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\begin{document}

%\pagestyle{empty}

\noindent
\parbox{2in}{\bf Math 3110} 
\hfill {\Large \bf Test \#1} \hfill
\parbox{2in}{\bf \hfill February $17^{\mathrm{th}}$, 2021}

\vspace{0.15in}

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

\vspace{0.1in}

\noindent {\bf\large 1. (20 points)} Definition and Basics

\begin{enumerate}[(a)]
   \item Suppose that $G$ is a non-empty set equipped an operation. What 4 things do I need to check to see if $G$ is a group? 

          \begin{enumerate}[1:] \setlength\itemsep{0em}
             \item Closure: $\forall x,y \in G$, we have $xy \in G$

             \item Associativity: $\forall x,y,z \in G$, we have $(xy)z = x(yz)$

             \item Identity: $\exists e \in G$  such that $\forall x \in G$, $xe = x = ex$

             \item Inverses:  $\forall x \in G$,  $\exists y \in G$ such that $xy = e = yx$

%\vspace{0.05in}

           \force\hspace{-0.35in} What additional property needs to hold for $G$ to be an {\bf Abelian} group?

%\vspace{0.05in}

           \item Commutivity:  $\forall x,y \in G$, $xy = yx$
\end{enumerate}   
      
   \item The positive real numbers $\mathbb{R}_{>0} = \{ r \in \mathbb{R} \;|\; r>0 \}$ do not form a group under division. Why not?
 
\vspace{0.05in}

Notice that $1 / (2/3) = 3/2$ whereas $(1/2)/3=1/6$. Therefore, division is not associative. Thus $\mathbb{R}_{>0}$ does not form a group under division. Alternatively, we could also see that the identity axiom fails: Suppose $x/e=x=e/x$ for all $x \in \mathbb{R}_{\not=0}$. Then, $x/e=x$ implies $x=ex$ so that $e=1$. But $1/x \not= x$ for most $x \not=1$. So there is a right identity but not a two sided identity.

%\vspace{0.05in}
   
   \item On the other hand, the positive real numbers $\mathbb{R}_{>0} = \{ r \in \mathbb{R} \;|\; r>0 \}$ do form a group if we select the right operation. Which operation turns this collection of numbers into a group: Addition or \fbox{Multiplication}? Then explain why the other operation does not yield a group.

\vspace{0.05in}

We have that $\mathbb{R}_{>0}$ is a group under \underline{multiplication} (in fact, an abelian group): positive times positive is positive, multiplication is associative, $1$ is a positive number and acts as the multiplicative identity, if $x>0$ then $x^{-1}>0$ is too, (and multiplication is commutative). On the other hand, positive real numbers cannot form a group under addition since they lack $0$ (the additive identity). Alternatively, we can see that they fail to form a group under addition since the additive inverse of a positive number is a negative number (i.e., they are not closed under inverses).

%\vspace{0.05in}

   \item The non-zero rational numbers $\mathbb{Q}_{\not=0}$ form a group under multiplication. On the other hand, the (non-zero) irrational numbers $\mathbb{I}=\mathbb{R}-\mathbb{Q} = \{ x \in \mathbb{R} \;|\; x \not\in \mathbb{Q}\}$ do not. Why?

\vspace{0.05in}

The irrational numbers do not form a group under multiplication since closure fails: $\sqrt{2} \in \mathbb{I}$, but $(\sqrt{2})^2=2 \not\in \mathbb{I}$ (i.e., $\sqrt{2}$ is irrational but $(\sqrt{2})^2=2$ is rational). Alternatively, they do not form a group since they do not have a multiplicative identity ($1$ is rational and so $1 \not\in \mathbb{I}$). 

\vspace{0.05in}

\end{enumerate}

\noindent {\bf\large 2. (20 points)} Some modular arithmetic.

   \begin{enumerate}[(a)]
      \item Make a list of all of the cyclic subgroups of $\mathbb{Z}_{10}$ along with their contents (for example: $\langle 0 \rangle = \{0\}$).
      
\vspace{0.05in}

\fbox{$\langle 0 \rangle = \{0\}$}, \fbox{$\langle 1 \rangle = \{0,1,\dots,9\}$} ($= \langle 3 \rangle = \langle 7 \rangle = \langle 9 \rangle$), \fbox{$\langle 2 \rangle = \{0,2,4,6,8\}$} ($=\langle 4 \rangle = \langle 6 \rangle = \langle 8 \rangle$), and \fbox{$\langle 5 \rangle = \{0,5\}$}.


\vspace{0.05in}

      \item Fill out the following table referring to the operations of addition and multiplication modulo 10:
      
      {\it Note:} Just put an {\large\bf X} if something is undefined / does not exist.

