\documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{enumerate} \usepackage{ifthen} \usepackage{graphicx} \usepackage{wasysym} \setlength{\unitlength}{0.1in} \setlength{\oddsidemargin}{-0.50in} \setlength{\evensidemargin}{-0.50in} \setlength{\topmargin}{-0.50in} \setlength{\headheight}{0.0in} \setlength{\headsep}{0.0in} \setlength{\topskip}{0.0in} \setlength{\textheight}{9.4in} \setlength{\textwidth}{7.5in} \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}}} \newcommand{\diagquotient}[2]{\displaystyle{{#1 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop #2}}} \newcommand{\force}{{$\mbox{\;}$}} \begin{document} \pagestyle{empty} \noindent \parbox{2in}{\bf Math 3110} \hfill {\Large \bf Test \#3} \hfill \parbox{2in}{\bf \hfill April $4^{\mathrm{th}}$, 2022} \vspace{0.3in} \noindent {\large\bf Name:} \underline{\hskip 3.0 truein} \hfill {\bf Be sure to show your work!} \vspace{0.2in} \noindent {\large 1. (15 points)} Getting things in order\dots \begin{enumerate}[(a)] \item Let $G = \mathbb{Z}_{10} \times S_3 \times Q$ where $S_3$ is the group of permutations on 3 elements and $Q = \{ \pm 1, \pm i, \pm j, \pm k\}$ is the quaternion group. \vspace{0.1in} The order of $G$ is $|G|=$ \underline{\hspace*{1in}} \vspace{0.1in} What is the largest element order in $\mathbb{Z}_{10} \times S_3 \times Q$? {\bf Give an example} of such an element. \vspace{1.5in} \item Let $H$ and $K$ be subgroups of a group $G$. In addition suppose that $|H|=15$ and $|K|=8$. Explain why $H \cap K$ must be the trivial subgroup. \vspace{1in} \end{enumerate} \noindent {\large 2. (15 points)} Let $\varphi:G \to H$ be a homomorphism, and let $G$, $H$, and $K$ denote groups.\\ \force \qquad \qquad $\longrightarrow$ State the definition a homomorphism.\\ \force \qquad \qquad $\longrightarrow$ State the definition of the kernel: $\mathrm{ker}(\varphi)$.\\ \force \qquad \qquad $\longrightarrow$ Then prove $\pi:K \times H \to H$ defined by $\pi(k,h)=h$ is a homomorphism and determine its kernel ($= ?? \times ??$).\\ \force \qquad \qquad $\longrightarrow$ [Bonus question I meant to ask:] Notice $\pi$ is onto. What does the first isomorphism theorem say? \vfill \newpage \noindent {\large 3. (15 points)} Consider $H = \{ (1), (12)(34), (13)(24), (14)(23) \}$ in $S_4$ be the group of permutations on 4 elements. {\it Note:} It can be shown that $H$ is a normal subgroup of $S_4$ (just accept this for now). \begin{enumerate}[(a)] \item Quick questions about $\diagquotient{S_4}{H}$. \qquad \qquad The order of $\diagquotient{S_4}{H}$ is \underline{\hspace*{0.5in}}. \vspace{0.15in} \hfill {\large True \ / \ False:\ } \quad $(123)H=H(123)$ \hfill \hfill {\large True \ / \ False:\ } \quad $(1234)H=(12)(34)H$ \hfill \force \vspace{0.25in} The identity of $\diagquotient{S_4}{H}$ is \underline{\hspace*{0.6in}}. \hfill $(12)H = \Big\{$ \underline{\hspace*{3.25in}} $\Big\}$\\ \force \hfill [list the contents] \qquad \force \vspace{0.15in} The order of $(1234)H$ in $\diagquotient{S_4}{H}$ is \underline{\hspace*{1in}}. \hfill $((123)H)^{-1}=$ \underline{\hspace*{2in}}.\\ \vspace{0.05in} Scratch work: \vspace{1.75in} \item Let $H = \{ 1, x^2, y, x^2y\}$ and note that $H$ is a subgroup of $D_4 = \langle x,y \;|\; x^4=1, y^2=1, (xy)^2=1 \rangle = \{1,x,x^2,x^3,y,xy,x^2y,x^3y\}$. Is this a normal subgroup? Justify your answer. \vspace{1in} \end{enumerate} \noindent {\large 4. (10 points)} Consider $\diagquotient{\mathbb{Z}_{24}}{H}$ where $H=\langle 4 \rangle = \{0,4,8,12,16,20\}$.\\ \force \qquad \qquad $\longrightarrow$ List all of the cosets of $H$ (and their contents) in $\mathbb{Z}_{24}$.\\ \force \qquad \qquad $\longrightarrow$ Then make a Cayley table for this quotient group. \vfill What is the order of $3+H$ in $\mathbb{Z}_{24}/H$? \newpage \noindent {\large 5. (15 points)} Not quite proofs. \begin{enumerate}[(a)] \item Let $H \triangleleft \mathbb{Z}_{10} \times \mathbb{Z}_{6}$. Why is $\diagquotient{\mathbb{Z}_{10} \times \mathbb{Z}_{6}}{H} \cong A_4$ impossible? \vfill \item Let $\varphi:D_4 \to \mathbb{Z}_{12}$ be a homomorphism. Why can't $\varphi$ be one-to-one? \vfill \item The center of $D_6 = \langle x,y \;|\; x^6=1,y^2=1,(xy)^2=1\rangle = \{1,x,\dots,x^5,y,xy,\dots,x^5y\}$ is $Z(D_6)=\{1,x^3\}$. Suppose a classmate thinks they found an element of order 6 in $\diagquotient{D_6}{Z(D_6)}$. How do we know they are wrong? \vfill \end{enumerate} \noindent {\large 6 (10 points)} Let $H$ and $K$ be normal subgroups of $G$. Prove that $H \cap K$ is a normal subgroup of $G$.\\ \force \hfill {\it Note:} Show {\bf both} $H \cap K$ is a subgroup {\bf and} that it's normal in $G$. \vfill \newpage \noindent {\large 7. (20 points)} Finite Abelian Groups \begin{enumerate}[(a)] \item List all of the non-isomorphic Abelian groups of order $200 =2^3 5^2$. Circle any that are cyclic. \vfill \item Suppose that $G$ is an Abelian group of order $200$ and we know it has an element of order 8 but $G$ is not cyclic. Using part (a), which group(s) could $G$ be isomorphic to? \vfill \item How many non-isomorphic Abelian groups of order 793,800 are there? {\it Note:} 793,800 $=2^3 \cdot 3^4 \cdot 5^2 \cdot 7^2$ and there are 5 non-isomorphic Abelian groups of order 81 $=3^4$. \smiley \vfill \item What is the largest order among elements of $\mathbb{Z}_{10} \times \mathbb{Z}_{4} \times \mathbb{Z}_{6}$? Explain you answer. \vspace{0.75in} \end{enumerate} \end{document}