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\begin{document}

\begin{center}
	\LARGE{Math 31 - Homework 3}
	
	\vspace{2mm}
	\large{Due Friday, July 13}
\end{center}

\subsection*{Easy}
\paragraph{1.} Let $G$ be a group of order $pq$, where $p$ and $q$ are prime numbers. Show that every \emph{proper} subgroup of $G$ is cyclic.

\paragraph{2.} We proved in class that every subgroup of a cyclic group is cyclic. The following statement is almost the converse of this:
	\begin{center}
		``Let $G$ be a group. If every \emph{proper} subgroup of $G$ is cyclic, then $G$ is cyclic.''
	\end{center}
Find a counterexample to the above statement.

\paragraph{3.} [Herstein, Section 2.4 \#1] Verify that the relation $\sim$ is an equivalence relation on the set $S$ given.
\begin{thmenum}
	\item $S = \R$, and $a \sim b$ if $a-b$ is rational.
	\item $S = \C$, and $a \sim b$ if $\abs{a} = \abs{b}$.
	\item $S = \{\text{straight lines in the plane}\}$, and $a \sim b$ if $a, b$ are parallel.
	\item $S = \{\text{all people}\}$, and $a \sim b$ if they have the same color eyes.
\end{thmenum}

\paragraph{4.} [Herstein, Section 2.4 \#2] The relation $\sim$ on the real numbers $\R$ defined by $a \sim b$ if both $a > b$ and $b > a$ is \emph{not} an equivalence relation. Why not? What properties of an equivalence relation does it satisfy?


\subsection*{Medium}
\paragraph{5.} Let $r$ and $s$ be positive integers, and define 
\[
	H = \left\{nr + ms : n, m \in \Z \right\}.
\]
\begin{thmenum}
	\item Show that $H$ is a subgroup of $\Z$.
	\item We saw in class that every subgroup of $\Z$ is cyclic. Therefore, $H = \la d \ra$ for some $d \in \Z$. What is this integer $d$? Prove that the 		$d$ you've found is in fact a generator for $H$.
\end{thmenum}

\paragraph{6.} Let $a$ and $b$ be elements of a group $G$. Show that if $ab$ has finite order $n$, then $ba$ also has order $n$.

\paragraph{7.} Let $H$ be a subgroup of a group $G$ and let $g \in G$. Define a one-to-one map of $H$ onto $Hg$. Prove that your map is one-to-one and onto.

\paragraph{8.} We will see in class that if $p$ is a prime number, then the cyclic group $\Z_p$ has no proper subgroups as a consequence of Lagrange's theorem. This problem will have you investigate a ``converse'' to this result.
\begin{thmenum}
	\item {[Herstein, Section 2.3 \#14]} If $G$ is a group which has no proper subgroups, prove that $G$ must be cyclic.
	\item {[Herstein, Section 2.3 \#15]} Extend the result of (a) by showing that if $G$ has no proper subgroups, then $G$ is not only cyclic, but 
		\[
			\abs{G} = p
		\]
		for some prime number $p$.
\end{thmenum}

\subsection*{Hard}
\paragraph{9.} Let $G = \la a \ra$ be a cyclic group of order $n$. Prove that for any positive divisor $m$ of $n$, $G$ has exactly one subgroup of order $m$. [Hint: You may want to use the formula that relates $\abs{a^j}$ to $\abs{a}$.]

\paragraph{10.} [Herstein, Section 2.4 \#8] Let $G$ be a group with $H \leq G$, and for $a \in G$ define
	\[
		aHa^{-1} = \left\{aha^{-1} : h \in H \right\}.
	\]
If every right coset of $H$ in $G$ is a left coset of $H$ in $G$, prove that $aHa^{-1} = H$ for all $a \in G$. [\textbf{Note:} To say that a left coset $aH$ is also a right coset does not necessarily mean that $aH = Ha$. It only means that $aH = Hb$ for some $b \in G$. However, you will be able to show that $Hb = Ha$ in this case.]





















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