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\noindent
\parbox{1.5in}{\bf Math 3110} 
\hfill {\Large \bf  Test \#2} \hfill
\parbox{1.5in}{\bf \hfill October $9^{\mathrm{th}}$, 2015}

\vspace{0.1in}

\noindent {\large\bf Name:} \underline{\Large\color{red}\sc \quad Answer Key \quad} \hfill {\bf Be sure to show your work!}

\vspace{0.1in}

\noindent{\bf\large 1. (20 points)} Random Group Stuff --- Fill out the following table:

\vspace{0.2in}

\mbox{} \hspace{-0.2in}
\begin{tabular}{|c|||c|c|c|l|c|} 
\hline
$G=$ & \parbox[c][0.5in]{0.85in}{What is the\\ identity of $G$?} & \parbox[c][0.5in]{0.85in}{Is $G$ abelian?} & \parbox[c][0.5in]{0.85in}{Is $G$ cyclic?} & \parbox[c][0.5in]{1.75in}{What is the\\ order of ...?} & \parbox[c][0.5in]{1.25in}{Does $G$ have an\\ element of order 5?} \\
\hline \hline \hline
$\mathbb{Z}_{99}$ & $0$ & Yes & Yes  & $|15|=\dfrac{99}{\mathrm{gcd}(99,15)}=\dfrac{99}{3}=33$ & No. $5 \not\Big| \  99$ \varspace{0.45in}{0.01in} \\ \hline
$U(10)$ & $1$ & Yes & Yes  & $|3|=4$ & No. $5 \not\Big| \  4=|U(10)|$ \varspace{0.45in}{0.01in}   \\ \hline
$D_{20}$ & $1=R_{0^\circ}$ & No & No & $|x^6|=\dfrac{20}{\mathrm{gcd}(20,6)}=\dfrac{20}{2}=10$ & Yes. $|x^4|=5$  \varspace{0.45in}{0.01in} \\ \hline
$S_9$ & $(1)$ & No & No  & $|(123)(4567)(89)|=\mathrm{lcm}(3,4,2)=12$ & Yes. $|(12345)|=5$ \varspace{0.45in}{0.01in} \\ \hline
\end{tabular} 

\vspace{0.05in}

{\bf Recall:} $D_{20} = \{ 1, x, \dots, x^{19}, y, xy, \dots, x^{19}y \}$ where $x^{20}=1$, $y^2=1$, and $xyxy=1$.

\vspace{0.1in} 

\underline{Scratch Work:}

\vspace{0.05in}

$\mathbb{Z}_{99}=\langle 1 \rangle$ so it is cyclic and thus also abelian. Notice that in $U(10)$: $3^1=3$, $3^2=9$, $3^3=27=7$, and $3^4=81=1$. Thus $|3|=4$ and $\langle 3 \rangle = \{ 1,3,7,9 \} = U(10)$ so $U(10)$ is cyclic and thus also abelian.
On the other hand $D_{20}$ and $S_9$ are not abelian. Thus they cannot be cyclic either.

\vspace{0.15in}

\noindent {\bf\large 2. (24 points)} Cyclic Stuff

   \begin{enumerate}[(a)]

      \item Let $G$ be a finite group and $g \in G$. Suppose that $|g|=66$. 
      
      \begin{enumerate}[i.]
      \item What is the order of $g^{55}$? List the distinct elements in $\langle g^{55} \rangle$.
      
\vspace{0.05in}      
      
      $\langle g^{55} \rangle = \left\{ e, g^{55}, g^{110}, g^{165}, \dots \right\} = \left\{ e, g^{55}, g^{44}, g^{33}, g^{22}, g^{11} \right\}$
          
\vspace{0.05in}      

Alternatively, notice that $\langle g^{55} \rangle = \langle g^{\mathrm{gcd}(66,55)} \rangle = \langle g^{11} \rangle = \{ e,g^{11},\dots,g^{55} \}$.

\vspace{0.05in}      
      
      \item  Is $g^{22} \in \langle g^{22} \rangle$? \quad {\large\bf \fbox{Yes} \ \  /  \ \  No}
      
\vspace{0.05in}      

And, yes, this was a very very silly typo on the test. 

