*, \cdot)\) and \((\mathbb{C}^*, \cdot)\) are groups (where \(\mathbb{R}^* = \mathbb{R}\setminus\{0\}\) and \(\mathbb{C}^* = \mathbb{C}\setminus\{0\})\)</li>
</ul></li>
<li>Ex 8)
<ul>
<li>\((\{1, i, -1, -1\}, \cdot)\) is a group, where \(i \in \mathbb{C}\) such that \(i^2 = -1\). Here:</li>
</ul></li>
</ul>
<table>
<thead>
<tr>
<th>\(\cdot\)</th>
<th>\(1\)</th>
<th>\(i\)</th>
<th>\(-1\)</th>
<th>\(-i\)</th>
</tr>
</thead>
<tbody>
<tr>
<td>\(1 \)</td>
<td>\( 1 \)</td>
<td>\( i\)</td>
<td>\(-1\)</td>
<td>\(-i\)</td>
</tr>
<tr>
<td>\(i \)</td>
<td>\( i \)</td>
<td>\( -1\)</td>
<td>\(-i\)</td>
<td>\(1 \)</td>
</tr>
<tr>
<td>\(-1 \)</td>
<td>\( -1 \)</td>
<td>\( -i\)</td>
<td>\(1 \)</td>
<td>\(i \)</td>
</tr>
<tr>
<td>\(-i\)</td>
<td>\( -i \)</td>
<td>\( 1 \)</td>
<td>\(i \)</td>
<td>\(1 \)</td>
</tr>
</tbody>
</table>
<p><em>2016/05/04</em></p>
<ul>
<li>Review from last lecture:
<ul>
<li>Def'n (<strong>Groups</strong>): Let \(G\) be a non-empty set with a binary operation, \(G\times G\implies G\), then \(G\) is a <strong>group</strong> if the binary operation has the following properties:
<ul>
<li>(i) (<strong>Associativity</strong>): \(a(bc) = (ab)c\), \(\forall a,b,c \in G\)</li>
<li>(ii) (<strong>Identity</strong>): \(\exists e \in G\) such that: \(ea = ae =a, \forall a\) in \(G\)</li>
<li>(iii) (<strong>Inverse</strong>): \(\forall a \in G\), \(\exists a^{-1} \in G\) such that: \(aa^{-1} = a^{-1} a = e\)</li>
</ul></li>
</ul></li>
<li>Note: Properties (i)-(iii) in the def'n are called the <strong>group axioms</strong></li>
<li>Ex 9) <strong>Integer module n</strong>: \(\mathbb{Z}_n\)
<ul>
<li>In \(\mathbb{Z}_n\), two integer \(r\) and \(s\) are equivalent if \(r = s+an\), for same \(a\) in \(\mathbb{Z}\), this is written as \(r \equiv s\) (mod \(n\))</li>
<li>e.g.:
<ul>
<li>\(\mathbb{Z}_3 = \{\overline{0}, \overline{1}, \overline{2}\}\)</li>
<li>\(\overline{0} \equiv \{0, 3, -3, 6, \cdots\} \equiv \overline{3}\)</li>
<li>\(\overline{1} \equiv \{1, 4, -2, 7, \cdots\} \equiv \overline{4}\)</li>
<li>\(\overline{2} \equiv \{2, 5, -1, 8, \cdots\} \equiv \overline{5}\)</li>
</ul></li>
<li>\(\forall F, F' \in \mathbb{Z}_n\), we set:
<ul>
<li>\(\overline{r} + \overline{r_1} = \overline{r+r_1}\), and</li>
<li>\(\overline{r} \cdot \overline{r_1} = \overline{r \cdot r1}\)</li>
</ul></li>
<li>\((\mathbb{Z}_n, +)\): since \(+\) is associative in \(\mathbb{Z}\), it is also associative with \(\mathbb{Z}_n\). Moreover, \(\overline{0}\) is the identity for \(+\). finally, \(\overline{-r}\) is <strong>inverse</strong> of \(F\) with respect to \(+\). We will denote \(\overline{-r}\) by \(-\overline{r} \implies (\mathbb{Z}_n, +)\) is a group</li>
