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\begin{document}

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\noindent
\parbox{1.95in}{\bf Math 3110}
\hfill {\Large \bf Homework \#1} \hfill
\parbox{1.95in}{\bf \hfill Due: Fri., Jan. 26${}^\mathbf{th}$, 2024}

%\vspace{0.2in}
%
%\noindent {\bf\large Name:} \quad \underline{\large\qquad This is my Name \qquad}

\vspace{0.05in}

\noindent
Please remember when submitting any work via email or in person to\dots\\
\force \hfill \fbox{\fbox{\bf\large\color{red}PUT YOUR NAME ON YOUR WORK!}} \hfill \force


\vspace{0.2in}

\begin{description}
\item[\#1.] Group axiom basics.
\begin{enumerate}[(a)]
\item We know how to add, subtract, multiply, and divide real numbers: $\mathbb{R}$. In addition, we now know that $\mathbb{R}$ is a group under addition. Explain why each of these other operations fails to make $\mathbb{R}$ into a group.\\ \force \hfill {\bf Give concrete counterexamples.}

{\it Example:} If I wanted to show $\mathbb{Z}$ is not a group under multiplication, I would say something like ``The integers, $\mathbb{Z}$, do not form a group under multiplication because of a lack of multiplicative inverses. For example, $2 \in \mathbb{Z}$ but $2^{-1}=\frac12 \not\in \mathbb{Z}$.''

\vspace{0.05in}

\item Let's equip $\mathbb{Z}$ with a really weird operation: $x \circ y = x+y+3$. For example, $-1 \circ 5 = -1+5+3=7$ and $(x\circ y)\circ z = (x+y+3)\circ z = (x+y+3)+z+3 = x+y+z+6$. 

Obviously, $\mathbb{Z}$ is closed under $\circ$. Show $\mathbb{Z}$ is a group under this operation.\\ \force \hfill You need to show $\circ$ is associative, has an identity, and elements have inverses. 

Also, is $\mathbb{Z}$ equipped with $\circ$ an abelian group?

\end{enumerate}
   
\vspace{0.05in}

\item[\#2.] Let $G$ be a group with identity $e \in G$. In general, the law of exponents $(ab)^n=a^nb^n$ may fail to hold.
   \begin{enumerate}[(a)]
   \item Give a concrete example of a group $G$ and elements $a,b \in G$ where $(ab)^2 \not= a^2b^2$.
   
   \vspace{0.05in}
   
   \item Prove $G$ is an abelian group if and only if for all $a,b \in G$, $(ab)^2=a^2b^2$.
   
   {\it Note:} This is an ``if and only if'' statement. You need to prove two implications.
   \end{enumerate}
   
\vspace{0.05in}

\item[\#3.] Consider the dihedral group $D_6 = \{ R_{0^\circ},R_{60^\circ},R_{120^\circ},R_{180^\circ},R_{240^\circ},R_{300^\circ},V_1,V_2,V_3,V_4,V_5,V_6 \}$ (symmetries of a regular hexagon). [Rotations are done counter-clockwise and reflections are labeled in the picture below.]

\force \hfill
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\draw (-0*360/6+30:1) node[vertex] (v0) {$\,$};
\draw (-1*360/6+30:1) node[vertex] (v1) {$\,$};
\draw (-2*360/6+30:1) node[vertex] (v2) {$\,$};
\draw (-3*360/6+30:1) node[vertex] (v3) {$\,$};
\draw (-4*360/6+30:1) node[vertex] (v4) {$\,$};
\draw (-5*360/6+30:1) node[vertex] (v5) {$\,$};

\draw (-0*360/12+30:2) node[vertex] (w0) {$\,$};
\draw (-1*360/12+30:2) node[vertex] (w1) {$\,$};
\draw (-2*360/12+30:2) node[vertex] (w2) {$\,$};
\draw (-3*360/12+30:2) node[vertex] (w3) {$\,$};
\draw (-4*360/12+30:2) node[vertex] (w4) {$\,$};
\draw (-5*360/12+30:2) node[vertex] (w5) {$\,$};
\draw (-6*360/12+30:2) node[vertex] (w6) {$\,$};
\draw (-7*360/12+30:2) node[vertex] (w7) {$\,$};
\draw (-8*360/12+30:2) node[vertex] (w8) {$\,$};
\draw (-9*360/12+30:2) node[vertex] (w9) {$\,$};
\draw (-10*360/12+30:2) node[vertex] (w10) {$\,$};
\draw (-11*360/12+30:2) node[vertex] (w11) {$\,$};

\draw (-1*360/12+30:2.5) node (l0) {$V_1$};
\draw (0*360/12+30:2.5) node (l1) {$V_2$};
\draw (1*360/12+30:2.5) node (l2) {$V_3$};
\draw (2*360/12+30:2.5) node (l3) {$V_4$};
\draw (3*360/12+30:2.5) node (l4) {$V_5$};
\draw (4*360/12+30:2.5) node (l5) {$V_6$};

\draw (v0) -- (v1);
\draw (v1) -- (v2);
\draw (v2) -- (v3);
\draw (v3) -- (v4);
\draw (v4) -- (v5);
\draw (v5) -- (v0);

\draw [dotted] (w0)--(w6);
\draw [dotted] (w1)--(w7);
\draw [dotted] (w2)--(w8);
\draw [dotted] (w3)--(w9);
\draw [dotted] (w4)--(w10);
\draw [dotted] (w5)--(w11);

\draw[->,thick] (1.5,-1.75) arc (-55:0:1.25) ;
\draw (1.75,-2.75) node (rot) {$R_{0^\circ},R_{60^\circ},R_{120^\circ},\dots$};

\end{tikzpicture}
\hfill \force

\begin{enumerate}[(a)]
\item Compute $V_1R_{60^\circ}$, $R_{180^\circ}V_4$, and $V_3V_6$. [Draw some pictures!]
\item Is $D_6$ Abelian? Why or why not?
\item Make a table of inverses and orders for each element:

\begin{tabular}{|rc||c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 
{\Large\strut} Element: & $g=$ & $R_{0^\circ}$ & $R_{60^\circ}$ & $R_{120^\circ}$ & $R_{180^\circ}$ & $R_{240^\circ}$ & $R_{300^\circ}$ & $V_1$ & $V_2$ & $V_3$ & $V_4$ & $V_5$  & $V_6$ \\ \hline \hline
{\Large\strut} Inverse: & $g^{-1}=$ & ??? & $\cdots$ & & & & & & & & & & \\ \hline
{\Large\strut} Order: & $|g|=$ & ??? & $\cdots$  & & & & & & & & & & \\ \hline
\end{tabular}

\vspace{0.05in}

{\it Note:} Recall that the order of an element $g$ is the smallest positive power $n$ such that $g^n$ is the identity.\\ \phantom{{\it Note:} }If no such power exists the order of $g$ is $\infty$.

\end{enumerate}

\end{description}

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