\vspace{0.05in}      
    
      \end{enumerate}

     \item How many elements of order 6 does $\mathbb{Z}_{120}$ have? What are they?
     
\vspace{0.05in}      

6 does divide 120 so there are elements of order 6. Recall that we can find the number of elements of various order in {\it any} cyclic group using our big theorem about the existence and uniqueness of subgroups of every divisor order. As discussed in class, there is 1 element of order 1, $2-1=1$ element of order 2, $3-1=2$ elements of order 3, and $6-2-1-1=$  \fbox{$2$ elements of order 6.} 

\vspace{0.05in}      

    \item List the orders of elements in $\mathbb{Z}_{55}$. Then determine the number of elements of each order.   

\vspace{0.05in}      

Use the same procedure as in part (b). The divisors of 55 are 1, 5, 11, and 55.

\begin{tabular}{c||c|c|c|c} 
Order = & 1 & 5 & 11 & 55 \Large\strut \\ \hline
Number of elements = & 1 & $5-1=4$ & $11-1=10$ & $55-10-4-1=40$ \Large\strut 
\end{tabular}

\vspace{0.05in}      

    \item List the orders of elements in $D_{55}$. Then determine the number of elements of each order.   

\vspace{0.05in}      

Part (c) accounts for the rotations, we just need to add in the reflections (all of which have order 2).

\begin{tabular}{c||c|c|c|c|c|c} 
Order = & 1 & 2 & 5 & 11 & 55 \Large\strut \\ \hline
Number of elements = & 1 & 55 & 4 & 10 & 40 
\end{tabular}
\end{enumerate}

\newpage

\noindent {\bf\large 3. (22 points)} Permutations
\begin{enumerate}[(a)]
   \item Let $G = \langle i \rangle = \{ 1,i,-1,-i \}$ where $i=\sqrt{-1}$.  \qquad [$G$ is a subgroup of $\mathbb{C}_{\not=0}$ (nonzero complex numbers).] 
   
   Label $1$ as $1$, $i$ as $2$, $-1$ as $3$, and $-i$ as $4$. Cayley's theorem says that $G$ is isomorphic to a subgroup of $S_4$. Find this subgroup [using left multiplication maps and the labels provided].
   
\vspace{0.05in}      

Carefully listing $G$ in the order prescribed, we have the following Cayley table\dots

\force \hfill
\begin{tabular}{c||c|c|c|c|} 
           &  1 & $i$ & $-1$ & $-i$ \\ \hline \hline
1         &  1 & $i$ & $-1$ & $-i$ \\ \hline 
$i$      &  $i$ & $-1$ & $-i$ & $1$ \\ \hline
$-1$   & $-1$ & $-i$ & $1$ & $i$ \\ \hline
$-i$   &  $-i$ & $1$ & $i$ & $-1$ \\ \hline
\end{tabular}
\hfill \force

Notice that the left multiplication maps give\dots

\vspace{-0.15in}

$$L_1: 1 \mapsto 1\cdot 1=1, i \mapsto 1 \cdot i=i, -1 \mapsto 1 \cdot (-1) = -1, -i \mapsto 1 \cdot (-i) = -i \qquad \Longleftrightarrow \qquad (1)(2)(3)(4) 
$$

\vspace{-0.25in}

$$L_i: 1 \mapsto i\cdot 1=i, i \mapsto i \cdot i=-1, -1 \mapsto i \cdot (-1) = -i, -i \mapsto i \cdot (-i) = 1 \qquad \Longleftrightarrow \qquad (1234)
$$     

\vspace{-0.25in}

$$L_{-1}: 1 \mapsto -1\cdot 1=-1, i \mapsto -1 \cdot i=-i, -1 \mapsto -1 \cdot (-1) = 1, -i \mapsto -1 \cdot (-i) = i \qquad \Longleftrightarrow \qquad (13)(24)
$$     

\vspace{-0.25in}

$$L_{-i}: 1 \mapsto -i\cdot 1=-i, i \mapsto -i \cdot i=1, -1 \mapsto -i \cdot (-1) = i, -i \mapsto -i \cdot (-i) = -1 \qquad \Longleftrightarrow \qquad (1432)
$$     

In more detail, for example, $L_{-1}$ maps $1$ to $-1$ and $-1$ to $1$ (that is element 1 maps to element 3 and then element  3 maps to 1). Likewise, $L_{-1}$ sends element 2 to element 4 and then 4 back to 2 (i.e. $i$ and $-i$ are interchanged). This gives us the permutation $(13)(24)$.


