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MATH4550 Modern Algebra课程简介

Class information:

• Syllabus

Notes:

• Notes on the dihedral group can be found here.
• Lemma about the identity element and inverses in subgroups.
• Lemma about what cyclic subgroups look like.
• The division algorithm statement and proof
• Classification of homomorphisms from cyclic groups: lemma and main theorem.
• A proof that the set of permutations of a set is a group under composition.



PREREQUISITES 

Homework:

• Homework #1: problems and solutions 
  See this for a better way to do the solutions for 2,3,13
  Typos: The solutions on problem 5 should say “Thus, e = 2 and e = 1.   This is a contradition.”
   
• Homework #2: problems and solutions 
  See this for a better way to do #2,9,10 and for two extra problems to do
   
• Homework #3: problems and solutions.
  Typo: At the end of the proof 5(a) it says “By the lemma, k divides the order of phi(x).”   It should say “By the lemma, the order of phi(x) divides k.”  
   
• Homework #4: problems and solutions.
  Typos: In the solutions for problem 6 I found the homomorphisms from Z_6 to Z_8.   It should have been Z_8 to Z_6 as stated in the problem set.   But my solution is correct for the Z_6 to Z_8 problem.   Try doing Z_8 to Z_6 in the same way.



MATH4550 Modern AlgebraHELP（EXAM HELP， ONLINE TUTOR）

(a) Let $G=\mathbb{Z}_{12}={\overline{0},\overline{1},\ldots,\overline{11}}$ and $H=\langle\overline{4}\rangle={\overline{0},\overline{4},\overline{8}}$. The left cosets of $H$ in $G$ are given by: \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 given by: \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) Let $G=\mathbb{Z}$ and $H=4\mathbb{Z}={\ldots,-8,-4,0,4,8,\ldots}$. The left cosets of $H$ in $G$ are given by: \begin{align*} 0+H &= H = {\ldots,-8,-4,0,4,8,\ldots}\ 1+H &= {\ldots,-7,-3,1,5,9,\ldots}\ 2+H &= {\ldots,-6,-2,2,6,10,\ldots}\ 3+H &= {\ldots,-5,-1,3,7,11,\ldots} \end{align*} The right cosets of $H$ in $G$ are given by: \begin{align*} H+0 &= H = {\ldots,-8,-4,0,4,8,\ldots}\ H+1 &= {\ldots,-7,-3,1,5,9,\ldots}\ H+2 &= {\ldots,-6,-2,2,6,10,\ldots}\ H+3 &= {\ldots,-5,-1,3,7,11,\ldots} \end{align*} The left cosets and the right cosets are equal.

(c) Let $G=S_3={(1),(1,2),(1,3),(2,3),(1,2,3),(1,3,2)}$ and $H=\langle(1,2)\rangle={(1),(1,2)}$. The left cosets of $H$ in $G$ are given by: \begin{align*} (1)+H &= H = {(1),(1

Suppose $aH = bH$. Then for any $h \in H$, there exist $h_1,h_2 \in H$ such that $a h = b h_1$ and $b h = a h_2$. Rearranging these equations gives $h_1 = a^{-1}b h$ and $h_2 = b^{-1}a h$. Since $h_1, h_2 \in H$, it follows that $a^{-1}b h, b^{-1}a h \in H$. Therefore, $H a^{-1}b \subseteq H$ and $H b^{-1}a \subseteq H$.

Similarly, we can show that $b^{-1}a H = H a^{-1}$. Let $x \in b^{-1}a H$, then there exists $h \in H$ such that $x = b^{-1}a h$. But then $h = a^{-1}b x \in H$, so $x \in H a^{-1}b$. Conversely, let $y \in H a^{-1}b$, then there exists $h \in H$ such that $y = h a^{-1}b$. But then $h = b^{-1}a y \in H$, so $y \in b^{-1}a H$. Therefore, $b^{-1}a H = H a^{-1}b$.

Putting it all together, we have $H a^{-1} = b^{-1}(b H a^{-1}) = b^{-1}(a H b^{-1}) = (b^{-1}a) (H b^{-1}) = (a^{-1}b)^{-1} (H b^{-1}) = (b^{-1}a)^{-1} (H b^{-1}) = H b^{-1}$, as desired.

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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