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\noindent
\parbox{2in}{\bf Math 3110} 
\hfill {\Large \bf Test \#3} \hfill
\parbox{2in}{\bf \hfill November $2^{\mathrm{nd}}$, 2020}

\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 $G = D_5 \times S_4$ where $D_5 = \langle x,y \;|\; x^5=1,y^2=1,xyxy=1 \rangle$ and $S_4$ is permutations on 4 things. 

\vspace{0.1in}

The order of $G$ is $|G|=$ \underline{\hspace*{1in}}

\vspace{0.1in}

What is the largest element order in $D_5 \times S_4$? Give an example of such an element and explain why it has the largest possible order.

\vspace{1in}

\item Let $G$ be a group of order 36 with subgroup $H$ of order 12. Let $K$ be some other subgroup of $G$.

\vspace{0.25in}

\noindent
Can $H \cap K$ have order 9? Why or why not? 


\vspace{0.5in}

\noindent
If $H \cap K$ has order 6, what are the possible orders of $K$?

\vspace{0.5in}

\end{enumerate}


\noindent {\large 2. (10 points)} Let $H$ and $K$ be normal subgroups of $G$. Prove that $H \cap K$ is a normal subgroup of $G$.


\vfill

\newpage

\newpage
\noindent {\large 3. (15 points)} Consider $H = \{ 1,x^2 \}$ in $D_4 = \langle x,y \;|\; x^4=1,y^2=1,xyxy=1 \rangle = \{ 1,x,x^2,x^3, y,xy,x^2y, x^3y\}$. Note that $H$ is a normal subgroup of $D_4$.

\begin{enumerate}[(a)]
   \item  Quick questions about $\diagquotient{D_4}{H}$.
   
\vspace{0.15in}   

   The order of $\diagquotient{D_4}{H}$ is \underline{\hspace*{2in}}.

\vspace{0.15in}   

   The identity of $\diagquotient{D_4}{H}$ is \underline{\hspace*{1in}}.
   \hfill
   $(xH)^{-1}=$ \underline{\hspace*{2in}}.

\vspace{0.15in}

  The order of $xH$ in $\diagquotient{D_4}{H}$ is \underline{\hspace*{1in}}. \hfill The size of the set $xH$ is \underline{\hspace*{1in}}.

\vspace{0.05in}

Scratch work:

\vspace{1.75in}   

\item (Still referring to part (a).) $H=Z(D_4)$ (our $H$ above is actually the center of $D_4$). Is $D_4/H$ cyclic? Abelian? Identify this quotient group (it has a special name).

\vspace{1.5in}

\end{enumerate}

 \noindent 
 {\large 4. (10 points)} Consider $\diagquotient{\mathbb{Z}_{20}}{H}$ where $H=\langle 4 \rangle = \{0,4,8,12,16\}$. List all of the cosets (and their contents) of $H$ in $\mathbb{Z}_{20}$. Then make a Cayley table for this quotient group. 
 
 
 \vfill
 
 What is the order of $2+H$ in $\mathbb{Z}_{20}/H$?
 
 




\newpage
\noindent {\large 5. (15 points)} Something is terribly, horribly wrong!

\begin{enumerate}[(a)]

\item Let $H \triangleleft S_4$. Why is $\diagquotient{S_4}{H} \cong \mathbb{Z}_{18}$ impossible?

\vfill

\item Let $\varphi:S_4 \to G$ be a homomorpihsm into some group $G$. Let $H=\langle (12) \rangle = \{(1),(12)\}$. Why is $\mathrm{ker}(\varphi)=H$ impossible?


\vfill

\item Why can't I have an epimorphism (i.e., onto homomorphism) $\psi:D_4 \times Q \to \mathbb{Z}_{15}$ (where $Q$ is the quaternion group)?

\vfill

\end{enumerate}


\noindent {\large 6 (10 points)} Let $H$ and $K$ be groups. Define $\pi:H \times K \to K$ by $\pi((h,k))=k$. Show $\pi$ is a homomorphism. What does the first isomorphism theorem tell us in this particular situation?

\vfill


\newpage
\noindent {\large 7. (25 points)} Finite Abelian Groups

\begin{enumerate}[(a)]
   \item List all of the non-isomorphic abelian groups of order $72 =2^3 3^2$. Circle any that are cyclic.

\vfill

   \item How many non-isomorphic abelian groups of order 55,635,800 are there?

   {\it Note:}  55,635,800 $=2^3  \cdot 5^2  \cdot 11^4 \cdot 19$ and there are 5 non-isomorphic abelian groups of order 14,641 $=11^4$. \smiley

\vfill

   \item Are the groups $\mathbb{Z}_{10} \times \mathbb{Z}_{25} \times \mathbb{Z}_{36}$ and $\mathbb{Z}_{100} \times \mathbb{Z}_{90}$ isomorphic? Explain your answer.

\vfill

   \item What is the largest order among elements of $\mathbb{Z}_{35} \times \mathbb{Z}_{10} \times \mathbb{Z}_{14}$? Explain you answer.

\vspace{0.75in}
\end{enumerate}

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