      %\hspace{-0.5in}      
      \begin{tabular}{|r|c|c|c|c|c|c|c|c|c|c|} \hline
                     Element $x=$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \large\strut \\ \hline
      Additive Inverse $-x=$ & 0 & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \large\strut \\ \hline
      Order (in $\mathbb{Z}_{10}$) $|x|=$ & 1 & 10 & 5 & 10 & 5 & 2 & 5 & 10 & 5 & 10 \large\strut \\ \hline
      Multiplicative Inverse $x^{-1}=$ & {\large\bf X} & 1  & {\large\bf X} & 7 & {\large\bf X} & {\large\bf X} & {\large\bf X} & 3 & {\large\bf X} & 9 \Large\strut \\ \hline
      Order (in $U(10)$) $|x|=$ & {\large\bf X} & 1  & {\large\bf X} & 4 & {\large\bf X} & {\large\bf X} & {\large\bf X} & 4 & {\large\bf X} & 2 \Large\strut \\ \hline
      \end{tabular}

\vspace{0.05in}

Briefly, additive inverses are just negatives and additive orders match the sizes the cyclic subgroups found in part (a) or, for example, $4\not=0$, $4+4=8\not=0$, $4+4+4=2\not=0$, $4+4+4+4=6\not=0$, but $4+4+4+4+4=0$ so the additive order of $4$ is $5$. Only things relatively prime to $10$ have a multiplicative inverse. Notice $1 \cdot 1 =1$, $3 \cdot 7=1$, and $9\cdot 9 =1$ so $1^{-1}=1$, $3^{-1}=7$, $7^{-1}=3$, and $9^{-1}=9$. Finally, multiplicative orders are computed by looking at successive powers. For example, $3^1=3 \not=1$, $3^2=9\not=1$, $3^3=7\not=1$, but $3^4=1$ so its multiplicative order is $4$.


\vspace{0.05in}
      
      \item Compute $2^{-1}(3-8)+7$ mod $10$ or explain why this is undefined.

\vspace{0.05in}

\fbox{Undefined} since $2^{-1}$ does not exist (because $\mathrm{gcd}(2,10)=2\not=1$).

\vspace{0.05in}

      \item Compute $3^{-1}(7-3)-11$ mod $10$ or explain why this is undefined.

\vspace{0.05in}

$3^{-1}(7-3)-11 = 7(4)-11=28-11=17=$ \fbox{$7$} (mod 10). 

\vspace{0.05in}

\end{enumerate}

\noindent {\bf\large 3. (20 points)} More Modular Arithmetic.

\begin{enumerate}[(a)]
      \item Write down a Cayley table for $U(8)$. Is $U(8)$ cyclic (circle the correct answer)? \quad {\Large Yes \quad / \quad \fbox{No}}
      
\vspace{0.05in}

\force\hfill
\begin{tabular}{r||c|c|c|c|} 
    & 1 & 3 & 5 & 7 \large\strut\\ \hline\hline
 1 & 1 & 3 & 5 & 7 \large\strut\\ \hline
 3 & 3 & 1 & 7 & 5 \large\strut\\ \hline
 5 & 5 & 7 & 1 & 3 \large\strut\\ \hline
 7 & 7 & 5 & 3 & 1 \large\strut\\ \hline
\end{tabular} \qquad \parbox[b]{3in}{$U(8)=\{ k \;|\; \mathrm{gcd}(k,8)=1\} = \{1,3,5,7\}$ is not cyclic since $\langle 1 \rangle = \{1\}$, $\langle 3 \rangle = \{1,3\}$, $\langle 5 \rangle = \{1,5\}$, and $\langle 7 \rangle = \{1,7\}$ (thus nothing generates the whole group).}
\hfill\force

\vspace{0.05in}

      \item Draw the subgroup lattice for $\mathbb{Z}_{20}$. [$20 = 2^2 \cdot 5$]

\vspace{-0.75in}

\force \hfill \hfill \hfill
\begin{tikzpicture}[node distance=1.5cm]
\node(12){$20$};
\node(6)[below left of = 12]{$10$};
\node(4)[below right of = 12]{$4$};
\node(3)[below of = 6]{$5$};
\node(2)[below of = 4]{$2$};
\node(1)[below right of = 3]{$1$};

\draw(12)--(6);
\draw(12)--(4);
\draw(6)--(3);
\draw(6)--(2);
\draw(4)--(2);
\draw(3)--(1);
\draw(2)--(1);
\end{tikzpicture}
\qquad \qquad
\begin{tikzpicture}[node distance=1.5cm]
\node(12){$\mathbb{Z}_{20} = \langle 1 \rangle$};
\node(6)[below left of = 12]{$\langle 2 \rangle$};
\node(4)[below right of = 12]{$\langle 5 \rangle$};
\node(3)[below of = 6]{$\langle 4 \rangle$};
\node(2)[below of = 4]{$\langle 10 \rangle$};
\node(1)[below right of = 3]{$\langle 0 \rangle$};

\draw(12)--(6);
\draw(12)--(4);
\draw(6)--(3);
\draw(6)--(2);
\draw(4)--(2);
\draw(3)--(1);
\draw(2)--(1);
\end{tikzpicture}
\hfill \force

\vspace{0.05in}

   \item Find $10^{-1}$ mod 77 using the extended Euclidean algorithm [Don't just guess and check].

