Exercise 4.2.4 (Left regular representation of $Q_8$)

Question

Use the left representation of $Q_8$ to produce two elements of $S_8$ which generate a subgroup of $S_8$ isomorphic to the quaternion group $Q_8$.

Richard Ganaye · Accepted Answer

(The table of $Q_8$ is given in Exercise 1.5.8.)
\begin{proof} 
We label the elements of $Q_8$ with the bijection $f$:
\begin{center}
\begin{tabular}{c|cccccccc}
$k$ & $1$ & $2$ & $3$ & $4$ &$5$ & $6$ & $7$ & $8$ \
\hline
$f(k)$ & $1$ &$-1$& $i$ &$ -i$& $j$ & $-j$ &$k$ & $-k$
\end{tabular}
\end{center}

Since $Q_8 = \langle i, j angle$, we compute $\sigma_i$ and $\sigma_j$:
\begin{align*}
i \cdot 1 = i &\Rightarrow \sigma_i(1) = 3,\
i \cdot -1 = -i &\Rightarrow \sigma_i(2) = 4,\
i \cdot i = -1&\Rightarrow \sigma_i(3) = 2,\
i \cdot -i = 1 &\Rightarrow \sigma_i(4) = 1,\
i \cdot j = k &\Rightarrow \sigma_i(5) = 7,\
i \cdot -j = -k &\Rightarrow \sigma_i(6) = 8,\
i \cdot k = -j&\Rightarrow \sigma_i(7) = 6,\
i \cdot -k = j &\Rightarrow \sigma_i(8) = 5,\
\end{align*}
so
$$\sigma_i = \begin{pmatrix}1 &2 & 3 & 4 & 5 & 6 &7 &8\ 3 & 4 & 2 & 1 & 7 & 8 & 6 & 5 \end{pmatrix} = (1\ 3\ 2\ 4)(5\ 7 \ 6\ 8).$$
Moreover,
\begin{align*}
j\cdot 1 = j&\Rightarrow \sigma_j(1) = 5,\
j \cdot -1 = -j &\Rightarrow \sigma_j(2) = 6,\
j\cdot i = -k&\Rightarrow \sigma_j(3) = 8,\
j\cdot -i = k &\Rightarrow \sigma_j(4) = 7,\
j \cdot j = -1 &\Rightarrow \sigma_j(5) = 2,\
j \cdot -j = 1 &\Rightarrow \sigma_j(6) = 1,\
j\cdot k = i&\Rightarrow \sigma_j(7) = 3,\
j \cdot -k = -i&\Rightarrow \sigma_j(8) = 4,\
\end{align*}
so
$$\sigma_j= \begin{pmatrix}1 &2 & 3 & 4 & 5 & 6 &7 &8\ 5 & 6 & 8 & 7 & 2 & 1 & 3 & 4 \end{pmatrix} = (1\ 5\ 2\ 6)(3\ 8 \ 4\ 7).$$

So 
\begin{align*}
Q_8 &\simeq \langle (1\ 3\ 2\ 4)(5\ 7 \ 6\ 8),  (1\ 5\ 2\ 6)(3\ 8 \ 4\ 7) angle\
\end{align*}
\end{proof}

Verification with Gap:
\begin{verbatim}
gap> i := (1,3,2,4)(5,7,6,8);
(1,3,2,4)(5,7,6,8)
gap> j := (1,5,2,6)(3,8,4,7);
(1,5,2,6)(3,8,4,7)
gap> G := Group(i,j);
Group([ (1,3,2,4)(5,7,6,8), (1,5,2,6)(3,8,4,7) ])
gap> Order(G);
8
gap> List(G);
[ (), (1,2)(3,4)(5,6)(7,8), (1,4,2,3)(5,8,6,7), (1,3,2,4)(5,7,6,8), (1,6,2,5)(3,7,4,8), 
  (1,5,2,6)(3,8,4,7), (1,8,2,7)(3,6,4,5), (1,7,2,8)(3,5,4,6) ]
gap> StructureDescription(G);
"Q8"
\end{verbatim}

Exercise 4.2.4 (Left regular representation of $Q_8$)

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$k$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$

$f (k)$	$1$	$- 1$	$i$	$- i$	$j$	$- j$	$k$	$- k$

Exercise 4.2.4 (Left regular representation of $Q_8$)

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