\vspace{0.05in}      

 $G \cong$ \fbox{$\left\{ (1),(1234),(13)(24),(1432) \large\strut \right\}$} $= \langle (1234) \rangle$

   \item Write $\sigma = (237)(1724)(27563)$ as a product of disjoint cycles. \qquad \fbox{$\sigma = (1234)(567)$}

   
\vspace{0.05in}      

  
             $\sigma^{-1}=(765)(4321)=$ \fbox{$(1432)(576)$} 

\vspace{0.05in}      
   
            The order of $\sigma$ is $|\sigma|=$ \underline{\quad $\mathrm{lcm}(4,3)=12$ \quad}.  
            
\vspace{0.05in}      
   
           Write $\sigma$ as a product of transpositions. \qquad $\sigma$ is \quad {\large\bf Even \ \  / \ \  \fbox{Odd}}

\vspace{0.05in}      

$\sigma = (27)(23)(14)(12)(17)(23)(26)(25)(27)$ \quad OR \quad \fbox{$(14)(13)(12)(57)(56)$}

\vspace{0.05in}      
    
         Compute $\sigma^{30}$.

\vspace{0.05in}      

$\sigma^{30} = ((1234)(567))^{30} = (1234)^{30}(567)^{30} = (1234)^2(567)^0 =$ \fbox{$(13)(24)$} where the second equality holds because disjoint cycles commute and the next equality holds since the order of a 4-cycle is 4 and a 3-cycle is 3 (so exponents can be reduced mod 4 and 3 respectively).

\vspace{0.05in}      

\end{enumerate}

\noindent {\bf\large 4. (18 points)} Explain why the following pairs of groups are not isomorphic.

\begin{enumerate}[(a)]

\item $\mathbb{Q} \not\cong \mathbb{Z}_{123}$ \quad [$\mathbb{Q}$ is the rational numbers.]

\vspace{0.05in}      

Both groups are abelian, but $\mathbb{Q}$ is infinite (countably infinite) whereas $\mathbb{Z}_{123}$ is finite (of order 123). Therefore, they cannot be isomorphic. 

Of course, there are other reasons as well. The order of any non-identity element of $\mathbb{Q}$ is infinite (if $nx=0$ for $n \in \mathbb{Z}_{>0}$ and $x \in \mathbb{Q}$, then $x=0$) whereas the orders of elements in $\mathbb{Z}_{123}$ are divisors of 123. Or  $\mathbb{Z}_{123}$ is cyclic and $\mathbb{Q}$ isn't cyclic (if $\mathbb{Q}=\langle x \rangle$ for some $x \in \mathbb{Q}$, then it can be shown that $x/2 \not\in \mathbb{Q}$ -- contradiction).


\vspace{0.05in}      

\item $Q \not\cong \mathrm{Aut}(\mathbb{Z}_{8})$  \quad [$Q = \{\pm 1, \pm i, \pm j, \pm k\}$ is the group of quaternions.]

\vspace{0.05in}      

Recall that $\mathrm{Aut}(\mathbb{Z}_n) \cong U(n)$ (the unit group of $\mathbb{Z}_n$). These groups cannot be isomorphic since $Q$ is not abelian whereas $\mathrm{Aut}(\mathbb{Z}_8) \cong U(8)$ is abelian. OR $|Q|=8 \not= |\mathrm{Aut}(\mathbb{Z}_8)|=|U(8)| = |\{1,3,5,7\}|=4$ (they are groups of different orders). Of course, yet again, there are other reasons as well.

\vspace{0.05in}      

\item $A_4 \not\cong D_{6}$

\vspace{0.05in}      

First, note that $|A_4|=\dfrac{4!}{2} = 12 = 2 \cdot 6 = |D_6|$, so these are both non-abelian groups of order 12 (no help here). 

Now $A_4 = \{ (1),(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}$ has elements of orders 1, 2, and 3. On the other hand, $D_6$ has elements of orders 1, 2, 3, and {\bf 6}. So since $A_4$ has no elements of order 6, these groups cannot be isomorphic. 