<li>\((\mathbb{Z}_n, \cdot)\): Again, since \(\cdot\) is associative in \(\mathbb{Z}\), it is also associative in \(\mathbb{Z}\),. And \(\overline{1}\) is the identity for \(\cdot\), however, not every element in \(\mathbb{Z}_n\) has an inverse with respect to \(\cdot\). In fact, recall that \(F \in \mathbb{Z}_n\) has a multiplicative inverse iff \(gcd(r,n) = 1\). I.e. \(\exists \overline s \in \mathbb{Z}_n\) such that \(\overline r \cdot \overline s = \overline s \cdot \overline r = \overline 1\) iff \(gcd(r, n) = 1\)</li>
<li>If an element \(\overline r \in \mathbb{Z}_n\) has an inverse in \(\mathbb{Z}_n\), it is called a <strong>unit</strong>.</li>
<li>Set \(\mathbb{Z}_n^* = \{\text{units in }\mathbb{Z}_n\} = \{F\in \mathbb{Z}_n \mid gcd(r,n)=1\}\). Then, \((\mathbb{Z}_n^*, \cdot)\) is a group
<ul>
<li>E.g. \(\mathbb{Z}_6 = \{\overline 1, \overline 2, \overline 3, \overline 4, \overline 5, \overline 6\}\) and \(\mathbb{Z}_6^* = \{\overline 1, \overline 5\}\)</li>
</ul></li>
<li>The Cayley table for \((\mathbb{Z}_6, \cdot)\) is:</li>
</ul></li>
</ul>
<table>
<thead>
<tr>
<th>\(\cdot\)</th>
<th>\(\overline 1\)</th>
<th>\(\overline 5\)</th>
</tr>
</thead>
<tbody>
<tr>
<td>\(\overline 1\)</td>
<td>\(\overline 1\)</td>
<td>\(\overline 5\)</td>
</tr>
<tr>
<td>\(\overline 5\)</td>
<td>\(\overline 5\)</td>
<td>\(\overline 1\)</td>
</tr>
</tbody>
</table>
<ul>
<li>Def'n: Let \((G, \cdot)\) be a group, the <strong>order of the group</strong> \(G\) is denoted \(\mid G \mid\), is the number of element in the group
<ul>
<li>Remark: If the binary operation is clear, we write \(G\) instead of \((G, \cdot)\)</li>
<li>E.g.:
<ul>
<li>\((\mathbb{Z}, +) \implies \mid\mathbb{Z}\mid = \infty\)</li>
<li>\((\mathbb{Z}_6,+) \implies \mid\mathbb{Z}_6\mid = 6\)</li>
<li>\((\mathbb{Z}_6^*,\cdot) \implies \mid\mathbb{Z}_6^* \mid = 2\)</li>
</ul></li>
</ul></li>
<li>Aside: There is another important set of objets in abstract algebra, which corresponds to rings (with identity):
<ul>
<li>Def'n: A <strong>ring</strong> (with identity) is a non-empty set \(G\) together with <strong>two</strong> binary operations \(+\), \(\cdot\), such that:
<ul>
<li>1) \((G,+)\) is a group (i.e. \(+\) satisfies properties (i)-(iii)</li>
<li>2) \((G,\cdot)\) is such that \(\cdot\) satisfies (i) and (iii) </li>
<li>3) \(+\) and \(\cdot\) satisfy the distributive lays: \((a+b)\cdot c = a\cdot c+b\cdot c, \forall a,b,c \in G\)</li>
</ul></li>
<li>E.g. \((\mathbb{Z}, +, \cdot)\), \((\mathbb{R}, +, \cdot)\), \((\mathbb{C}, +, \cdot)\), \((\mathbb{Q}, +, \cdot)\) , \((\mathbb{Z}_6, +, \cdot)\) etc.</li>
<li>In a ring, the elements that have inverses with respect to multiplication are called <strong>units</strong></li>
</ul></li>
<li>Def'n: A group \((G, \cdot)\) is called <strong>abelian</strong> if the binary operation is <strong>commutative</strong>: \(a \cdot b = b \cdot a, \forall a,b \in G\)
<ul>
<li>E.g. \((\mathbb{Z}, +)\), \((\mathbb{R}, +)\), \((\mathbb{C}, +)\), \((\mathbb{Q}, +)\) , \((\mathbb{Z}_6, +)\) etc.</li>
<li>However, not every group is abelian!
