\documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{enumerate} \usepackage{ifthen} \usepackage{graphicx} \setlength{\unitlength}{0.1in} \setlength{\oddsidemargin}{0.0in} \setlength{\evensidemargin}{0.0in} \setlength{\topmargin}{0.0in} \setlength{\headheight}{0.0in} \setlength{\headsep}{0.0in} \setlength{\topskip}{0.0in} \setlength{\textheight}{8.9in} \setlength{\textwidth}{6.5in} \newcommand{\comp}{ \,{\scriptstyle \stackrel{\circ}{}}\, } \newcommand{\nullset}{\mathrm{O}\!\!\!\!\big/\,} \newcommand{\divides}{\,\Big|\,} \newcommand{\myspace}{\mbox{\parbox[c][0.35in]{0.35in}{\hfill}}} \newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}} \begin{document} \pagestyle{empty} \noindent \parbox{2in}{\bf Math 3110} \hfill {\Large \bf Final Exam} \hfill \parbox{2in}{\bf \hfill December 9$^{th}$, 2009} \vspace{0.3in} \noindent {\large\bf Name:} \underline{\hskip 3.0 truein}\\ Don't merely state answers, prove your statements. {\bf Be sure to show your work!} \noindent {\LARGE \bf The last page is a copy of Cayley tables for $D_4$ and $Q$.} \begin{description} \item[\large 1. (\underline{\hskip 0.35 truein}/12 points)] For each of the following pairs of groups, if the groups are isomorphic, circle $G_1 \cong G_2$ and explain why they are isomorphic. If the groups aren't isomorphic, circle $G_1 \not\cong G_2$ and explain why not. \begin{enumerate}[(a)] \item \qquad $\mathbb{Q} \cong Q$ \qquad\qquad OR \qquad\qquad $\mathbb{Q} \not\cong Q$ \hfill \hfill [$\mathbb{Q}$ = rationals, $Q$ = quaternions] \vspace{2in} \item \qquad $A_4 \cong D_6$ \qquad\qquad OR \qquad\qquad $A_4 \not\cong D_6$ \hfill \hfill [$A_4$ = even permutations in $S_4$] \vspace{2in} \item \qquad $U(5) \cong \mathbb{Z}_4$ \qquad\qquad OR \qquad\qquad $U(5) \not\cong \mathbb{Z}_4$ \end{enumerate} \newpage \item[\large 2. (\underline{\hskip 0.35 truein}/12 points)] Sub-Things \begin{enumerate}[(a)] \item Let $G$ be a group. Recall that $Z(G) = \{ x \in G \,|\, gx=xg \mbox{ for all } g \in G \}$ is the ``center'' of $G$. Prove that $Z(G)$ is a normal subgroup of $G$. [Hint: You may either {\it show it is a subgroup} {\bf and} {\it show it is normal}, or you can find a homomorphism with kernel $Z(G)$.] \vspace{3in} \item Let $S = \{ m+n\sqrt{5} \,|\, m,n \in \mathbb{Z} \}$. Show $S$ is a sub{\bf{}ring} of $\mathbb{R}$. \vspace{3in} \item Let $T = \{ m+n\sqrt{5} \,|\, m,n \in \mathbb{Z} \mbox{ and } m \mbox{ is \bf EVEN}\}$. Show $T$ is a {\bf subgroup} of $\mathbb{R}$ (under addition of course) and then give a concrete counter-example which shows why $T$ is {\bf not} a sub{\bf{}ring} of $\mathbb{R}$. \end{enumerate} \newpage \item[\large 3. (\underline{\hskip 0.35 truein}/13 points)] Calculatin' mod 50. \quad [${\bf 50 = 2 \cdot 5^2}$] \begin{enumerate}[(a)] \item Is $3$ a unit, zero divisor, or neither in $\mathbb{Z}_{50}$? If $3$ is a unit, find its inverse. If $3$ is zero divisor, show this by finding some $0 \not= m \in \mathbb{Z}_{50}$ such that $3m=0$ (mod 50). \vspace{2in} \item Is $20$ a unit, zero divisor, or neither in $\mathbb{Z}_{50}$? If $20$ is a unit, find its inverse. If $20$ is zero divisor, show this by finding some $0 \not= m \in \mathbb{Z}_{50}$ such that $20m=0$ (mod 50). \vspace{2in} \item Find all of the principle ideals of $\mathbb{Z}_{50}$. \vspace{2in} \item How many elements generate $\mathbb{Z}_{50}$?