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\noindent
\parbox{2in}{\bf Math 3110} 
\hfill {\Large \bf Test \#1} \hfill
\parbox{2in}{\bf \hfill September $16^{\mathrm{th}}$, 2020}

\vspace{0.3in}

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

\vspace{0.2in}

\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:]
             \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 odd integers $\mathbb{O} = 2\mathbb{Z}+1 = \{ 2k+1 \;|\; k \in \mathbb{Z} \}$ do not form a group under either addition or multiplication. Explain why they fail be a group under both of these operations. 

The odd integers do not form a group under addition since closure fails. For example, $1 \in \mathbb{O}$ but $1+1=2 \not\in\mathbb{O}$. 
      Alternatively, they don't form a group under addition because they have no additive identity ($0$ is even). 
      
      On the other hand, associativity holds and the additive inverse of an odd number is odd so those axioms kind of hold.

The odd integers do not form a group under multiplication since the multiplicative inverse of an odd integer is not necessarily an integer at all. For example, $3 \in \mathbb{O}$ but $3^{-1} = 1/3 \not\in \mathbb{O}$.
   
   On the other hand, we do have closure (odd times odd is odd), associativity, and a multiplicative identity ($1$ is odd). 
   
   \item On the other hand, the even integers $\mathbb{E}=2\mathbb{Z} = \{ 2k \;|\; k \in\mathbb{Z}\}$ do form a group under one of these operations. Briefly explain why it is a group under that operation and why it isn't a group under the other operation.

\vspace{0.05in}

The even integers form a subgroup of the group of integers under addition since they form a non-empty subset of $\mathbb{Z}$, they are closed under addition (even plus even is even), and they are closed under additive inverses (the negative of an even number is even). 

However, the even integers are not a group under multiplication since the multiplicative inverse of an even integer may not even be an integer. For example, $2 \in \mathbb{E}$ but $2^{-1} = 1/2 \not\in \mathbb{E}$.  Alternatively, the even integers lack a multiplicative identity ($1$ is odd not even). On the other hand, the even integers are closed under multiplication (even times even is even) and multiplication is associative.

\vspace{0.05in}

   \item Why aren't the integers $\mathbb{Z}$ a group under subtraction?

\vspace{0.05in}

We do have closure under subtraction, but the associativity and identity axioms do not hold (and without a clear identity we really can't sensibly discuss inverses). For example, $1-(2-3)=1-(-1)=2$ while $(1-2)-3=-1-3=-4$ (associtvitiy fails). Alternatively, given some $n \in \mathbb{Z}$, if we have $n-x=n$ then $x=0$ but $0-n=-n$ (which doesn't always equal $n$). Thus we have a right identity (i.e., $0$) but not a two-sided identity. 

\vspace{0.05in}

\end{enumerate}

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

   \begin{enumerate}[(a)]

      \item Make a table listing the elements of $\mathbb{Z}_{9}$, their inverses, and their orders.

\vspace{0.05in}

First, recall that $\mathbb{Z}_9$ is a group under addition modulo $9$. Thus we refer to an {\it additive} identity (i.e., $0$) and {\it additive} inverses (i.e., negatives).
Next, if $k$ is relatively prime to $9$, then $\underbrace{k+k+\cdots+k}_{n-\mathrm{times}} \not=0$ mod $9$ unless $n$ is a multiple of $9$. Thus the order of such a $k$ will be $9$. On the other hand, the order of the identity is $1$ (i.e., $|0|=1$) always. Finally, $3+3+3=9=0$ mod $9$ while $3+3=6 \not=0$ mod $9$, so $|3|=3$. Likewise, $6+6+6=18=0$ mod $9$ whereas $6+6=12=3 \not=0$ mod $9$, so $|6|=3$ as well.


\vspace{0.05in}

\force \hfill
\begin{tabular}{|c||c|c|c|c|c|c|c|c|c|} \hline
$x=$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8  {\large\strut} \\ \hline \hline
$-x=$ & $-0=0$ & $-1=8$ & $-2=7$ & $-3=6$ & $-4=5$ & $-5=4$ & $-6=3$ & $-7=2$ & $-8=1$  {\large\strut} \\ \hline
$|x|=$ & $1$ & $9$ & $9$ & $3$ & $9$ & $9$ & $3$ & $9$ & $9$  {\large\strut} \\ \hline
\end{tabular}
\hfill \force

\vspace{0.05in}

      \item Do the same for $U(9)$.

