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\noindent
\parbox{2in}{\bf Math 3110} 
\hfill {\Large \bf Test \#2} \hfill
\parbox{2in}{\bf \hfill October 28$^{th}$, 2009}

\vspace{-0.1in}

\begin{center} \Large ANSWER KEY \end{center}

\vspace{0.2in}

   \noindent {\bf \large 1. (16 points)} Either prove $G$ is a group
        or explain why it is not a group.
\begin{enumerate}[(a)]
      \item $G = \{ x \in D_5 \,|\, x \mbox{ is a reflection} \}$ (the operation is composition of symmetries).

\vspace{0.1in}

\noindent {\bf Answer:} No. $G$ is not a group. 

Two easy ways to see that $G$ is not a group are: \\
(1) $G$ lacks an identity. The identity symmetry is not a reflection (it's a rotation of zero degrees).\\
(2) The operation lacks closure. A reflection composed with a reflection is a rotation.
      So given any $r_1,r_2 \in G$ we have $r_1r_2 \not\in G$.

\vspace{0.1in}
                
      \item $\mathbb{R}_{>0}$ (positive real numbers) where the operation is multiplication.

\vspace{0.1in}

\noindent {\bf Answer:} Yes. $\mathbb{R}_{>0}$ is a group under multiplication.

\vspace{0.1in}

Notice that $\mathbb{R}_{>0}$ is a subset of $\mathbb{R}_{\not=0}$ (which is a group under multiplication).
So to show that $\mathbb{R}_{>0}$ is a group we may just show it is a subgroup of $\mathbb{R}_{\not=0}$.
To do this we need to check closure under multiplication and closure under inversion.
\begin{itemize}
   \item $x>0$ and $y>0$ $\Rightarrow$ $xy>0$ so we have closure under multiplication.
   \item $x >0$ $\Rightarrow$ $1/x >0$ so we have closure under inversion.
\end{itemize}
Therefore, $\mathbb{R}_{>0}$ is a subgroup of $\mathbb{R}_{\not=0}$ (and so it is a group).

\vspace{0.1in}
                
\end{enumerate}

   \noindent {\bf \large 2. (16 points)} 
\begin{enumerate}[(a)]
      \item Show $\displaystyle{ H = \left\{ \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \,\Bigg|\, a,b \in \mathbb{R} \mbox{ and } a,b \not=0 \right\}}$ is a subgroup of $\mathrm{GL}_2(\mathbb{R})$ 

\vspace{0.1in}

(Recall $\mathrm{GL}_2(\mathbb{R})$ --- the set of $2 \times 2$ invertible matrices ---
 is a group under matrix multiplication).

\vspace{0.1in}

First, we'll set up some notation.
Let $A,B \in H$. Thus there exists some non-zero real numbers $a,b,c,d$ such that
$\displaystyle{A = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}}$ and 
$\displaystyle{B = \begin{bmatrix} c & 0 \\ 0 & d \end{bmatrix}}$. 

Notice that $\mathrm{det}(A) = ab \not=0$ because $a\not=0$ and $b\not=0$.
Therefore, for all $A \in H$, we have $A \in \mathrm{GL}_2(\mathbb{R})$. Thus $H \subset \mathrm{GL}_2(\mathbb{R})$.

Obviously $H$ is not empty.

$\displaystyle{AB = \begin{bmatrix} ac & 0 \\ 0 & bd \end{bmatrix}}$ (which looks right -- it's diagonal). Since $a,b,c,d$ are non-zero, we have that $ac$ and $bd$
are non-zero as well. Therefore, $AB \in H$.

$\displaystyle{A^{-1} = \frac{1}{ab-0^2} \begin{bmatrix} b & -0 \\ -0 & a \end{bmatrix} = \begin{bmatrix} 1/a & 0 \\ 0 & 1/b \end{bmatrix}}$ (which looks right -- it's diagonal).
Since $a,b$ are non-zero, we have that $1/a$ and $1/b$ are non-zero as well. Therefore, $A^{-1} \in H$.

Therefore, $H$ is a non-empty subset of $\mathrm{GL}_2(\mathbb{R})$ which is closed under matrix multiplication and inversion.
Thus $H$ is a subgroup of $\mathrm{GL}_2(\mathbb{R})$.

\vspace{0.1in}
                
      \item Quickly (in a few words) why are the even integers a subgroup of the integers?

\vspace{0.1in}

Even plus even is even and the negative of an even number is even.

