Exercise 1.6.26 (Matrix representation of $Q_8$)

Question

Let $i$ and $j$ be the generators of $Q_8$ described in Section 5. Prove that the map $\varphi$ from $Q_8$ to $\mathrm{GL}_2(\mathbb{C})$ defined on generators by
$$\varphi(i) = \begin{pmatrix} \sqrt{-1} & 0\ 0 & - \sqrt{-1} \end{pmatrix} \qquad 	ext{and} \qquad \varphi(j) = \begin{pmatrix} 0 & -1\ 1 & 0\end{pmatrix} $$
extends to a homomorphism. Prove that $\varphi$ is injective.

Richard Ganaye · Accepted Answer

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\begin{proof} 
Let $\iota$ be one of the two square roots of $-1$ in $\C$ ( I get goosebumps when I write $\sqrt{-1}$).
Put
$$I = \begin{pmatrix} \iota & 0\ 0 & - \iota \end{pmatrix},\qquad J = \begin{pmatrix} 0 & -1\ 1 & 0\end{pmatrix}.$$
Then $I,J \in \mathrm{GL}_2(\C)$.

We know a presentation of the group $Q_8$ (see Exercises 1.5.3 and 6.3.7):
$$Q_8 \simeq \langle i,j \mid i^2 = j^2,\ i^{-1} j  i  = j^{-1}angle.$$
Note that 
\begin{align*}
I^2 = \begin{pmatrix} \iota^2 & 0\ 0 & - \iota^2 \end{pmatrix} = \begin{pmatrix} -1 & 0\ 0 & -1 \end{pmatrix},\
J^2 = \begin{pmatrix} 0 & -1\ 1 & 0\end{pmatrix}^2 =  \begin{pmatrix} -1 & 0\ 0 & -1 \end{pmatrix}.
\end{align*}
Therefore $I^2 = J^2$. Moreover
\begin{align*}
I^{-1} J I &= \begin{pmatrix} -\iota & 0\ 0 &  \iota \end{pmatrix}  \begin{pmatrix} 0 & -1\ 1 & 0\end{pmatrix}  \begin{pmatrix} \iota & 0\ 0 & - \iota \end{pmatrix}\
&=\begin{pmatrix} 0 & 1\ -1 & 0\end{pmatrix}\
&= J^{-1}.
\end{align*}
Since $I^2 = J^2,\ I^{-1} J I = J^{-1}$, there exists a homomorphism $\varphi$ from $Q_8 \simeq \langle i,j \mid i^2 = j^2,\ i^{-1} j  i  = j^{-1}angle$ to $\mathrm{GL}_2(\C)$ such that $\varphi(i) = I,\ \varphi(j) = J$.

We show that $\varphi$ is injective. Note that $\varphi(-1) = \varphi(i)^2  = I^2 = -I_2$, and $K = \varphi(k) = \varphi(i) \varphi(j) = I J =  \begin{pmatrix} 0 & -\iota\ -\iota & 0 \end{pmatrix}$. This gives all the values of $\varphi$:

\begin{center}
\begin{tabular}{c|cccccccc|}
\hline
$ x$ & $1$  &$ i$ & $j$ & $k$ &$-1$ & $-i$ & $-j$ & $-k$\
\hline
 $\varphi(x)$ & $\begin{pmatrix}1 & 0 \0 & 1 \end{pmatrix}$  &$ \begin{pmatrix} \iota & 0\ 0 & - \iota \end{pmatrix}$ & $\begin{pmatrix} 0 & -1\ 1 & 0\end{pmatrix}$  &$\begin{pmatrix} 0 & -\iota\ -\iota & 0 \end{pmatrix}$ & $\begin{pmatrix}-1 & 0 \0 & -1 \end{pmatrix}$ & $ \begin{pmatrix} -\iota & 0\ 0 &  \iota \end{pmatrix}$& $\begin{pmatrix} 0 & 1\ -1 & 0\end{pmatrix}$& $\begin{pmatrix} 0 & \iota\ \iota & 0 \end{pmatrix}$ \
\hline
\end{tabular}
\end{center}
This shows that $\varphi(x) = I_2$ if and only if $x = 1$. Therefore $\varphi$ is injective.

($\varphi$ is a faithful matrix representation of the group $Q_8$.)
\end{proof}


$x$	$1$	$i$	$j$	$k$	$- 1$	$- i$	$- j$	$- k$

$φ (x)$	$(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$	$(\begin{matrix} ι & 0 \\ 0 & - ι \end{matrix})$	$(\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$	$(\begin{matrix} 0 & - ι \\ - ι & 0 \end{matrix})$	$(\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix})$	$(\begin{matrix} - ι & 0 \\ 0 & ι \end{matrix})$	$(\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})$	$(\begin{matrix} 0 & ι \\ ι & 0 \end{matrix})$

Exercise 1.6.26 (Matrix representation of $Q_8$)

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Exercise 1.6.26 (Matrix representation of $Q_8$)

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