\vspace{0.05in}

First, recall that $U(9) = \{ k \in \mathbb{Z}_9 \;|\; k \text{ and } 9 \text{ are relatively prime} \} = \{1,2,4,5,7,8\}$ is a group under {\it multiplication} modulo $9$. Thus we refer to a {\it multiplicative} identity (i.e., $1$) and {\it multiplicative} inverse (i.e., $x^{-1}$ is a number such that $xx^{-1}=1$ mod $9$).  We can figure out inverses and orders by brute force. 
Note that $1^{-1}=1$ and $|1|=1$ since it's the identity. Next, $2^1=2$, $2^2=4$, $2^3=8$, $2^4=16=7$, $2^5=14=5$, $2^6=10=1$ (mod $9$) so that $|2|=6$ and $2^{-1}=2^5=5$ (since $2\cdot 5=10=1$ mod $9$). Then $4^1=4$, $4^2=16=7$, $4^3=28=1$ (mod $9$) so that $|4|=3$ and $4^{-1}=4^2=7$. As a shortcut, we also have $5^{-1}=2$ since $2^{-1}=5$ as well as $|5|=|2^{-1}|=|2|=6$. Likewise, $7^{-1}=4$ and $|7|=|4^{-1}|=|4|=3$. Finally, $8^1=8$, $8^2=64=1$ (mod $9$) or we could notice that $8=-1$ (mod $9$) so of course $8^2=(-1)^2=1$ mod $9$. Thus $8^{-1}=8$ and $|8|=2$.

\vspace{-0.1in}

\force\hfill
\begin{tabular}{|c||c|c|c|c|c|c|} \hline
$x=$ & 1 & 2 & 4 & 5 & 7 & 8  {\large\strut} \\ \hline \hline
$x^{-1}=$ & $1$ & $5$ & $7$ & $2$ & $4$ & $8$ {\large\strut} \\ \hline
$|x|=$ & $1$ & $6$ & $3$ & $6$ & $3$ & $2$ {\large\strut} \\ \hline
\end{tabular}
\hfill\force

\vspace{0.05in}

      \item Compute $2^{-1}(3-8)+7$ mod $9$ or explain why this is undefined.

\vspace{0.05in}

$$2^{-1}(3-8)+7=5(3-8)+7 = 5(-5)+7=-25+7=-18= \fbox{$0$} \text{ mod }9$$

\vspace{0.05in}

      \item Compute $3^{-1}(7-3)-11$ mod $9$ or explain why this is undefined.

\vspace{0.05in}

Since $\mathrm{gcd}(3,9)=3 \not= 1$, $3^{-1}$ (mod $9$) does not exist (or noting that $3 \not\in U(9)$). Thus this result is undefined.

\vspace{0.05in}

\end{enumerate}


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

\begin{enumerate}[(a)]


      \item List all of the distinct cyclic subgroups of $\mathbb{Z}_{12}$. Show each subgroup's contents (e.g., $\langle 0 \rangle = \{0\}$).
      
\vspace{0.05in}

\begin{itemize}
\item $\langle 0 \rangle = \{0\}$
\item $\langle 1 \rangle = \{0,1,2,3,4,5,6,7,8,9,10,11 \}$ \quad ($=\langle 5 \rangle=\langle 7 \rangle=\langle 11 \rangle$) 
\item $\langle 2 \rangle = \{0,2,4,6,8,10\}$ \quad ($=\langle 10 \rangle$)
\item $\langle 3 \rangle=\{0,3,6,9\}$ \quad ($=\langle 9 \rangle$)
\item $\langle 4 \rangle=\{0,4,8\}$ \quad ($=\langle 8 \rangle$)
\item $\langle 6 \rangle=\{0,6\}$
\end{itemize}

\vspace{0.05in}

      \item Draw the subgroup lattice for $\mathbb{Z}_{12}$.

