\documentclass[11pt]{article}
\usepackage{latexsym,amsmath,amssymb}
\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{8.0in}
\begin{document}
\centerline{Mathematics 243, section 3 -- Algebraic Structures}
\centerline{Problem Set 2}
\centerline{{\bf due:} September 14, 2012}
\vskip 10pt
\noindent
{\it `A' Section}
\bigskip
\begin{enumerate}
\item[1.]  Let $f : \mathbb{Z} \to \mathbb{Z}$ and $g : \mathbb{Z} \to \mathbb{Z}$ be 
the indicated functions.  In each case, say whether $f, g, f\circ g, g\circ f$ are
one-to-one (injective) or onto (surjective), both, or neither.  Justify your answers.
\begin{enumerate}
\item[a.]  $f(x) = 3x$, $g(x) = 4 - x$
\item[b.]  $f(x) = |x|$, $g(x) = \begin{cases} x & \text{ if } x \text{ is even }\\
                                              x-1 & \text{ if } x \text{ is odd }\end{cases}$.
\end{enumerate}
\item[2.]  Consider the binary operation on $\mathbb{Z}$ given by 
$$x * y = x + y + 5$$
\begin{enumerate}
\item[a.]  Is $*$ commutative?  Why or why not?
\item[b.]  Is $*$ associative?  Why or why not?
\item[c.]  Is there an identity element for $*$.  What is the identity, or why not?
\item[d.]  Are there any elements of $\mathbb{Z}$ that have inverses under this
operation?  What are they, and what are the inverses?  
\end{enumerate}
\item[3.]  Let $A = \{x,y,z,w\}$ and let $*$ be the binary operation on $A$
given by the following table:
\[
\begin{array}{rrrrrr}
          *& \vline & x & y & z & w \\ \hline
          x& \vline & x & y & z & w \\
          y& \vline & y & y & w & w \\
          z& \vline & z & w & z & w \\
          w& \vline & w & w & w & w
\end{array}
\]
\begin{enumerate}
\item[a.] Explain how you can tell this operation is commutative.
\item[b.] Explain why $x$ is an identity element for $*$.
\item[c.] Which elements have inverses and what are the inverses?
\item[d.] What is $(y * z) * z$?  Is that the same as $y * (z * z)$?
\end{enumerate}
\end{enumerate}

\vskip 10pt
\noindent
{\it `B' Section}
\vskip 10pt
\begin{enumerate}
\item[1.]  Let $f : A \to B$ and $g : B \to A$ be mappings.  Prove
that if $f \circ g$ is onto and $g \circ f$ is one-to-one, then 
$f$ is one-to-one \emph{and} onto.
\item[2.]  Let $*$ be an associative binary operation on 
a set $A$ and assume there is an identity element $e$ for $*$.
If $a \in A$ has inverses $b_1$ and $b_2$, show that
$b_1 = b_2$.  Hint: Consider the ``product'' $(b_1 * a) * b_2$.  
\item[3.]  Let $A$ be a set and let $\mathcal{P}(A)$ be the power
set of $A$ as defined in \S 1 of the text and on Problem Set 1.  Let 
$*$ be the binary operation on $\mathcal{P}(A)$ defined by 
$S * T = S \cup T$.  Answer the following questions and prove 
your assertions.
\begin{enumerate}
\item[a.]  Is $*$ associative?  Is $*$ commutative?  
\item[b.]  Is there an identity element in $\mathcal{P}(A)$ for this
operation?  
\item[c.]  What elements of $\mathcal{P}(A)$ have inverses for this operation?  
\item[d.]  Make a table like the one in problem 3 of the `A' section 
for the operation in this problem, when $A = \{a,b\}$.  List the elements of
$\mathcal{P}(A)$ in this order on the borders of the table:
$$\emptyset, \{a\}, \{b\}, \{a,b\}.$$ 
Do you notice something?
\end{enumerate}
\item[4.]  Let $\times : \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$
be the vector \emph{cross product} from multivariable calculus (MATH 241).  
Recall that this operation is defined by the following formula:
$$(a_1,a_2,a_3) \times (b_1,b_2,b_3) = (a_2b_3 - a_3b_2, -(a_1b_3 - a_3b_1), a_1b_2 - a_2b_1).$$
\begin{enumerate}
\item[a.]  Show that $\times$ is not associative and not commutative.
\item[b.]  Show that $\times$ does satisfy the \emph{Jacobi identity}:
$$(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) + \mathbf{b} \times (\mathbf{c} \times \mathbf{a})$$
for all $\mathbf{a},\mathbf{b}$, and $\mathbf{c}$ in $\mathbb{R}^3$.  The $+$ in this
formula means the vector sum in $\mathbb{R}^3$, defined for vectors $\mathbf{d} = (d_1,d_2,d_3)$ and 
$\mathbf{e} = (e_1,e_2,e_3)$ by the rule
$$\mathbf{d} + \mathbf{e} = (d_1,d_2,d_3) + (e_1,e_2,e_3) = (d_1 + e_1, d_2 + e_2, d_3 + e_3).$$
\item[c.]  {\it Extra Credit} In a sense, the additional term $\mathbf{b} \times (\mathbf{c} \times \mathbf{a})$ 
on the right in the Jacobi identity measures the failure of associativity.
Using that idea, is $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$ \emph{ever} equal to 
$(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}$ when all three of the vectors are 
nonzero?  Explain.  Hint:  One way to approach this is to think about the geometric
conditions on the three vectors under which it will be true that
$$\mathbf{b} \times (\mathbf{c} \times \mathbf{a}) = \mathbf{0}.$$
\end{enumerate}
\end{enumerate}

\end{document}
