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QUESTION


\begin{description}

\item[(a)]
Show that the set $\{x\in\mathbf{R}|x\neq-1\}$ is a group under
the operation * defined by $a*b=a+b+ab$.

\item[(b)]The following table in the (incomplete) Cayley table of
a group $G$ of order 8.

\begin{tabular}{c|cccccccc}
*&$p$&$q$&$r$&$s$&$t$&$u$&$v$&$w$\\ \hline $p$&$t$&$w$\\
$q$&$u$&$t$\\ $r$&&&$t$\\ $s$&&&&$t$\\ $t$&&&&&$t$\\
$u$&&&&&&&&$t$\\ $v$&&&&&&&$t$\\ $w$&&&&&&$t$
\end{tabular}

\begin{description}

\item[(i)]
State the classification of groups of order 8.

\item[(ii)]
decide, giving your reasons, which of the groups in your
classification is isomorphic to the one defined by the Cayley
table above.

\end{description}

\item[(c)]
Write down all the possible cycle structures for elements of
$S_7$, and use this to find all the possible orders for elements
of $S_7$, giving one example of an element for each possible
order. Explain the relationship between cycle structures and
conjugacy classes there are in $S_7$.

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ANSWER


\begin{description}

\item[(a)]
* is a binary operation since $a*b\in\mathbf{R}$ and
$a*b=-1\Leftrightarrow
a+b+ab=-1\Leftrightarrow-(a+1)b(a+1)\Leftrightarrow a=-1$ or
$b=-1$.

* is associative since

\begin{eqnarray*}
(a*b)*c&=&(a+b+ab)+c+ac+bc+abc\\ &=&a(b+c+bc)+ab+ac_+abc\\
&=&a*(b*c)
\end{eqnarray*}

The identity is 0 and the inverse of $a$, defined by
$\frac{-a}{1+a}$ is an element of the set.

\item[(b)]

\begin{description}

\item[(i)]
There are 3 abelian groups of order 8, $C_2\times C_2\times C_2;
C_2\times C_2$ and $C_8$.

There are 2 non-abelian groups of order 8, $D_8$ and the
Quaternians.

\item[(ii)]
Since $pq\neq qp$ the group is non-abelian. Since $t^2=t$, $t$ is
the identity element and the group has 5 elements of order 2. It
is therefore $D_8$.

\end{description}

\item[(c)]

\begin{tabular}{ccc}
cycle structure&order&example\\ $\left[7\right]$&7&(1234567)\\
$\left[6\right]$&6&(123456)\\ $\left[5,2\right]$&10&(12345)(67)\\
$\left[5\right]$&5&(12345)\\ $\left[4,3\right]$&12&(1234)(567)\\
$\left[4,2\right]$&4&(1234)(56)\\ $\left[4\right]$&4&(1234)\\
$\left[3,3\right]$&3&(123)(456)\\
$\left[3,2,2\right]$&6&(123)(45)(67)\\
$\left[3,2\right]$&6&(123)(45)\\ $\left[3\right]$&3&(123)\\
$\left[2,2,2\right]$&2&(12)(34)(56)\\
$\left[2,2\right]$&2&(12)(34)\\ $\left[2\right]$&2&(12)\\
$\left[1\right]$&1&(1)(2)(3)(4)(5)(6)(7)
\end{tabular}

There is exactly one cycle structure for each conjugacy class so
$S_7$ has 15 conjugacy classes. For example the two elements
(1234)(567) and $(abcd)(efg) $ conjugate via the element
$(1a)(2b)(3c)(4d)(5e)(6f)(7g)$.

\end{description}




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