\vspace{-0.75in}

\force \hfill \hfill
\begin{tikzpicture}[node distance=1.5cm]
\node(12){$12$};
\node(6)[below left of = 12]{$6$};
\node(4)[below right of = 12]{$4$};
\node(3)[below of = 6]{$3$};
\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){$\mathbb{Z}_{12} = \langle 1 \rangle$};
\node(6)[below left of = 12]{$\langle 2 \rangle$};
\node(4)[below right of = 12]{$\langle 3 \rangle$};
\node(3)[below of = 6]{$\langle 4 \rangle$};
\node(2)[below of = 4]{$\langle 6 \rangle$};
\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

\force \hfill \hfill Divisibility Lattice \hspace{0.55in} Subgroup Lattice \hfill \force


\vspace{0.05in}

    \item Find $\displaystyle \left\langle A \right\rangle$  in $\mathrm{GL}_2(\mathbb{Z}_6)$ where $\displaystyle A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$. What is the order of $A$? What is $A^{-1}$?

\noindent
{\color{red}\bf Embarrassing Error:} I changed $A$ at the last moment to its current definition. However, $\det(A) = 1(1)-2(2) = -3 =3$ (mod $6$) so $\det(A)^{-1}=3^{-1}$ does not exist in $\mathbb{Z}_6$. Thus $A \not\in \mathrm{GL}_2(\mathbb{Z}_6)$ since $A^{-1}$ (working mod 6) does not exist. So this part of the problem is \underline{\bf\color{red}nonsense}! 

Instead let's change to $A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$ and rework this part.

\vspace{0.05in}

$$A^1 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}, \quad A^2 =\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}, \quad A^3=A^2A=\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 6 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Therefore, $\langle A \rangle = \{ I_2, A, A^2\} = \left\{ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix},\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \right\}$, $|A|=3$, and $A^{-1}=A^2 = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}$.

\vspace{0.05in}

   \item Find $10^{-1}$ mod 27 using the extended Euclidean algorithm [Don't just guess and check].

%\vspace{0.05in}

\begin{itemize}
\item $27$ divided by $10$ yields: $27=10(2)+7$ so that $(-2)10+(1)27=7$
\item $10$ divided by $7$ yields: $10=7(1)+3$ so that $(-1)7+(1)10=3$
\item $7$ divided by $3$ yields: $7=3(2)+1$ so that $(-2)3+(1)7=1$
\item $3$ divided by $1$ yields: $3=3(1)+0$
\end{itemize}

\vspace{-0.05in}

Our last non-zero remainder is $1$. Thus $\mathrm{gcd}(10,27)=1$ (they are relatively prime) and so $10 \in U(27)$ (i.e., $10^{-1}$ does in fact exist).
Going backwards\dots

\vspace{-0.05in}

\begin{itemize}
\item $(-2)3+(1)7=1$
\item Plugging $(-1)7+(1)10=3$ into $3$ in the previous step: $(-2)[(-1)7+(1)10]+(1)7=1$ so $(3)7+(-2)10=1$
\item Plugging $(-2)10+(1)27=7$ into $7$ in the previous step: $(3)[(-2)10+(1)27]+(-2)10=1$ so $(-8)10+(3)27=1$
\end{itemize}

Therefore, $10^{-1} = -8=$ \fbox{$19$} (mod $27$).

\vspace{0.05in}

\end{enumerate}

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

\vspace{0.05in}

It will be helpful to keep in mind that in $D_n$ exponents of $x$ work modulo $n$, exponents of $y$ work modulo $2$ (in particular, $y^{-1}=y$), and $xy=yx^{-1}$ as well as $yx=x^{-1}y$ (i.e., moving a power of $x$ past a single $y$ flips the sign on $x$'s exponent).

\vspace{0.05in}

\begin{enumerate}[(a)]
   \item Use the relations for $D_{6}$ to simplify $y^{-2}x^{11}y^3x^{-4}y^{124}x$
      
\vspace{0.05in}

Let's simplify exponents ($x$'s mod 6 and $y$'s mod 2) and then push $y$'s to the right (flipping $x$'s exponents as needed):

$$ y^{-2}x^{11}y^3x^{-4}y^{124}x = 1x^{5}yx^{2}1x = x^5yx^3=x^5x^{-3}y= \fbox{$x^2y$}$$

   \item Make a table listing the elements of $D_6$, their inverses, and their orders.
   
\vspace{0.05in}

Is it easy to see why the orders of $1$, $x$, $x^2$, and $x^3$ are 1, 6, 3, and 2 respectively. For example: $x^2\not=1$, $(x^2)^2=x^4 \not=1$, but $(x^2)^3=x^6=1$.
Notice that $(x^k)^{-1}= x^{-k}=x^{6-k}$ since exponents of $x$ work mod 6. So, for example, $x^{-1}=x^5$. Next, the order of $g$ and $g^{-1}$ are the same so we get that
the orders of $x^5$ and $x$ match. Likewise, the orders of $x^4$ and $x^2$ match. Finally, the rest of the elements represent reflections. They are their own inverses and have order 2. For example, $x^3yx^3y=x^3x^{-3}yy=x^0y^2=1$ so $(x^3y)^{-1}=x^3y$ and $|x^3y|=2$.

