\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}