<ul>
<li>E.g. Consider \(G = GL(n, \mathbb{R}) = \{A\in M_{n\times n}(\mathbb{R}) \mid A\text{ is invertible}\} = \{A\in M_{n\times n}(\mathbb{R}) | \det A \neq 0\} \)</li>
</ul></li>
<li>Note that \(\exists\) natural operations on \(n\times n\) matrices: addition and multiplication. Let \(A, B \in M_{n\times n}(\mathbb{R})\). Then:
<ul>
<li>\(A+B,\, A\cdot B \in M_{n\times n}(\mathbb{R})\)</li>
<li>\(\implies +\), \(\cdot\) are binary operations for \(M_{n\times n}(\mathbb{R})\)</li>
<li>However, if \(A, B \in G\), although \(A\cdot B \in G\), may not have \(A+B \in G\) (e.g. \(a\in G\) so that \(0\in G\), but \(A+(-A)=0 \notin G\)) \(\implies\) \(+\) is not a binary operation with for \(G\)</li>
<li>Consider \((G, \cdot)\). Then it is a group (exercise) where the identity is the nxn identity matrix \(I_{n\times n}\). But, \((G, \cdot)\) is <strong>not</strong> abelian because \(AB \neq BA\) for most \(A,B \in G\)</li>
</ul></li>
<li>A group \((G, \cdot)\) is called <strong>non-abelian</strong> if it is not abelian.</li>
</ul></li>
<li><strong>Some basic properties</strong>:
<ul>
<li>1) The identity element \(e\) in a group \((G, \cdot)\) is unique</li>
<li>2) The inverse of an element in a group \((G, \cdot)\) is unique</li>
</ul></li>
</ul>
<p><em>2016/05/06</em></p>
<ul>
<li>Review:
<ul>
<li>Def'n (<strong>Groups</strong>): Let \(G\) be a non-empty set with a binary operation, \(G\times G\implies G\), then \(G\) is a <strong>group</strong> if the binary operation has the following properties:
<ul>
<li>(i) (<strong>Associativity</strong>): \(a(bc) = (ab)c\), \(\forall a,b,c \in G\)</li>
<li>(ii) (<strong>Identity</strong>): \(\exists e \in G\) such that: \(ea = ae =a, \forall a\) in \(G\)</li>
<li>(iii) (<strong>Inverse</strong>): \(\forall a \in G\), \(\exists a^{-1} \in G\) such that: \(aa^{-1} = a^{-1} a = e\)</li>
</ul></li>
<li><strong>Some basic properties</strong>:
<ul>
<li>1) (<strong>Uniqueness of identity</strong>) The identity element \(e\) in a group \((G, \cdot)\) is unique</li>
<li>2) (<strong>Uniqueness of inverses</strong>) The inverse of an element in a group \((G, \cdot)\) is unique</li>
</ul></li>
</ul></li>
<li>Proof of above identities:
<ul>
<li>1) (Uniqueness of identity)
<ul>
<li>Suppose \(\exists e, f \in G\) such that \(ae = ea = a\) and \(af = fa = a, \forall a \in G\).</li>
<li>In particular, since \(f\in G\), we have that \(f = fe\) because \(e\) is an identity</li>
<li>Similarly, because \(e\in G\) and \(f\) is an identity, \(fe = e\)</li>
<li>\(\implies f=fe=e\)</li>
</ul></li>
<li>2) (Uniqueness of inverses)
<ul>
<li>Let \(a\in G\). Suppose that a has <strong>two</strong> inverses in \(G\), say \(a^{-1}\) and \(b\). Therefore, by property (iii), we have: \(aa^{-1} = a^{-1} a = e\) and \(ab = ba = e\)</li>
<li>Then \(a^{-1} = a^{-1}e = a^{-1}(ab) = (a^{-1}a)b = eb = b\)</li>
<li>\(\implies a^{-1}=b\)</li>
</ul></li>
</ul></li>
<li>Using these properties, we obtain:
<ul>
<li>3) \(e{-1}=e\)</li>
<li>4) \((ab)^{-1}=b^{-1}a^{-1},\forall a,b \in G\)</li>
<li>5) \((a^{-1})^{-1}=a,\forall a \in G\)</li>
<li>6) \((a_1a_2\cdots a_n)^{-1} = a_n^{-1}\cdots a_2^{-1}a_1^{-1}\)</li>
<li>7) (Cancellation Property) Let \(a,b,c\in G\) with \(G\) a group. If ab = ac, then b=c. If ba=ca, then b = c.</li>
</ul></li>
<li>Proof:
<ul>
<li>3)
<ul>
<li>By uniqueness of the inverse, it is enough to check that \(ee = ee = e\) to proof that \(e^{-1}=e\). But this is true by property (ii).</li>
</ul></li>