\vspace{0.05in}

Repeated division gives: $77=(7)10+7$, $10=(1)7+3$, $7=(2)3+1$, $3=(3)1+0$. Thus $\mathrm{gcd}(77,10)=\mathrm{gcd}(10,7)=\mathrm{gcd}(7,3)=\mathrm{gcd}(3,1)=\mathrm{gcd}(1,0)=1$. We now run the extended portion of the algorithm using our division facts: $(1)7+(-2)3=1$, $(1)10+(-1)7=3$, and $(1)77+(-7)10=7$.

Plugging, $(1)10+(-1)7=3$ into 3 in $(1)7+(-2)3=1$ yields $(1)7+(-2)[(1)10+(-1)7]=1$ which gives $(-2)10+(3)7=1$. Next, we plug $(1)77+(-7)10=7$ into 7 in our previously obtained fact and get $(-2)10+(3)[(1)77+(-7)10]=1$. Simplifying yields, $(3)77+(-23)10=1$. [{\it Note:} See my \href{https://billcookmath.com/sage/algebra/Euclidean_algorithm.html}{Euclidean Algorithm} Sage interactive webpage to automate this computation.] Therefore, this equation says $(-23)10=1$ (mod 77). Thus $10^{-1}=-23=-23+77=$ \fbox{54} in $U(77)$.

\vspace{0.05in}

\end{enumerate}

\noindent {\bf\large 4. (20 points)} Recall $D_n = \{1,x,\dots,x^{n-1},y,xy,\dots,x^{n-1}y \} = \langle x,y \;|\; x^n=1, y^2=1, (xy)^2=1 \rangle$.

\begin{enumerate}[(a)]
   \item Use the relations for $D_{8}$ to simplify $x^{13}y^3x^{-2}y^{888}x$
      
\vspace{0.05in}

Recall that $x$'s exponents work mod 8 and $y$'s exponents work mod 2 -- odd powers of $y$ are just $y$ and even powers of $y$ are just the identity $y^0=1$. Finally, remember that $yx^\ell = x^{-\ell}y$ and then the computation follows:
$x^{13}y^3x^{-2}y^{888}x = x^5yx^{6}1x = x^5yx^{7}=x^5x^{-7}y=x^{-2}y=$ \fbox{$x^6y$}  

\vspace{0.05in}
   
   \item Make a table listing the elements of $D_8$, their inverses, and their orders.
   
\vspace{0.05in}

\begin{tabular}{|r|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
$g=$ & $1$ & $x$ & $x^2$ & $x^3$ & $x^4$ & $x^5$ & $x^6$ & $x^7$ & $y$ & $xy$ & $x^2y$ & $x^3y$ & $x^4y$ & $x^5y$ & $x^6y$ & $x^7y$ \Large\strut \\ \hline
$g^{-1}=$ & $1$ & $x^7$ & $x^6$ & $x^5$ & $x^4$ & $x^3$ & $x^2$ & $x$ & $y$ & $xy$ & $x^2y$ & $x^3y$ & $x^4y$ & $x^5y$ & $x^6y$ & $x^7y$ \Large\strut \\ \hline
$|g|=$ & $1$ & $8$ & $4$ & $8$ & $2$ & $8$ & $4$ & $8$ & $2$ & $2$ & $2$ & $2$ & $2$ & $2$ & $2$ & $2$ \Large\strut \\ \hline
\end{tabular}

\vspace{0.05in}
   
   
   \item What is $\langle x^6 \rangle$ in $D_8$?
   
\vspace{0.05in}

Notice $(x^6)^1=x^6$, $(x^6)^2=x^{12}=x^4$, $(x^6)^3=x^{18}=x^2$, $(x^6)^4=x^{24}=1$ so \fbox{$\langle x^6 \rangle = \{ 1,x^2,x^4,x^6 \}$}.

