\documentclass[11pt]{article} % Define the document type. The 11pt refers to font size.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Include packages required to compile document:
\usepackage{amssymb, amsthm, amsmath} % These packages are used for the basic math symbols and environments.
\usepackage{enumerate} % This package is used to provide various types of numbering for lists.
\usepackage[margin=1.25in]{geometry} % This is to set wide margins, so there is room to leave comments
\usepackage[parfill]{parskip} % This provides formatting for paragraphs so that there are no indents and paragraphs are separated by a line skip.
\usepackage[pdftex,colorlinks=true,urlcolor=blue]{hyperref}
% You may need to add additional packages depending on the commands you want to use.
%%%%%%%%%%%%%%%%%%%%
% User defined commands:
\newcommand{\br}{\mathbb R} % this is the symbol for the real numbers. mathbb is the "blackboard bold" font
\newcommand{\bz}{\mathbb Z} % integers
\newcommand{\bq}{\mathbb Q} % rationals
\newcommand{\bn}{\mathbb N} % natural numbers
\newcommand{\co}{\colon\thinspace} % this is a good way to write functions, i.e instead of $f: X \to Y$, you would write $f\co X \to Y$. It looks nicer.
\newcommand{\ssm}{\smallsetminus} % A nice way of writing set subtraction
% You may add shortcuts for any commands you find yourself using frequently.
\DeclareMathOperator{\ord}{ord} % This command lets you declare math operators. Some built in operators are \sin for sine and \cos for cosine. This makes the operators appear nicely in the math environment. If I wanted to make the sine operator myself, it would look like \DeclareMathOperator{\sin}{sine}.
% Ender user defined commands
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% These establish different environments for stating Theorems, Lemmas, Remarks, etc.
\newtheorem{Thm}{Theorem}
\newtheorem{Prop}[Thm]{Proposition}
\newtheorem{Lem}[Thm]{Lemma}
\newtheorem{Cor}[Thm]{Corollary}
\theoremstyle{definition}
\newtheorem{Def}[Thm]{Definition}
\theoremstyle{remark}
\newtheorem{Rem}[Thm]{Remark}
\newtheorem{Ex}[Thm]{Example}
\theoremstyle{definition}
\newtheorem{Exercise}{Exercise}
\newtheorem{ExGraded}[Exercise]{*Exercise}
\newtheorem{ExChallenge}[Exercise]{**Exercise}
\newenvironment{Solution}{\noindent{\it Solution.}}{\qed}
%End environments
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Now we're ready to start
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
% This is for the title
{\Large {\bf Homework 5}} \hfill MATH 301/601 \\
{Solutions to Graded Problems\\[-2mm]
\line(1,0){470}
\medskip
\setcounter{Exercise}2
\bigskip
\begin{Exercise}
Suppose \( H \) is a nonempty finite subset of a group \( G \) and that \( H \) is closed under multiplication (that is, \( ab \in H \) for all \( a, b \in H \)).
Prove that \( H \) is a subgroup of \( G \).
\end{Exercise}
\begin{Solution}
As \( H \) is closed under multiplication, it is enough to show that \( H \) contains the identity of \( G \) and that \( H \) is closed under inversion.
We know \( H \) is nonempty, so let \( h \in H \).
Using the fact that \( H \) is finite, we can find \( j,k \in \bz \) such that \( j < k \) and \( h^j = h^k \).
It follows that \( h^{k-j} = e \).
Now, as \( H \) is closed under multiplication and \( k-j \in \bn \), we have that \( e = h^{k-j} \in H \).
Also, observe that \( h^{k-j-1} = h^{-1} \), and as \( k-j-1 \in \bn \cup \{0\} \), we have that \( h^{-1} \in H \).
As \( h \) was an arbitrary element of \( H \), we see that \( H \) is closed under taking inverses.
Therefore, \( H \) is a subgroup of \( G \).
\end{Solution}
\setcounter{Exercise}5
\medskip
\begin{Exercise}
Suppose \( G \) is a nontrivial group in which the only two subgroups of \( G \) are itself and the trivial subgroup.
\begin{enumerate}[(a)]
\item Prove that \( G \) is cyclic.
\item Using part (a), prove that \( G \) is a finite group of prime order.
\end{enumerate}
\end{Exercise}
\begin{Solution}
(a) As \( G \) is nontrivial, there exists \( g \in G \ssm \{e\} \).
Since \( g \) is not the identity, the cyclic subgroup generated by \( g \) is nontrivial, and hence, as \( G \) has only two subgroups, the cyclic subgroup generated by \( g \) must be all of \( G \).
Therefore, \( G \) is cyclic.
In fact, we have shown that every element of \( G \ssm \{e\} \) is a generator of \( G \).
(b) We have already established that \( G \) is cyclic, so we may choose a generator of \( G \), call it \( a \).
If \( a^2 = e \), then \( |G| = 2 \), and hence \( G \) is a finite group of prime order.
We can therefore assume that \( a^2 \neq e \), and hence, the cyclic subgroup generated by \( a^2 \) is equal to \( G \).
Therefore, there exists \( k \in \bz \) such that \( (a^2)^k = a \), which implies that \( a^{2k-1} = e \).
Hence, \( G \) has finite order.
Now, suppose that \( n \in \bn \ssm \{1\} \) divides \( |G| \).
We want to show that \( n = |G| \).
Let \( q \in \bz \) such that \( |G| = nq \).
As \( n > 1 \), we must have that \( q < |G| \); hence, \( a^q \neq e \) and \( |a^q| = |G| \).
But, \( (a^q)^n = (a^{qn}) = a^{|G|} = e \), which implies that \( n \geq |G| \).
Therefore, as \( n \mid |G| \), we must have that \( n = |G| \).
We have shown that the only divisors of \( |G| \) are 1 and itself, and so \( |G| \) is prime.
\end{Solution}
\end{document}