\documentclass{article}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{enumerate}
\usepackage{ifthen}
\usepackage{ulem} % for strikethrough \sout{...} command
\usepackage{color}

\usepackage{wasysym}

\setlength{\unitlength}{0.1in}

\setlength{\oddsidemargin}{-0.25in}
\setlength{\evensidemargin}{-0.25in}
\setlength{\topmargin}{-0.25in}
\setlength{\headheight}{0.0in}
\setlength{\headsep}{0.0in}
\setlength{\topskip}{0.0in}
\setlength{\textheight}{9.75in}
\setlength{\textwidth}{7in}

\newcommand{\bdot}{\, {\scriptstyle{\bullet}} \,}
\newcommand{\B}[1]{{\bf {#1}}}
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}

\newcommand{\myspace}{\parbox[c][0.25in]{0.25in}{\hfill}}
\newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}}
\newcommand{\force}{{\noindent $\mbox{\;}$}}

\newcommand{\diagquotient}[2]{\displaystyle{{#1 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop #2}}}

\begin{document}

\pagestyle{empty}

\noindent
\parbox{2.25in}{\bf Math 4010-101}
\hfill {\Large \bf Homework \#1} \hfill
\parbox{2.25in}{\bf \hfill Due: Tue., Jan. $21^\mathrm{st}$, 2014}

\vspace{0.2in}

\begin{enumerate}
\item Consider $G = \langle x,y,z \;|\; x^3=1,y^2=1,z=1,xyxy=1 \rangle$. What is the order of $G$? Write down a Cayley table for $G$. What familiar group is $G$ isomorphic to?

\vspace{0.1in}

\item Prove $G = \langle x,y \;|\; xy = yx \rangle$ is an infinite group. [You must use the universal property.]

\vspace{0.1in}

\item {}[Grad Problem] Let $X_1$ and $X_2$ be arbitrary sets (they could be empty, finite, or infinite). Prove that $|X_1|=|X_2|$ (i.e. there is a bijection between $X_1$ and $X_2$) implies that $F(X_1) \cong F(X_2)$.

This means that sets of the same ``size'' yield isomorphic free groups.  The converse of this statement is also true: If $F(X_1) \cong F(X_2)$ then $|X_1| = |X_2|$. However, this implication is a bit harder to prove (challenge problem). 

We know that subgroups of free groups are themselves free (this is the difficult to prove Nielsen-Schreier theorem). So if $H$ is a subgroup of $F(X)$, then there is some set $Y$ such that $H=F(Y)$. It is interesting to note that if $F(Y)$ is a subgroup of $F(X)$, then it is {\bf not necessarily true} that $|Y| \leq |X|$!

\end{enumerate}

\end{document}