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QUESTION


The following table is the Cayley table of a group $G$ of order 8.

\begin{tabular}{c|cccccccc}
&$e$&$g$&$g^2$&$g^3$&$h$&$hg$&$hg^2$&$hg^3$\\ \hline
$e$&$e$&$g$&$g^2$&$g^3$&$h$&$hg$&$hg^2$&$hg^3$\\
$g$&$g$&$g^2$&$g^3$&$e$&$hg^3$&$h$&$hg$&$hg^2$\\
$g^2$&$g^2$&$g^3$&$e$&$g$&$hg^2$&$hg^3$&$h$&$hg$\\
$g^3$&$g^3$&$e$&$g$&$g^2$&$hg$&$hg^2$&$hg^3$&$h$\\
$h$&$h$&$hg$&$hg^2$&$hg^3$&$g^2$&$g^3$&$e$&$g$\\
$hg$&$hg$&$hg^2$&$hg^3$&$h$&$g$&$g^2$&$g^3$&$e$\\
$hg^2$&$hg^2$&$hg^3$&$h$&$hg$&$e$&$g$&$g^2$&$g^3$\\
$hg^3$&$hg^3$&$h$&$hg$&$hg^2$&$g^3$&$e$&$g$&$g^2$\\ \end{tabular}

\begin{description}

\item[(i)]
Write each element of the group as a permutation of the set of
elements, in disjoint circle notation.

\item[(ii)]
Give the order of each element.

\item[(iii)]
Give the sign of each element.

\item[(iv)]
Find all of the subgroups of $G$ of size 2 and of size 4, giving,
for each subgroup $H$, the set of elements and a generating set.

\item[(v)]
Show that no two elements of $G$ generate a subgroup isomorphic to
Klein's 4-group.

\end{description}



ANSWER


\begin{description}

\item[(i)]

\begin{eqnarray*}
e&=&(e)(g)(g^2)\ldots(hg^3)\\ g&=&(e,g,g^2,g^3)(h,hg^3,hg^2,hg)\\
g^2&=&(e,g^2)(g,g^3)(h,hg^2)(hg,hg^3)\\
g^2&=&(e,g^3,g^2,g)(h,hg,hg^2,hg^3)\\
h&=&(e,h,g^2,hg^2)(g,hg,g^3,hg^3)\\
hg&=&(e,hg,g^2,hg^3)(g,hg^2,g^3,h)\\
hg^2&=&(e,hg^2,g^2,h)(g,hg^3,g^3,hg)\\
hg^3&=&(e,hg^3,g^2,hg)(g,h,g^3,hg^2)
\end{eqnarray*}

\item[(ii)]
order 1$\Leftrightarrow e$

order 2$\Leftrightarrow g^2$

order 4$\Leftrightarrow g,g^3,h,hg,hg^2$ or $hg^3$.

\item[(iii)]
Each element of order 4 is a product of 2 4-cycles (each of which
is odd) so it is even $g^2$ is a product of 4 2-cycles so it too
is even. $e$ is always even so sgn$(x)=0\forall x\in G$

\item[(iv)]
$|H|=2\Rightarrow H$ is cyclic of order 2$\Rightarrow
H=\left<g\right>=\{e,g^2\}$

$|H|=4\Rightarrow H$ is cyclic of order 4 ( since $G$ has only one
element of order 2) so
$H+\left<g\right>=\left<g^3\right>=\{e,g,g^2,g^3\},\
H=\left<h\right>=\left<hg^2\right>=\{e,h,g^2,hg^2\}$ or
$H=\left<hg\right>=\left<hg^3\right>\{e,hg,g^2,hg^3\}$

\item[(v)]
Klein's 4-group is an abelian group of order 4 generated by 2
elements of order 2.

$G$ has only one element of order 2 so it has no subgroups
isomorphic to Klein's 4-group.

\end{description}




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