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\noindent
\parbox{2in}{\bf Math 3110} 
\hfill {\Large \bf Test \#3} \hfill
\parbox{2in}{\bf \hfill November $18^{\mathrm{th}}$, 2011}

\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. (\underline{\hskip 0.35 truein}/25 points)} 3-2-1...Go!

\begin{enumerate}[(a)]
   \item How many elements of order 3 are in $\mathbb{Z}_{9000} \oplus \mathbb{Z}_{3333333}$?

\vspace{1in}

   \item Let $H$ be a subgroup of $G$ where $G$ is abelian. Quickly explain why $H$ is a normal subgroup of $G$.
   
   \vspace{0.5in}
   
  
   \item Consider for example: $\diagquotient{\mathbb{Z}_{12}}{\langle 4 \rangle} \cong \mathbb{Z}_4$. Now let $k\ell=n$ (where $k,\ell,n$ are positive integers) and let $G=\diagquotient{\mathbb{Z}_n}{\langle k \rangle}$. Briefly, explain why the $G$ has order $k$, why $G$ is cyclic, and why $G \cong \mathbb{Z}_k$.

   \vspace{1in}

   \item Let $G$ and $H$ be group. Prove that $\pi:G \oplus H \rightarrow G$ defined by $\pi((g,h))=g$ is a homomorphism which is onto. Find the kernel of $\pi$. Finally, show that $\diagquotient{G \oplus H}{\{e\}\oplus H} \cong G$.

\end{enumerate}

\newpage
\noindent {\large 2. (\underline{\hskip 0.35 truein}/25 points)} Quotients

\begin{enumerate}[(a)]
   \item Given: $K=\{R_0, R_{180} \}$ is a normal subgroup of $D_6$. 
   
\vspace{0.15in}   

   The order of $\diagquotient{D_6}{K}$ is \underline{\hspace*{2in}}.

\vspace{0.15in}   

   The identity of $\diagquotient{D_6}{K}$ is \underline{\hspace*{1in}}.
   \hfill
   $(R_{60}K)^{-1}=$ \underline{\hspace*{2in}}.

\vspace{0.15in}

  The order of $R_{60}K$ in $\diagquotient{D_6}{K}$ is \underline{\hspace*{1in}}. \hfill The size of the set $R_{60}K$ is \underline{\hspace*{1in}}.

\vspace{0.05in}

Scratch work:

\vspace{1.25in}   

  %\item Let $G$ be a group. A group $H$ is called a {\it homomorphic image} of $G$ if there exists a homomorphism $\varphi : G \rightarrow H$ which is onto (i.e. the image of $\varphi$ is $H$). Explain why homomorphic images and quotients of $G$ share the same group properties. Specifically why is it that if $G$ has an abelian quotient of order 123 if and only if $G$ has an abelian homomorphic image of order 123).

%\vspace{1in}

   \item Let $H$ be a (normal) subgroup of $G$ where $G$ is abelian. Prove that $\diagquotient{G}{H}$ is abelian.
   
   \vspace{0.75in}


 \item Consider $\diagquotient{\mathbb{Z}_{12}}{H}$ where $H=\langle 4 \rangle = \{0,4,8\}$. List all of the cosets (and their contents). Then make a Cayley table for this quotient group.
 
 



\end{enumerate}

\newpage
\noindent {\large 3. (\underline{\hskip 0.35 truein}/25 points)} Oh no! I've made a mistake. $2^2 \cdot 5 \not= 50$

\vspace{0.1in}

\noindent
In each of the following situations, explain \underline{why} we know {\bf a mistake has been made}.\\ (Why are these statements wrong?)

\begin{enumerate}[(a)]
    \item We found a homomorphism $\varphi: \mathbb{Z}_{50} \oplus \mathbb{Z}_{10} \rightarrow D_{5}$  which is onto.

\vfill

    \item Let $H = Z(D_6) = \{ R_0, R_{180} \}$ (the center of $D_6$). After some shoddy computations, I've found that $\diagquotient{D_6}{H} \cong \mathbb{Z}_6$. \hfill [{\it Hint:} No computations needed to shoot this down.]

\vfill

   \item I just found a normal subgroup $H \triangleleft S_{4}$ such that $\diagquotient{S_4}{H}$ is an abelian group of order $18$.

\vfill

   \item My friend Herbert found an element $(x,y) \in D_6 \oplus \mathbb{Z}_{10}$ whose order is $60$.

\vspace{0.75in}

\end{enumerate}

\newpage
\noindent {\large 4. (\underline{\hskip 0.35 truein}/25 points)} Finite Abelian Groups

\begin{enumerate}[(a)]
   \item List all of the non-isomorphic abelian groups of order $100=2^25^2$. ({\it Note}: $4 \cdot 25 = 100$ $\leftarrow$ I can multiply!)

\vfill

   \item How many non-isomorphic abelian groups of order 449,878,000 are there?

   {\it Note:} 449,878,000 $=2^45^311^313^2$ and there are 5 non-isomorphic abelian groups of order $2^4=16$. \smiley

\vfill

   \item Are the groups $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12} \oplus \mathbb{Z}_{25}$ and $\mathbb{Z}_{180} \oplus \mathbb{Z}_{30}$ isomorphic? Explain your answer.

\vfill

   \item Is the group $\mathbb{Z}_{15} \oplus \mathbb{Z}_{7} \oplus \mathbb{Z}_{121}$ cyclic? (Note: $121=11^2$) Explain you answer.

\vspace{0.75in}
\end{enumerate}

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