\vspace{0.1in}
                
\end{enumerate}

   \noindent {\bf \large 3. (20 points)} Workin' mod 5.

\begin{enumerate}[(a)]
\item Fill out the following tables (don't worry about brackets for equivalence classes.)

\begin{center} \hspace*{-0.6in}
\begin{tabular}{c|c|c|c|c|c|}
$(\mathbb{Z}_5,+)$ & 0 & 1 & 2 & 3 & 4 \\ \hline
0 & 0 & 1 & 2 & 3 & 4 \\ \hline 
1 & 1 & 2 & 3 & 4 & 0 \\ \hline 
2 & 2 & 3 & 4 & 0 & 1 \\ \hline 
3 & 3 & 4 & 0 & 1 & 2 \\ \hline 
4 & 4 & 0 & 1 & 2 & 3 \\ \hline 
\multicolumn{6}{c}{$\mathbb{Z}_5$ Addition Table}
\end{tabular} \hspace{0.3in}
\begin{tabular}{c|c|c|c|c|c|}
$(\mathbb{Z}_5,\times)$ & 0 & 1 & 2 & 3 & 4 \\ \hline
 0 & 0 & 0 & 0 & 0 & 0 \\ \hline
 1 & 0 & 1 & 2 & 3 & 4 \\ \hline
 2 & 0 & 2 & 4 & 1 & 3 \\ \hline
 3 & 0 & 3 & 1 & 4 & 2 \\ \hline
 4 & 0 & 4 & 3 & 2 & 1 \\ \hline
\multicolumn{6}{c}{$\mathbb{Z}_5$ Multiplication Table}
\end{tabular}
\end{center}

\item Compute $2^{-1}(4+3)-2$ (mod 5).

\vspace{0.1in}

Notice that $2(3)=1$ (mod 5) so $2^{-1}=3$.

Thus $2^{-1}(4+3)-2 = 3(4+3)-2 = 3(7)-2 = 3(2)-2 = 6-2 = 1-2 = -1 = 4$ (mod 5)

Of course, we could wait until the very end to
reduce mod 5: $2^{-1}(4+3)-2=3(4+3)-2 = 3(7)-2=21-2=19=4$ (mod 5). The important thing to 
remember is that addition, subtraction and multiplication of integers haven't changed. When
working mod 5, we've just changed the idea of what ``='' means. The only operation that requires
special care is ``division''. To compute inverses we can either guess/look up the answer on the multiplication
table OR run the Euclidean algorithm.

\vspace{0.1in}

\item Find $\langle 3 \rangle$ (the subgroup generated by $3$) in $U(5)$ (NOT $\mathbb{Z}_5$!!!).

\vspace{0.1in}

Remember $U(5)$ contains (congruence classes) of integer which are relatively prime with 5. So $U(5) = \{ 1,2,3,4 \}$ 
(Also, remember that the operation is multiplication mod 5 NOT addition).

Note: $3^1=3$, $3^2=4$, $3^3=3^2(3)=4(3)=2$, and $3^4=3^3(3)=2(3)=1$ so the order of $3$ (in $U(5)$) is 4.

\vspace{-0.15in}

\[ \langle 3 \rangle = \{ \dots 3^{-1}, 3^0, 3^1, 3^2, \dots \} = \{ 1, 3, 4, 2 \} = U(5) \]
             
\item Find the orders of elements of $U(5)$. Is $U(5)$ cyclic? Why or why not?

\begin{minipage}[c]{3in}
\begin{itemize} \setlength{\itemsep}{1pt}
\item $1^1 = 1$
\item $2^1 = 2$, $2^2 = 4$, $2^3 = 3$, $2^4 = 1$
\item $3^1 = 3$, $3^2 = 4$, $3^3 = 2$, $3^4 = 1$
\item $4^1 = 4$, $4^2 = 1$
\end{itemize}
\end{minipage} \hfill \hfill
\begin{minipage}[c]{3in}
\begin{tabular}{r|c|c|c|c}
element = & 1 & 2 & 3 & 4 \\ \hline
order = & 1 & 4 & 4 & 2 
\end{tabular}
\end{minipage}

%\vspace{0.1in}

{\bf YES} $U(5)$ is cyclic because the order of $U(5)$ is 4 and we have an element (in fact 2 elements) of order 4.
...OR... $U(5)$ is cyclic because $U(5)=\langle 3\rangle$ ...OR... $U(5)$ is cyclic because $U(5) = \langle 2 \rangle$.

\vspace{0.1in}
                
\end{enumerate}

   \noindent {\bf \large 4. (16 points)} Quick proofs
\begin{enumerate}[(a)]
      \item Let $G$ be a group and suppose that $x=x^{-1}$ for all $x \in G$. Prove that $G$ is Abelian.\\
       Hint: $xy=(xy)^{-1}=\;\;??$

\vspace{0.05in}

Suppose that $x,y \in G$. Then $x^{-1}=x$, $y^{-1}=y$ and $(xy)=(xy)^{-1}$ (by assumption). 
Recall that in any group we have $(xy)^{-1}=y^{-1}x^{-1}$ (the ``socks shoes principle'').
Therefore, $xy=(xy)^{-1} = y^{-1}x^{-1} = yx$. Thus $G$ is Abelian.

\vspace{0.05in}
                
      \item Let $n$ be an integer such that $n \geq 3$. Consider $(12), (13) \in S_n$.
                Compute $(12)(13)$ and $(13)(12)$. Is $S_n$ cyclic? Why or why not?