\vspace{0.05in}

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

\vspace{-0.15in}
   
   \item What is $\langle x^4 \rangle$ in $D_6$?
   
\fbox{$\langle x^4 \rangle = \{ 1,x^2,x^4\}$} \quad since \quad $(x^4)^2=x^8=x^2$ and $(x^4)^3=x^{12}=1$ so powers of $x^4$ just cycle through these elements.
  
   \item Fill in the following rows of the Cayley table for $D_6$:
  
For example, the third entry in the first row is $x^5\cdot x^2= x^7=x$ and in the second row it's $y\cdot x^2=x^{-2}y=x^4y$. As another example, the last element in the
first row is $x^5\cdot x^5y=x^{10}y=x^4y$ and in the second row it's $y\cdot x^5y=x^{-5}yy=x$.

  \force\hfill 
   \begin{tabular}{c||c|c|c|c|c|c|c|c|c|c|c|c|} 
              &  $1$ & $x$ & $x^2$ & $x^3$ & $x^4$ & $x^5$ & $y$ & $xy$ & $x^2y$ & $x^3y$ & $x^4y$ & $x^5y$ \\ \hline \hline  
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\ \hline
$x^5$ & $x^5$ & $1$ & $x$ & $x^2$ & $x^3$ & $x^4$ & $x^5y$ & $y$ & $xy$ & $x^2y$ & $x^3y$ & $x^4y$  {\Large\strut} \\ \hline
$y$ & $y$ & $x^5y$ & $x^4y$ & $x^3y$ & $x^2y$ & $xy$ & $1$ & $x^5$ & $x^4$ & $x^3$ & $x^2$ & $x$ {\Large\strut} \\ \hline
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\ \hline
   \end{tabular}
\hfill\force  

\vspace{0.05in}
  
 \item Is $H = \{ 1,x^2,y,x^2y \}$ a subgroup of $D_6$? Why or why not?
 
\vspace{0.05in}

To check if a non-empty {\bf finite} subset of a group is a subgroup we {\bf only} need to {\bf check closure} under the operation. Notice that $x^2,x^2y \in H$ but $x^2 \cdot x^2y = x^4y \not\in H$, so \fbox{$H$ is {\bf not a subgroup} of $D_6$}.

\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 $g^2=1$ for all $g \in G$. Prove that $G$ is abelian.  

\vspace{0.05in}

First, notice that for every $g \in G$ we have that $g^2=1$ so $g^{-1}=g$ (every element is its own inverse). Now suppose that $a,b \in G$. Then $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$ where the first and third equalities following from the fact that every element is its own inverse and the middle equality is the sock-shoes property of inverses: $(ab)^{-1}=b^{-1}a^{-1}$ (always). 
Therefore, since $ab=ba$ for all $a,b\in G$, we have that $G$ is Abelian.

\vspace{0.05in}

   \item For all $g \in G$. Show that $|g|=|g^{-1}|$.

\vspace{0.05in}

Suppose that $g \in G$ and that $k \in \mathbb{Z}$. Next, assume that $g^k=1$, then taking inverses we have $(g^{-1})^k = g^{-k} = (g^k)^{-1}=1^{-1}=1$. So $g^k=1$ implies that $(g^{-1})^k=1$. Conversely, suppose that $(g^{-1})^k=1$, then again taking inverses we have $g^k = (g^{-1})^{-k} = ((g^{-1})^k)^{-1} = 1^{-1}=1$. So $(g^{-1})^k=1$ implies that $g^k=1$.

We have shown that any power of $g$ that equals the identity is also a power of $g^{-1}$ that equals the identity (and vice-versa). Therefore, the smallest positive power to send one to the identity is also the smallest positive power to do the same for the other. Thus they have the same order: $|g|=|g^{-1}|$.