<li>4)
<ul>
<li>By uniqueness of the inverse, it is enough to check that \(ab(b^{-1}a^{-1}) = (b^{-1}a^{-1})ab = e\)</li>
<li>By associatitivy, we have that \(ab(b^{-1}a^{-1} = a(bb^{-1})a^{-1} = aea^{-1} = (ae)a^{-1} = aa^{-1} = e\)</li>
<li>Similarly, \((b^{-1}a^{-1})ab=b^{-1}(a^{-1}a)b=b^{-1}eb=(b^{-1}e)b=b^{-1}b=e\)</li>
<li>\(\implies (ab)^{-1}=b^{-1}a^{-1}\)</li>
<li>Note:
<ul>
<li>\(GL(n,\mathbb{R})=\) all general linear group </li>
<li>One would be tempted to think that \((ab)^{-1}=a^{-1}b^{-1}\) but this is false in general.</li>
<li>Example:
<ul>
<li>\((G,\cdot) = (GL(2,\mathbb{R}),\cdot)\)</li>
<li>A = \(\begin{pmatrix}1&2\\0&1\end{pmatrix}\) and B = \(\begin{pmatrix}2&0\\0&3\end{pmatrix}\)</li>
<li>\(\implies AB = \begin{pmatrix}2&6\\0&3\end{pmatrix}\) and \((AB)^{-1}=\begin{pmatrix}\frac{1}{2}&-1\\ 0&\frac{1}{3} \end{pmatrix}\)</li>
<li>But, \(A^{-1}B^{-1} = \begin{pmatrix}\frac 1 2&-\frac 2 3\\0&\frac 1 3\end{pmatrix} \neq (AB)^{-1}\)</li>
<li>Nonetheless, \(B^{-1}A^{-1} = \begin{pmatrix}\frac{1}{2}&-1\\ 0&\frac{1}{3} \end{pmatrix} = (AB)^{-1}\)</li>
</ul></li>
<li>In fact, one can show that: \((ab)^{-1}=a^{-1}b^{-1},\forall a,b,\in G \Leftrightarrow G\) is abelian (Exercise).</li>
</ul></li>
</ul></li>
<li>5) and 6)
<ul>
<li>Exercise in assignment 1</li>
</ul></li>
<li>7)
<ul>
<li>Suppose that \(ab = ac\). Then, since \(G\) is a group, \(\exists a^{-1}\in G\) such that \(aa^{-1}=a^{-1}a=e\)</li>
<li>Therefore \(a^{-1}(ab)=(a^{-1}a)b=eb=b\)</li>
<li>Similarly, \(a^{-1}(ac)=c\).</li>
<li>But, \(ab=ac\), so that \(a^{-1}(ab)=a^{-1}(ac)=c\)</li>
<li>The proof of 2nd cancellation property is similar</li>
</ul></li>
</ul></li>
</ul>
<h2 id="toc_4">1.2 Subgroups</h2>
<ul>
<li>Def'n: Let \(G\) e a group. If a subset \(H\) of \(G\) is itself a group under the operation of \(G\), we say that H is a <strong>subgroup</strong> of \(G\).</li>
<li>Note: the definition of a subgroup implies that subset it <strong>is closed under the operation in G</strong>. I.e. \(\forall a,b\in H\), then \(ab\in H\). This ensures that restriction of the binary operation \(G\times G\implies G\) is again a binary operation taking \(H\times H\implies H\). However, a subset of a group of \(G\) may not be closed under the operation on \(G\).
<ul>
<li>E.g. \((M_{n\times n}(\mathbb{R}), +)\) is a group and \(GL(n,\mathbb R)=\{A\in M_{n\times n}(\mathbb R) \mid \det A\ne 0\} \subset M_{nxn}(\mathbb R)\), but \(GL(n, \mathbb R)\) is not closed under \(+\) (e.g. \(\forall A \in GL(n,\mathbb R), -A \in GL(n,\mathbb R)\)) but \(A+(-A) \notin GL(n,\mathbb R)\)</li>
</ul></li>
<li>We use the notation \(H \vartriangleleft G\) (or \(H<G\)) to mean that H is a subgroup of \(G\). (some authors use \(H\trianglelefteq G\) and \(H\vartriangleleft G\) indicates that \(H \leq G\)). If \(H \leq G\), then H is called a <strong>proper subgroup of G</strong>. The singleton \(\{e\}\) is a subgroup of \(G\) called the <strong>trivial subgroup</strong>; a subgroup of \(G\) that is not \(\{e\}\) is called a <strong>non-trivial subtroup of G</strong>.</li>
<li>How does one determine whether or not a subset \(H\) of \(G\) is a subgroup? There are 2 tests:
<ul>
<li>Theorem (<strong>One-step subgroup test</strong>): Let \(H\) be a non-empty subset of a group \(G\). If \(ab^{-1}\in H,\forall a,b\in H\), then \(H\vartriangleleft G\).</li>
</ul></li>
</ul>
<p><em>2016/05/09</em></p>
<ul>
<li><p>Review from previous lectures:</p>
<ul>
<li>\((G,\cdot)\) is a group if:
<ul>