\vspace{0.05in}

\begin{minipage}[c]{1.5in}
\begin{itemize}
\item $(12)(13)=(132)$
\item $(13)(12)=(123)$
\end{itemize}
\end{minipage} \hfill \hfill
\begin{minipage}[c]{4.5in}
Since $(12)(13) \not= (13)(12)$, we can conclude that $S_n$ is {\bf not Abelian}.
And since $S_n$ is not Abelian, it is {\bf not cyclic} (because cyclic implies Abelian).
\end{minipage}

\vspace{0.05in}
                
\end{enumerate}

   \noindent {\bf \large 5. (16 points)} $G = \{1,a,a^2,b,ab,a^2b \}$ is a group. 
   
   Finish filling out $G$'s Cayley table then answer some questions.
   
 \begin{center}
\begin{tabular}{|c|||c|c|c|c|c|c|}
\hline
$G$      & $1$         & $a$         & $a^2$   & $b$       & $ab$      & $a^2b$ \\ \hline 
\hline \hline
$1$       & $1$        & $a$         & $a^2$    & $b$      & $ab$      & $a^2b$ \\ \hline 
$a$       & $a$         &  $a^2$   & $1$        & $a  b$  & $a^2b$ & $   b$ \\ \hline 
$a^2$   & $a^2$    & $1   $     & $a$        & $a^2b$ & $   b$    & $a  b$ \\ \hline 
$b$       & $b$         & $a^2b$ & $a  b$    & $1   $   & $a^2 $   & $a$ \\ \hline 
$ab$     & $ab$      & $   b$     & $a^2b$ & $a$       & $1$        & $a^2$ \\ \hline 
$a^2b$ & $a^2b$ & $a  b$    & $b$        & $a^2 $ & $a$        & $1   $ \\ \hline 
\end{tabular}
\end{center}   
   
\begin{enumerate}[(a)]
   \item What is the order of $a^2$? Determine $\langle a^2 \rangle$ (the subgroup generated by $a^2$).

\vspace{0.05in}

$(a^2)^1 = a^2$, $(a^2)^2 = a$, and $(a^2)^3 = a(a^2) = 1$. Thus the order of $a^2$ is 3.\\
Also, $\langle a^2 \rangle = \{ a^2, a,1 \} = \{1,a,a^2 \}$.

\vspace{0.05in}
                
   \item Is $G$ Abelian? Is $G$ cyclic? Why or why not?

\vspace{0.05in}

{\bf NO} $G$ is {\bf not Abelian}. Just look at the Cayley table and notice that $ab \not= ba = a^2b$.
Also, since $G$ is not Abelian it is {\bf not cyclic} (since all cyclic groups are automatically Abelian).

\vspace{0.05in}

\end{enumerate}   
   
   \noindent {\bf \large 6. (16 points)} Permutations! 
\begin{enumerate}[(a)]
      \item Write $\sigma = (125)(35)(24)(264)$ as a product of disjoint cycles.

\vspace{0.05in}

$\sigma = (125)(35)(24)(264) = (12653)(4) = (12563)$

\vspace{0.05in}
                
      \item What is the order of $\sigma$? 

\vspace{0.05in}

The order of $\sigma$ is 5 (the least common multiple of the lengths of the cycles when
$\sigma$ is written as a product of {\bf disjoint} cycles).

\vspace{0.05in}
                
      \item Write $\sigma$ as a product of transpositions. Is $\sigma$ even or odd?

\vspace{0.05in}

There are infinitely many answers to this question here are two possibilities:\\
$\sigma = (15)(12)(35)(24)(24)(26) = (13)(16)(15)(12)$.

Either way we have an even number of transpositions so $\sigma$ {\bf is even}.

\vspace{0.05in}
                
      \item Let $\tau = (1452)(367)(89)$. What is the order of $\tau$? 

\vspace{0.05in}

$\tau$ is written as the product of disjoint cycle so we can just take the least common multiple of the
lengths of its cycles. $\mbox{lcm}(4,3,2) = 12$. So the order of $\tau$ is 12.

\vspace{0.05in}
                
      \item Write $\tau^{26}$ as the product of disjoint cycles.

\vspace{0.05in}

Since the order of $\tau$ is 12, we have that $\tau^{12} = (1)$ so
$\tau^{26} = \tau^{12}\tau^{12}\tau^2 = \tau^2 = (15)(24)(376)$.

\vspace{0.05in}
                
      \item What is the order of $\tau^6$?\\
       {}[Hint: You shouldn't need to compute any powers of $\tau$ to answer this question.]

\vspace{0.05in}

Quick way: $\tau^6 \not= (1)$ since the order of $\tau$ is greater than 6. But $(\tau^6)^2 = \tau^{12} = (1)$ since the
order of $\tau$ is 12. Thus the second power of $\tau^6$ is the identity so $\tau^6$ has order 2.

Calculatin' way: $\tau^6 = \left( (1452)(367)(89) \right)^6 = (1452)^6 (367)^6 (89)^6$ since disjoint cycles commute.
Continuing... $= (1456)^4(1456)^2((367)^3)^2((89)^2)^3 = (1)(15)(24)(1)^2(1)^3 = (15)(24)$ so $\tau^6$ has order
$\mbox{lcm}(2,2)=2$.

   \end{enumerate}

\end{document}
