## MATH067 Modern Algebra课程简介

This course is an introduction to abstract algebra and will survey basic algebraic systems-groups, rings, and fields. Although these concepts will be illustrated by concrete examples, the emphasis will be on abstract theorems, proofs, and rigorous mathematical reasoning. There is a strong emphasis on good mathematical writing, especially on mathematical proofs. This course includes a required additional weekly problem session.

## PREREQUISITES

That sounds like a challenging and rewarding course! Abstract algebra is a fascinating branch of mathematics that studies algebraic structures such as groups, rings, and fields, and their properties. It is a highly abstract and theoretical subject, but also has many important applications in areas such as cryptography, coding theory, and computer science.

In this course, you can expect to learn about the fundamental concepts and theorems of abstract algebra, and to develop your skills in mathematical reasoning and proof-writing. The additional weekly problem sessions will give you the opportunity to practice applying the concepts you’ve learned to solve challenging problems.

To succeed in this course, it will be important to have a solid foundation in algebra, including topics such as functions, equations, and matrices. You’ll also need to be comfortable with abstract reasoning and able to think creatively about mathematical problems.

Overall, this course promises to be a challenging but rewarding journey into the fascinating world of abstract algebra!

## MATH067 Modern Algebra HELP（EXAM HELP， ONLINE TUTOR）

1. For the following groups $G$ and subgroups $H$ compute the left cosets and the right cosets. Are they equal?
(a) $G=\mathbb{Z}_{12}$ and $H=\langle\overline{4}\rangle$.
(b) $G=\mathbb{Z}$ and $H=4 \mathbb{Z}$
(c) $G=S_3$ and $H=\langle(1,2)\rangle$.
(d) $G=S_3$ and $H=\langle(1,2,3)\rangle$.
(e) $G=D_8$ and $H=\langle r\rangle$.
(f) $G=D_8$ and $H=\langle s\rangle$.

(a) $G=\mathbb{Z}_{12}$ and $H=\langle\overline{4}\rangle$.

The left cosets of $H$ in $G$ are: \begin{align*} \overline{0}+H &= {\overline{0},\overline{4},\overline{8}}\ \overline{1}+H &= {\overline{1},\overline{5},\overline{9}}\ \overline{2}+H &= {\overline{2},\overline{6},\overline{10}}\ \overline{3}+H &= {\overline{3},\overline{7},\overline{11}} \end{align*} The right cosets of $H$ in $G$ are: \begin{align*} H+\overline{0} &= {\overline{0},\overline{4},\overline{8}}\ H+\overline{1} &= {\overline{1},\overline{5},\overline{9}}\ H+\overline{2} &= {\overline{2},\overline{6},\overline{10}}\ H+\overline{3} &= {\overline{3},\overline{7},\overline{11}} \end{align*} The left cosets and the right cosets are equal.

(b) $G=\mathbb{Z}$ and $H=4 \mathbb{Z}$.

The left cosets of $H$ in $G$ are: \begin{align*} 0+H &= {0,4,-4,8,-8,\ldots}\ 1+H &= {1,5,-3,9,-7,\ldots}\ 2+H &= {2,6,-2,10,-6,\ldots}\ 3+H &= {3,7,-1,11,-5,\ldots} \end{align*} The right cosets of $H$ in $G$ are: \begin{align*} H+0 &= {0,4,-4,8,-8,\ldots}\ H+1 &= {1,5,-3,9,-7,\ldots}\ H+2 &= {2,6,-2,10,-6,\ldots}\ H+3 &= {3,7,-1,11,-5,\ldots} \end{align*} The left cosets and the right cosets are equal.

(c) $G=S_3$ and $H=\langle(1,2)\rangle$.

The left cosets of $H$ in $G$ are: \begin{align*} {(),(1,2)}+H &= {(),(1,2)}+\langle(1,2)\rangle = {(),(1,2)}\ {(1,3),(2,3),(1,3)(2,3)}+H &= {(1,3),(2,3),(1,3)(2,3)}+\langle(1,2)\rangle = {(1,3),(2,3),(1,3)(2,3)} \end{align*} The right cosets of $H$ in $G$ are: \begin{align*} H+{(),(1,2)} &= {\langle(

Let $G$ be a group and $H$ be a subgroup of $G$. Let $a, b \in G$. Prove that if $a H=b H$, then $H a^{-1}=H b^{-1}$.

Suppose that $aH=bH$. Then there exists $h\in H$ such that $b=ah$. It follows that $b^{-1}=h^{-1}a^{-1}$.

Let $x\in H a^{-1}$. Then $x=ha^{-1}$ for some $h\in H$. We have

x=b^{-1}(bh^{-1}a^{-1})=b^{-1}(h^{-1}ah)a^{-1}=b^{-1}h^{-1}(ah)a^{-1}=b^{-1}h^{-1}h=a^{-1}h,x=b−1(bh−1a−1)=b−1(h−1ah)a−1=b−1h−1(ah)a−1=b−1h−1h=a−1h,

which shows that $x\in Hb^{-1}$.

Conversely, let $x\in Hb^{-1}$. Then $x=h’b^{-1}$ for some $h’\in H$. We have

x=(h’h^{-1})h^{-1}b^{-1}=a^{-1}(ah’h^{-1})h^{-1}b^{-1}.x=(h′h−1)h−1b−1=a−1(ah′h−1)h−1b−1.

Since $h’h^{-1}\in H$ and $ah’h^{-1}\in aH=bH$, there exists $h”\in H$ such that $ah’h^{-1}=bh”$. We have

x=a^{-1}bh”h^{-1}b^{-1}\in Ha^{-1},x=a−1bh′′h−1b−1∈Ha−1,

which shows that $x\in Ha^{-1}$.

Therefore, we have shown that $x\in Ha^{-1}$ if and only if $x\in Hb^{-1}$, which implies that $Ha^{-1}=Hb^{-1}$.

## Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

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