\\ {}[That is: How many $x \in \mathbb{Z}_{50}$ are there such that $(x)=\langle x \rangle = \mathbb{Z}_{50}$?] \end{enumerate} \newpage \item[\large 4. (\underline{\hskip 0.35 truein}/15 points)] The set $I = (4) = \{ 0,4,8,12,16 \}$ is an ideal of $\mathbb{Z}_{20}$. Let $\displaystyle{R = {\mathbb{Z}_{20} \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop I}}$ \begin{enumerate}[(a)] \item Write down all of the distinct cosets of $I$ in $\mathbb{Z}_{20}$. \vspace{1in} \item Finish filling out the following addition and multiplication tables for $R$: \vspace{0.1in} \noindent \mbox{} \hspace{-0.6in} \begin{tabular}{|c||c|c|c|c|} \hline $+$ & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I \\ \hline \hline \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I \\ \hline \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I \\ \hline \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I \\ \hline \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I \\ \hline \end{tabular} \hspace{0.2in} \begin{tabular}{|c||c|c|c|c|} \hline $\times$ & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I \\ \hline \hline \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I \\ \hline \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I \\ \hline \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I \\ \hline \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I & \varspace{0.3in}{0.2in} + I \\ \hline \end{tabular} \vspace{0.1in} \item Fill out the following table of information about $R$. \vspace{0.1in} \noindent \mbox{} \hspace{-0.65in} \begin{tabular}{|c|c|c|} \hline Property & \parbox[c][0.5in]{0.75in}{Is this true\\ about $R$?} & Briefly Explain \\ \hline \hline $R$ has a ``unity'' & & \varspace{0.5in}{3.5in} \\ \hline $R$ is commutative & & \varspace{0.5in}{3.5in} \\ \hline $R$ is an integral domain & & \varspace{0.5in}{3.5in} \\ \hline $R$ is a field & & \varspace{0.5in}{3.5in} \\ \hline \end{tabular} \vspace{0.1in} \item Fill out the following table concerning $R$. {\it Hint:} These tables are longer than they need to be. \vspace{0.1in} \mbox{} \hspace{-0.65in} \begin{tabular}{|c|c|} \hline Zero Divisors & \parbox[c][0.4in]{1.25in}{I am a zero divisor\\ because...} \\ \hline \hline \parbox[c]{1in}{\hfill} & \varspace{0.35in}{1.75in} \\ \hline \myspace & \myspace \\ \hline \myspace & \myspace \\ \hline \myspace & \myspace \\ \hline \end{tabular} \hspace{0.2in} \begin{tabular}{|c|c|} \hline Units & \parbox[c][0.4in]{1.25in}{My multiplicative\\ inverse is...} \\ \hline \hline \parbox[c]{1in}{\hfill} & \varspace{0.35in}{1.75in} \\ \hline \myspace & \myspace \\ \hline \myspace & \myspace \\ \hline \myspace & \myspace \\ \hline \end{tabular} \end{enumerate} \newpage \item[\large 5. (\underline{\hskip 0.35 truein}/12 points)] Homomorphisms. \begin{enumerate}[(a)] \item Show $\varphi : \mathbb{Z}_3 \rightarrow \mathbb{Z}_6$ defined by $\varphi(x)=2x$ is a {\bf well-defined} homomorphism. \vspace{2in} \item Find the kernel and image of $\varphi$. Is $\varphi$ one-to-one? onto? an isomorphism? \vspace{1.5in} \item $\displaystyle{S = \left\{ \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \,\Big|\, a,b \in \mathbb{R} \right\}}$ is a subring of $M_2(\mathbb{R}) = \mathbb{R}^{2 \times 2}$. Define $\psi : S \rightarrow \mathbb{R}$ by $\displaystyle{\psi \left(\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \right) = a}$. Show that $\psi$ is a ring homomorphism. \vspace{2in} \item Find the kernel and image of $\psi$. Is $\psi$ one-to-one? onto? an isomorphism? \end{enumerate} \newpage \item[\large 6. (\underline{\hskip 0.35 truein}/12 points)] Some random proofs. \begin{enumerate}[(a)] \item Let $R$ be a ring with $1$. Let $u$ be a unit of $R$. Show that $u$ cannot be a zero divisor. (This shows that all fields are integral domains.) \vspace{2.5in} \item Let $R$ be a commutative ring with $1$. Suppose $a \in R$ and recall that $(a) = \{ ra \,|\, r \in R \}$ is the principle ideal generated by $a$. Prove that $(a)$ is an ideal of $R$. \vspace{3.5in} \item Let $G$ be a group such that $(xy)^{-1}=x^{-1}y^{-1}$ for all $x,y \in G$. Show that $G$ is abelian. \end{enumerate} \newpage \item[\large 7. (\underline{\hskip 0.35 truein}/12 points)] Just a little harmless permuting. \begin{enumerate}[(a)] \item Let $H = \{ (1),(12) \}$. Find all of the {\bf left and right} cosets of $H$ in $S_3 = \{ (1),(12),(13),(23),(123),(132) \}$. Is $H$ a normal subgroup? \vspace{3in} \item Explain why $A_n$ (the even permutations) is a normal subgroup of $S_n$. [Hint: You may either briefly explain {\it why it is a subgroup} {\bf and} {\it why it is normal}, or you can find a homomorphism with kernel $A_n$.] \vspace{3in} \item Consider the quotient group $\displaystyle{{S_n \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop A_n}}$. It this a cyclic group? Why or why not? \end{enumerate} \newpage \item[\large 8. (\underline{\hskip 0.35 truein}/12 points)] $D_{10}$. $n=10$ really? why oh why? \begin{enumerate}[(a)] \item Let $G$ be a {\bf non-abelian} group of order 20. What are possible orders of elements of $G$? What does ``non-abelian'' rule out and why? \vspace{2in} \item Half of the elements of $D_{10}$ are reflections and half are rotations. In fact, the {\bf rotations} in $D_{10}$ form a {\bf cyclic subgroup}. Without worrying about the cyclic part, {\bf briefly} explain why the rotations do form a subgroup and why the reflections do not. \vspace{2in} \item Use the description of $D_{10}$ given in part (b) to determine the number of elements of each order in $D_{10}$. \vspace{0.1in} {} \hspace*{-0.5in} \begin{tabular}{r} Element order = \\ {} \\ Number of elements with that order = \\ \end{tabular} \end{enumerate} \end{description} \newpage \LARGE \begin{tabular}{|c|||c|c|c|c|c|c|c|c|} \hline $D_4$ & $1$ & $x$ & $x^2$ & $x^3$ & $y$ & $xy$ & $x^2y$ & $x^3y$ \\ \hline \hline \hline $1$ & $1$ & $x$ & $x^2$ & $x^3$ & $y$ & $xy$ & $x^2y$ & $x^3y$ \\ \hline $x$ & $x$ & $x^2 $ & $x^3 $ & $1 $ & $x y$ & $x^2y$ & $x^3y$ & $ y$ \\ \hline $x^2$ & $x^2$ & $x^3 $ & $1 $ & $x $ & $x^2y$ & $x^3y$ & $ y$ & $x y$ \\ \hline $x^3$ & $x^3$ & $1 $ & $x $ & $x^2 $ & $x^3y$ & $ y$ & $x y$ & $x^2y$ \\ \hline $y$ & $y$ & $x^3y$ & $x^2y$ & $x y$ & $1 $ & $x^3 $ & $x^2 $ & $x $ \\ \hline $xy$ & $xy$ & $ y$ & $x^3y$ & $x^2y$ & $x $ & $1 $ & $x^3 $ & $x^2 $ \\ \hline $x^2y$ & $x^2y$ & $x y$ & $ y$ & $x^3y$ & $x^2 $ & $x $ & $1 $ & $x^3 $ \\ \hline $x^3y$ & $x^3y$ & $x^2y$ & $x y$ & $ y$ & $x^3 $ & $x^2 $ & $x $ & $1 $ \\ \hline \end{tabular} \vspace{1in} \begin{tabular}{|r|||r|r|r|r|r|r|r|r|} \hline $Q$ & $1$ & $-1$ & $i$ & $-i$ & $j$ & $-j$ & $k$ & $-k$ \\ \hline \hline \hline $1$ & $1$ & $-1$ & $i$ & $-i$ & $j$ & $-j$ & $k$ & $-k$ \\ \hline $-1$ & $-1$ & $1$ & $-i$ & $i$ & $-j$ & $j$ & $-k$ & $k$ \\ \hline $i$ & $i$ & $-i$ & $-1$ & $1$ & $k$ & $-k$ & $-j$ & $j$ \\ \hline $-i$ & $-i$ & $i$ & $1$ & $-1$ & $-k$ & $k$ & $j$ & $-j$ \\ \hline $j$ & $j$ & $-j$ & $-k$ & $k$ & $-1$ & $1$ & $i$ & $-i$ \\ \hline $-j$ & $-j$ & $j$ & $k$ & $-k$ & $1$ & $-1$ & $-i$ & $i$ \\ \hline $k$ & $k$ & $-k$ & $j$ & $-j$ & $-i$ & $i$ & $-1$ & $1$ \\ \hline $-k$ & $-k$ & $k$ & $-j$ & $j$ & $i$ & $-i$ & $1$ & $-1$ \\ \hline \end{tabular} \end{document}