\vspace{0.05in}
  
   \item Fill in the following rows of the Cayley table for $D_4$:
  
  \force \hfill
   \begin{tabular}{c||c|c|c|c|c|c|c|c|c|c|c|c|} 
              &  $1$ & $x$ & $x^2$ & $x^3$ & $y$ & $xy$ & $x^2y$ & $x^3y$ \\ \hline \hline  
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\ \hline
$x^3$ & $x^3$  & $1$  & $x$  & $x^2$  & $x^3y$  & $y$  & $xy$  & $x^2y$ \Large\strut \\ \hline
$y$ & $y$  & $x^3y$  & $x^2y$  & $xy$  & $1$  & $x^3$  & $x^2$  & $x$ \Large\strut \\ \hline
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\ \hline
   \end{tabular}
  \hfill \force
  
  
 \item Is $H = \{ 1, y, x^2y \}$ a subgroup of $D_4$? Why or why not?
 
\vspace{0.05in}

\fbox{No.} This is not a subgroup since closure fails. Notice that $x^2y \cdot y = x^2y^2=x^2 \cdot 1=x^2 \not\in H$.

\vspace{0.05in}
 
\end{enumerate}

\noindent {\bf\large 5. (20 points)} Proofs!

\begin{enumerate}[(a)]
   \item Choose one of the following:  \qquad \underline{Assume $G$ is a group under multiplication with identity $1$.}
   \begin{enumerate}[I.] 
   \item Suppose that $g=g^{-1}$ for all $g \in G$. Prove that $G$ is abelian.  

\vspace{0.05in}

Suppose $a,b \in G$. Then $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$ where $ab=(ab)^{-1}$, $b^{-1}=b$, and $a^{-1}=a$ follow from our hypothesis that every element is its own inverse and $(ab)^{-1}=b^{-1}a^{-1}$ by the so-called socks shoes principle. Since $ab=ba$ for all $a,b \in G$, we have that $G$ is abelian.

\vspace{0.05in}

   \item Suppose that $G=\langle g \rangle$ is a cyclic group. Prove that $G$ is abelian.

\vspace{0.05in}

Suppose $a,b \in G=\langle g \rangle$. Then there exists $k,\ell \in \mathbb{Z}$ such that $a=g^k$ and $b=g^\ell$. Therefore, $ab=g^kg^\ell=g^{k+\ell}=g^{\ell+k}=g^\ell g^k=ba$. Thus $G$ is abelian.

\vspace{0.05in}

   \end{enumerate}   
   
   \item Choose one of the following:  \qquad (You {\bf must} use a subgroup test in your proof.)
   \begin{enumerate}[I.] 
   \item Prove that $H = 8\mathbb{Z} = \{ 8k \;|\; k \in \mathbb{Z} \}$ is a subgroup of $\mathbb{Z}$.

\vspace{0.05in}

That $H$ is a non-empty subset is obvious ($8$ times an integer is an integer and for example, $0=8(0) \in H$). We need to show closure and closure under inverses.
Suppose $x,y \in H$. Then there exists $k, \ell \in \mathbb{Z}$ such that $x=8k$ and $y=8\ell$. Notice that $x+y=8k+8\ell=8(k+\ell) \in H$ since $k+\ell \in \mathbb{Z}$. Also, $-x =-8k=8(-k) \in H$ since $-k \in \mathbb{Z}$. Therefore, $H$ is closed under addition and under (additive) inverses. Thus $H$ is a subgroup of $\mathbb{Z}$.

Alternatively, we could have just check closure under subtraction (i.e., the one step test): $x-y=8k-8\ell=8(k-\ell) \in H$ since $k-\ell \in \mathbb{Z}$.

\vspace{0.05in}

   \item Prove that $K = \left\{ \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \;\Bigg|\; a,b \in \mathbb{R}_{\not=0} \right\}$ is a subgroup of $\mathrm{GL}_2(\mathbb{R})$.

\vspace{0.05in}

Let $A \in K$ then $A = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$ where $a,b \in \mathbb{R}_{\not=0}$. Notice that $\det(A)=ab\not=0$ since $a,b \not=0$. Therefore, $A$ is invertible and so $A \in \mathrm{GL}_2(\mathbb{R})$. This shows that $K$ is a subset of $\mathrm{GL}_2(\mathbb{R})$. Obviously, $K$ is not empty. For example, the identity matrix belongs to $K$.

We need to check if $K$ is closed under matrix multiplication and inverses. Let $A,B \in K$. Then there are $a,b,x,y \in \mathbb{R}_{\not=0}$ such that $A = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$ and $B= \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix}$. We have that $AB = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix} = \begin{bmatrix} ax & 0 \\ 0 & by \end{bmatrix} \in K$ since $ax \not=0$ and $by \not=0$ (and $AB$ is diagonal). Also, $A^{-1} =  \begin{bmatrix} 1/a & 0 \\ 0 & 1/b \end{bmatrix} \in K$ since $1/a \not=0$ and $1/b \not=0$ (and $A^{-1}$ is diagonal). Therefore, $K$ is a subgroup of $\mathrm{GL}_2(\mathbb{R})$.


\vspace{0.05in}

   \end{enumerate}
   
\end{enumerate}


\end{document}


