\documentclass[12pt]{article}
\newcommand{\set}[1]{\lbrace #1 \rbrace}
\newcommand{\setc}[2]{\lbrace #1 \mid #2 \rbrace}
\usepackage{amssymb,amsfonts,amsmath}
\pagestyle{myheadings} \markright{Battaglino}
\newenvironment{hwproblem}[1]
{\noindent{\bf{Problem #1:}} }
{\vspace{0.2 in}}

\newenvironment{solution}
{\noindent{\underline{Solution:}} }
{\vspace{0.2 in}}
\begin{document}
\begin{center}
\Large
21-373: Algebraic Structures\\
Extra Credit Assignment
\end{center}

\begin{flushright}
Peter Battaglino\\ 18. February, 2005
\end{flushright}

\noindent
Let a group $G$ be given with the following property:
\[
\forall a,b\in G,\; \langle a \rangle \subset \langle b \rangle \quad
{\mathrm{or}} \quad \langle b \rangle \subset \langle a \rangle; \quad (*)
\]
in other words, the subgroups of $G$ generated by single
elements of $G$ are linearly ordered by inclusion.
\\\\
\begin{hwproblem}{1}
If $G$ is a finite group, show that $G$ is cyclic (and, hence, that every subgroup of $G$ is cyclic). Establish
also that there exist positive integers $m$ and $n$ such that
\[
\setc{|H|}{H\leqslant G} = \setc{n^k}{k=1,\cdots,m}.
\]
Hint: Show first that for each positive integer $q$ there is at most one subgroup of the form $\langle a \rangle$
of order $q$. Then choose an element of $G$ (other than $e$) of lowest order.
\end{hwproblem}

\begin{solution}
Suppose that the group $G$ is finite. Then we can represent $G$ as the set
\begin{equation}
G = \set{g_1, g_2, \cdots, g_n},
\end{equation}
where
\begin{equation}
\langle g_1 \rangle \subset \langle g_2 \rangle \subset \cdots \subset \langle g_n \rangle.
\end{equation}
Since $g_i\in \langle g_i \rangle$ for all $i\in \set{1,\cdots,n}$, we see that the group $\langle g_n \rangle$
contains every element of $G$ and is generated by the element $g_n\in G$. Thus we see that $G$ is cyclic.
All the subgroups of $G$ are cyclic, so each one is generated by an element of $G$. In addition, the orders
of subgroups of $G$ form a bounded set of positive integers. This bounded set must have a smallest element $n$
besides 1 (1 corresponding to the identity  $e$ of $G$). In addition, according to theorem 4.3 of our text, the
element $n$ corresponds to a unique subgroup $\langle a \rangle \in G$ with order $n$. It is also a direct
consequence of this theorem that every subgroup of $G$ has an order that divides the order of $G$. We see
immediately that there is an element $b\in G$ with order $|G|/n$. Now, since the element $a$ generates the
unique subgroup of $G$ with smallest order, and by property (*), we know that there are no elements of $G$
with order between $|b|$ and $|G|$. That is, the order of $b$ is the next-largest order, besides $|G|$ of all
possible orders of subgroups of $G$. In addition, since $a$ is the unique element with smallest order,
$\langle a \rangle \subset \langle b \rangle$. This means, once again by theorem 4.3, that $|a| \mid |b|$, so
we can find an element $c\in G$ such that $|a||c|=|b|$. But $|b|=|G|/n$ and $|a|=n$, so this means that
$|c|=|G/n^2$. This order is the third-largest order of any subgroup of $G$. We can see that if we continue
this process, we will exhaust the subgroups of $G$ from largest order to smallest. We will eventually arrive
at an element of $G$ whose order is as small as possible without being as small as $|a|=n$. If we call this
element $f$, then we still satisfy $\langle a \rangle \subset \langle f \rangle$, so we have $n\mid |f|$.
By construction, there exists a positive integer $m$ such that $|f|=|G|/n^{m-2}$. But since $n\mid |f|$, there
exists a positive integer $q$ such that $nq=|f|$, so there must exist yet a smaller subgroup of $G$ with
order $q$. But since there are no subgroups contained in $\langle f \rangle$ and containing $\langle a \rangle$
by supposition, we must have that $q=n$, so $|f|=n^2=|G|/n^{m-2}$, which means that $|G|=n^m$. Since we have
exhausted all of the subgroups of $G$, and each of them has order of the form $n^k$, where $k$ is an integer
between 1 and $m$ inclusive, we have shown what we set out to prove.
\end{solution}
\clearpage
\begin{hwproblem}{2}
If $G$ is an infinite group, show that $G$ has no proper infinite subgroup and that every finite
subgroup of $G$ is cyclic. Hint: It helps first to show that every element of $G$ has finite order.
\end{hwproblem}

\begin{solution}

\end{solution}
\clearpage
\begin{hwproblem}{3}
Find an example of an infinite group $G$ with the property (*). Is $G$ cyclic?
\end{hwproblem}

\begin{solution}
Consider the group $G$ of rotations about angles of the form $m\pi/2^n$, where $m,n\in\mathbb{Z}$. It is
simple to show that this is a group. Upon successive rotations of $m\pi/2^n$ and $p\pi/2^q$, the total
rotation is $(2^q m+2^n p)\pi/2^{n+q}$, which is a rotation in our group. The identity is the zero rotation,
and every rotation has an inverse.
We see that each subgroup of $G$ has an order of the form $2^k$, where $k\in\set{1,\cdots,m}$, and
$m$ is a positive integer, and there is only one unique subgroup of any given order. The (*) property is exhibited
by the following fact: any rotation of $m\pi/2^n$ is in the subgroup generated by the rotation of $\pi/2^n$, and
is also in the subgroup generated by $\pi/2^{n+1}$, etc. Every element of the group has finite order (as would
any rotation of a rational multiple of $\pi$),
so it follows that none of the elements of the group is capable of generating the entire (infinite) group. Thus,
$G$ cannot be cyclic.
\end{solution}
\end{document}
