\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][0.4in]{0.4in}{\hfill}}} \begin{document} \pagestyle{empty} \noindent \parbox{2in}{\bf Math 3110} \hfill {\Large \bf Test \#3} \hfill \parbox{2in}{\bf \hfill November 20$^{th}$, 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}/16 points)] The following groups are {\bf NOT} isomorphic. Explain why. \begin{enumerate}[(a)] \item $D_5 \not\cong \mathbb{Z}_{10}$ \vspace{1in} \item $D_4 \not\cong Q$ where $Q = \{ \pm 1, \pm i, \pm j, \pm k \}$ (the quaternions). \vspace{1.5in} \end{enumerate} \item[\large 2. (\underline{\hskip 0.35 truein}/14 points)] Is it possible? \begin{enumerate}[(a)] \item Let $G_1$ and $G_2$ be groups of order 7. Is it possible that $G_1 \not\cong G_2$? If so, give examples of non-isomorphic groups of order 7. If $G_1$ is always isomorphic to $G_2$, explain why. \vspace{2in} \item Let $G_1$ and $G_2$ be groups of order 6. Is it possible that $G_1 \not\cong G_2$? If so, give examples of non-isomorphic groups of order 6. If $G_1$ is always isomorphic to $G_2$, explain why. \end{enumerate} \newpage \item[\large 3. (\underline{\hskip 0.35 truein}/18 points)] Homomorphisms \begin{enumerate}[(a)] \item Quickly, explain why $\psi : \mathbb{Z}_3 \rightarrow \mathbb{Z}_7$ defined by $\psi(x)=x$ is {\bf NOT} a well-defined function. \vspace{1in} \item Let $\varphi : \mathbb{Z}_8 \rightarrow \mathbb{Z}_{12}$ be defined by $\varphi(x)=6x$. Show that $\varphi$ is a well-defined homomorphism\\ (i.e. show that $\varphi$ is well-defined {\it and} $\varphi$ is a homomorphism). \vspace{3in} \item Using the homomorphism $\varphi$ from part (b), find $\mbox{Ker}(\varphi)$ and $\varphi(\mathbb{Z}_8)$. \vspace{0.2in} Is $\varphi$ one-to-one? \underline{\hskip 0.5 truein} \qquad Is $\varphi$ onto? \underline{\hskip 0.5 truein} \qquad Is $\varphi$ an isomorphism? \underline{\hskip 0.5 truein} \end{enumerate} \newpage \item[\large 4. (\underline{\hskip 0.35 truein}/16 points)] Write down the left multiplication operators for $U(8) = \{ 1,5,3,7 \}$ then translate them into permuations (in $S_4$) labeling the elements of $U(8)$ as follows: \[ 1 \rightsquigarrow 1, \qquad 5 \rightsquigarrow 2, \qquad 3 \rightsquigarrow 3, \qquad \mbox{and} \qquad 7 \rightsquigarrow 4. \] Finally, write down a subgroup of $S_4$ which isomorphic to $U(8)$ (using Cayley's theorem). \newpage \item[\large 5. (\underline{\hskip 0.35 truein}/18 points)] Quotient Groups \begin{enumerate}[(a)] \item Let $H = \{0,4,8\} \lhd \mathbb{Z}_{12}$. Write out all of the cosets of $H$. \\ Then fill out a Cayley table for the quotient $\displaystyle{{\mathbb{Z}_{12} \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop H}}$. \vspace{4in} \item Consider $K = \{ 1, x^2 \}$ in $D_4$. Then $xK \in \displaystyle{{D_4 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop K}}$.\\ Compute $(xK)^2 = xK \, xK$. What is the order of $xK$ (as an element of this quotient group)? \end{enumerate} \newpage \item[\large 6. (\underline{\hskip 0.35 truein}/18 points)] Other stuff. \begin{enumerate}[(a)] \item Let $H$ be a subgroup of $G$ and $K$ be a subgroup of $H$ ($K \stackrel{\subseteq\,}{\mbox{\tiny s.g.}} H \stackrel{\subseteq\,}{\mbox{\tiny s.g.}} G$). \vspace{0.1in} If $|G|=18$, then what are the possible orders of $H$? \underline{\hskip 2.5 truein} \vspace{0.05in} Suppose that $H \not= G$ and $H$ has more than 7 elements. Also, $K$ contains more than just the identity. What are the possible orders of $K$?\\ {}[Hint: determine the order of $H$ first, then remember that $K$ is a subgroup of $H$.] \vspace{3in} \item Let $G$ be an {\bf Abelian} group and $H \stackrel{\subseteq\,}{\mbox{\tiny s.g.}} G$. Give a quick explanation why $H$ is {\bf normal} in $G$ and then prove that $\displaystyle{{G \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop H}}$ is Abelian. \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}