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QUESTION


\begin{description}

\item[(a)]
For real numbers $a$ and $bn$ let $a*b$ denote the number
$a+b+2ab$.

\begin{description}

\item[(i)]
Show that the set $X=\{x\in R|x\neq-\frac{1}{2}\}$ is closed under
the binary operation *.

\item[(ii)]
Say what it means for a binary operation to be associative and
show that the binary operation * is associative.

\item[(iii)]
Show that there is a unique real number $r$ such that $r*s=s=s*r$
for every real number $s$, and that this lies in $X$.

\item[(iv)]
Show that every element of $X$ has an inverse under the binary
operation *.

\end{description}

\item[(b)]
For each positive integer $n$ less than or equal to 8 give a list
of groups of order $n$ so that no two groups on your list in your
list are isomorphic, but every group of order $n$ is isomorphic to
one of the groups on your list. For each order where there is more
than one isomorphism class of group of that order, indicate how
you distinguish the different isomorphism classes.

\end{description}



ANSWER


\begin{description}

\item[(a)]

\begin{description}

\item[(i)]
For any $a,b\in X,\ a+b+2ab\in R$.

Suppose $a+b+2ab=-\frac{1}{2}$. Then $a+b(1+2a)=-\frac{1}{2}$ so
$b(1+2a)=\frac{-2a-1}{2}$ or, as $1+2a\neq0,\
b=\frac{-2a-1}{2(2a+1)}=-\frac{1}{2}$.

So, as $b\in X,\ a*b\neq-\frac{1}{2}$ as required.

\item[(ii)]
* is associative $\Leftrightarrow a*(b*c)=(a*b)*c$ for all
$a,b,c\in X$.

\begin{eqnarray*}
a*(b*c)&=&a*(b+c+2bc)\\ &=&a+b+c+2bc+2(a(b+c+2bc))\\
&=&a+b+c+2bc+2ab+2ac+4abc
\end{eqnarray*}

\begin{eqnarray*}
(a*b)*c&=&(a+b+2ab)*c\\ &=&a+b+c+2ab+2(a+b+2ab)c\\
&=&a+b+c+2ab+2ac+2bc+4abc
\end{eqnarray*}

so $(a*b)*c=a*(b*c)$ for any $a,b,c\in X$

\item[(iii)]
$r*s=s\Leftrightarrow r+s+2rs=3\Leftrightarrow
r(1+2s)=0\Leftrightarrow r=0$ or $s=-\frac{1}{2}$.

Now putting $s=-1$ we see that $r=0$ is the only real number such
that $r*s=4\forall s\in R$, and $0\in X$. Furthermore $s*0=s$.

\item[(iv)]
$a*b=0\Leftrightarrow a+b+2ab=0\Leftrightarrow
b(1+2a)=a\Leftrightarrow b=\frac{a}{1+2a}\Leftrightarrow b*a=0$

\end{description}

\item[(b)]
\begin{tabular}{c|c|l}
Group&order&distinguishing features\\ \hline $\{e\}$&1\\ $Z_2$&2\\
$Z_3$&3\\ $Z_4$&4&contains an element of order 4,\\ &&and two
elements of order 2\\ $z_2\times Z_2$&2&3 elements of order 2\\
$Z)_5$&5\\ $Z_6$&6&abelian\\ $D_3$&6&non-abelian\\ $Z_7$&7\\
$Z_8$&8&abelian with one element of order 2\\ $Z_4\times
Z_2$&8&abelian with three elements of order 2\\ $Z_2\times
Z_2\times Z_2$&8&abelian with seven elements of order 2\\
$D_4$&8&non-abelian with five elements of order 2\\
$\mathcal{Q}$&8&non-abelian with one element of order 2\\
\end{tabular}

\end{description}




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