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\begin{center}
{{\Large\bf MODERN ALGEBRA II}\\
\bigskip

{\Large\bf HOMEWORK ASSIGNMENT I}\\
\bigskip
{\bf (Turn in underlined problems.)}}
\end{center}

\vspace{.5in}

\noindent
{\bf 
\begin{tabbing}
\underline{\large\bf CHAPTER}\makebox[2in]{}\=\underline{\large\bf PROBLEMS}\\

\bigskip
\noindent
2 \> \underline{2.26},
\underline{2.27},
\underline{2.28}
\end{tabbing}

}
\vspace{.5in}

\noindent
\underline{\large\bf ADDITIONAL PROBLEMS}
\vspace{.25in}

{\bf 

\underline{1.}
Describe all groups of size $1, 2$ and $3$ up to
isomorphism.

\bigskip

2. True or False: 

\smallskip

(a) The additive group $\langle {\BR};+,-,0\rangle$
is isomorphic to the additive group $\langle {\BC};+,-,0\rangle$.

\smallskip

(b) The additive group $\langle {\BR};+,-,0\rangle$
is isomorphic to the multiplicative group 
$\langle {\BR}^+;\cdot,{}^{-1},1\rangle$.

\bigskip

\underline{3.}
Let ${\bf G} = \langle G; \cdot, {}^{-1}, 1\rangle$
be a group. Define an algebra with the ``opposite
multiplication'': ${\bf G}^{op}$ has the same set $G$
of elements, but its multiplication ``$*$'' is
defined by $x*y = y\cdot x$.

\smallskip

(a) What must the ``opposite inverse'' and ``opposite identity''
be, if ${\bf G}^{op}$ is to be a group?

\smallskip

(b) Show that with these definitions the function
$\varphi:{\bf G}\rightarrow {\bf G}^{op}:x\mapsto x^{-1}$
is an isomorphism.

\bigskip

For Graduate Credit:

A. Show that a finite group has an element of order two if and
only if the size of the group is divisible by two.
(Hint: if $a^2 = 1$, then show that the sets of the form
$\{x, ax\}$ partition the group into two-element subsets.
For the other direction, read Problem 2.29.)

}
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