Another way to see that $A_4$ and $D_6$ aren't isomorphic is to count the number of elements of order 2. $A_4$ has 3 elements of order 2: $(12)(34), (13)(24),$ and $(14)(23)$. On the other hand, $D_6$ has 7 elements of order 2: the $180^\circ$ rotation and 6 reflections.


\vspace{0.05in}      

\end{enumerate}


\noindent {\bf\large 5. (16 points)} A few proofs

\begin{enumerate}[(a)]

\item Explain why $A_3 \cong \mathbb{Z}_3$ but $A_n \not\cong \mathbb{Z}_n$ for $n>3$.

\vspace{0.05in}      

$A_3 = \{ (1),(123),(132) \} = \langle (123) \rangle$ is a cyclic group of order 3. Since any two cyclic groups of the same order are isomorphic, $A_3 \cong \mathbb{Z}_3$.
On the other hand, $A_n$ is not abelian for $n>3$. Therefore, $A_n$ (for $n>3$) isn't cyclic. Thus $A_n \not\cong \mathbb{Z}_n$ for $n>3$.


\vspace{0.05in}      

\item Pick {\bf ONE} of the following\dots

   \begin{enumerate}[I.]
   \item Let $G$ be an abelian group. Show that $\varphi:G \to G$ defined by $\varphi(x)=x^{-1}$ is an automorphism of $G$.

\vspace{0.05in}      

First, notice that $\varphi(\varphi(x))=\varphi(x^{-1})=(x^{-1})^{-1}=x$. Therefore, $\varphi = \varphi^{-1}$ ($\varphi$ is its own inverse) and so $\varphi$ is bijective (one-to-one and onto). [Alternatively, suppose $x,y \in G$ such that $\varphi(x)=\varphi(y)$. Then $x^{-1}=y^{-1}$ so that $x=y$. Thus $\varphi$ is one-to-one. Also, suppose $x \in G$. Then $\varphi(x^{-1})=(x^{-1})^{-1}=x$. Therefore, $\varphi$ is onto.]

Next, let $x,y \in G$. $\varphi(xy)=(xy)^{-1}=y^{-1}x^{-1}=x^{-1}y^{-1}=\varphi(x)\varphi(y)$ (the third equality holds because $G$ is abelian). Therefore, $\varphi$ is a homomorphism.

We have shown that $\varphi$ is a bijective homomorphism from $G$ to itself. This means that $\varphi$ is an automorphism.

{\it Note:} The converse is also true: If $\varphi(x)=x^{-1}$ is an automorphism, then $G$ is abelian. Why? Suppose $\varphi$ is an automorphism and let $x,y \in G$. Then $\varphi(x^{-1}y^{-1})=\varphi(x^{-1})\varphi(y^{-1})$ ($\varphi$ is operation preserving) so that $(x^{-1}y^{-1})^{-1}=(x^{-1})^{-1}(y^{-1})^{-1}$. Therefore, by socks-shoes, $(y^{-1})^{-1}(x^{-1})^{-1}=(x^{-1})^{-1}(y^{-1})^{-1}$ and so $yx=xy$ ($G$ is abelian). 

\vspace{0.05in}      

   \item Let $\varphi, \psi$ be automorphisms of $G$. Prove that $H = \{ x \in G \;|\; \varphi(x)=\psi(x) \}$ is a subgroup of $G$.

\vspace{0.05in}      

First, note that $\varphi(e)=e=\psi(e)$. Therefore, $e \in H$. Thus $H$ is a non-empty subset of $G$.

Next, let $x,y \in H$. Then by definition: $\varphi(x)=\psi(x)$ and $\varphi(y)=\psi(y)$. This implies that $\varphi(xy)=\varphi(x)\varphi(y)=\psi(x)\psi(y)=\psi(xy)$, so $xy \in H$. This also implies that $\varphi(x)^{-1}=\psi(x)^{-1}$ so that $\varphi(x^{-1})=\psi(x^{-1})$. Therefore, $x^{-1} \in H$. 

Thus by the 2-step subgroup test, $H$ is a subgroup of $G$. 


\vspace{0.05in}      

   \end{enumerate}

\end{enumerate}


\end{document}


