= Solution
Let $J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Since $J^2=-I_2$ and $J^4=I_2$, its powers define the <representation> of the cyclic group of order four:
$$
\boxed{\rho(1)=I_2,\qquad\rho(-1)=-I_2,\qquad\rho(-i)=-J.}
$$
For the <quaternion group>, denote the three displayed complex <matrices> by $A,B,C$. Direct multiplication gives $A^2=B^2=C^2=-I_2$, $AB=C$, $BA=-C$, and $ABC=-I_2$. Together with the images of $\pm1$, these are exactly the quaternion multiplication relations, proving that the assignment defines a <group homomorphism> into $\operatorname{GL}(2,\mathbb C)$.
Use the <realification of a complex representation>: for $M=X+iY$, define $\mathcal R(M)=\begin{pmatrix}X&-Y\\Y&X\end{pmatrix}$. It acts on grouped real and imaginary coordinates and satisfies $\mathcal R(MN)=\mathcal R(M)\mathcal R(N)$. The explicit generators are
$$
\boxed{D(i)=\begin{pmatrix}0&0&0&-1\\0&0&-1&0\\0&1&0&0\\1&0&0&0\end{pmatrix},\quad
D(j)=\begin{pmatrix}0&0&-1&0\\0&0&0&1\\1&0&0&0\\0&-1&0&0\end{pmatrix},\quad
D(k)=\begin{pmatrix}0&1&0&0\\-1&0&0&0\\0&0&0&1\\0&0&-1&0\end{pmatrix}.}
$$
Set $D(1)=I_4$ and $D(-q)=-D(q)$ to specify all eight elements.
For an invertible real $S$, the <equivalent representation> $\widetilde D(q)=SD(q)S^{-1}$ obeys $\widetilde D(q_1)\widetilde D(q_2)=SD(q_1q_2)S^{-1}=\widetilde D(q_1q_2)$. The <trace> is invariant under similarity, so their <characters> agree. Explicitly,
$$
\boxed{\chi(1)=4,\qquad\chi(-1)=-4,\qquad\chi(\pm i)=\chi(\pm j)=\chi(\pm k)=0.}
$$
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