\vspace{0.05in}

{\bf Faster Proof:} Notice that $\langle g \rangle = \{ g^k \;|\; k \in \mathbb{Z} \} = \{ (g^{-1})^{-k} \;|\; k \in \mathbb{Z} \} = \{ (g^{-1})^\ell \;|\; \ell \in \mathbb{Z} \} = \langle g^{-1} \rangle$ since if $k$ ranges over all integers, then so does $\ell=-k$. Therefore, $|g|=|\langle g \rangle|= |\langle g^{-1} \rangle | = |g^{-1}|$.

\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 $H = \{ 10k+6\ell \;|\; k,\ell \in \mathbb{Z} \}$ is a subgroup of $\mathbb{Z}$.

\vspace{0.05in}

First, clearly $H$ is a non-empty subset of $\mathbb{Z}$ (for example, $0=10(0)+6(0) \in H$). Next, let $x,y \in H$. Then there exists some $k,\ell,k',\ell' \in \mathbb{Z}$ such that $x=10k+6\ell$ and $y=10k'+6\ell'$ (this is what is means to belong to $H$). Now $x+y = (10k+6\ell)+(10k'+6\ell') = 10(k+k')+6(\ell+\ell') \in H$ since $k+k'$ and $\ell+\ell'$ are integers. Likewise, $-x=-(10k+6\ell) = 10(-k)+6(-\ell) \in H$ since $-k$ and $-\ell$ are integers. Therefore, $H$ is closed under addition and negation (i.e., additive inverses). Thus $H$ passes the (two-step) subgroup test -- it's a subgroup of $\mathbb{Z}$.

\vspace{0.05in}

{\bf Secret Solution:} Notice that $H=10\mathbb{Z}+6\mathbb{Z}=\mathrm{gcd}(10,6)\mathbb{Z}=2\mathbb{Z}$ (by a basic number theory result flowing from the extended Euclidean algorithm). Thus $H$ is just the even integers. So since even minus even is even, $H$ is a subgroup by the one-step test. Or even better (if we don't have to use a test) $H=2\mathbb{Z} = \langle 2 \rangle$ is a subgroup because cyclic subgroups are subgroups!

\vspace{0.05in}

   \item Prove that $K = \left\{ \begin{bmatrix} 1 & 0 \\ a & 1 \end{bmatrix} \;\Bigg|\; a \in \mathbb{R} \right\}$ is a subgroup of $\mathrm{GL}_2(\mathbb{R})$.

\vspace{0.05in}

Again, it is clear that $K$ is non-empty (just pick a value for $a$ and you'll get something in $K$). Also, notice that for any $a\in\mathbb{R}$, we have $\det\left(\begin{bmatrix} 1 & 0 \\ a & 1 \end{bmatrix}\right)=1\cdot 1-a \cdot 0 =1 \not=0$ so $K$ is a subset of $\mathrm{GL}_2(\mathbb{R})$ (in fact -- not that it matters -- it's a subset of $\mathrm{SL}_2(\mathbb{R})$).

Next, suppose that $A,B \in K$. Then there exists $a,b \in\mathbb{R}$ such that $A=\begin{bmatrix} 1 & 0 \\ a & 1 \end{bmatrix}$ and $B=\begin{bmatrix} 1 & 0 \\ b & 1 \end{bmatrix}$.
Notice that $AB = \begin{bmatrix} 1 & 0 \\ a & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ b & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ a+b & 1 \end{bmatrix} \in K$ since $a+b \in \mathbb{R}$ and we have 1's on the diagonal and a 0 in the upper-right hand corner (i.e., $AB$ ``looks right''). Finally, $A^{-1} = (\det(A))^{-1}\begin{bmatrix} 1 & -0 \\ -a & 1 \end{bmatrix} = (1(1)-0(a))^{-1}\begin{bmatrix} 1 & 0 \\ -a & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ -a & 1 \end{bmatrix} \in K$ since $-a \in\mathbb{R}$ and the rest of the entries ``look right''. Therefore, $K$ is a subgroup of $\mathrm{GL}_2(\mathbb{R})$ by our two-step subgroup test.

\vspace{0.05in}

{\bf Forward Looking Observation:} It it interesting to note that multiplying elements of $K$ just amounts to adding the $(2,1)$-entries of the matrices and finding inverses just amounts to negating the $(2,1)$-entry. We can see that the {\it group structure} of $K$ is the same as the group structure of $\mathbb{R}$ (under addition). Formally we could write: $K \cong \mathbb{R}$ (read: $K$ is isomorphic to the real numbers under addition). We will discuss isomorphisms in the near future.

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