= Solution
Distinguish the <quaternion group> generators $\mathbf i,\mathbf j,\mathbf k$ from the complex scalar $i$. Denote their proposed image matrices by
$$
A=\begin{pmatrix}0&1\\-1&0\end{pmatrix},\qquad
B=\begin{pmatrix}0&i\\i&0\end{pmatrix},\qquad
C=\begin{pmatrix}i&0\\0&-i\end{pmatrix}.
$$
Direct multiplication gives
$$
A^2=B^2=C^2=-I_2,\qquad
AB=C,\quad BC=A,\quad CA=B,\qquad
BA=-C,\quad CB=-A,\quad AC=-B.
$$
For example $AB=\operatorname{diag}(i,-i)=C$ and $ABC=C^2=-I_2$. Sending $I$ to $I_2$ and each negative quaternion element to the negative of its positive image now reproduces every quaternion product. The matrices all have <determinant> one and are invertible, so this is a <group representation> in $GL(2,\mathbb C)$.
The eight matrices $\pm I_2,\pm A,\pm B,\pm C$ are distinct. Therefore the <group homomorphism> is injective, and the representation is \b[faithful]; equivalently its <kernel> is $\{I\}$.
The <character of a representation> is the <trace> of its representing matrix. Consequently
$$
\boxed{\chi(I)=2,\qquad\chi(-I)=-2,\qquad
\chi(\pm\mathbf i)=\chi(\pm\mathbf j)=\chi(\pm\mathbf k)=0.}
$$
These are constant on the five <conjugacy classes> $\{I\}$, $\{-I\}$, $\{\pm\mathbf i\}$, $\{\pm\mathbf j\}$, $\{\pm\mathbf k\}$.
The complete list of <subgroups> is
$$
\boxed{\begin{array}{c|c}
\text{subgroup}&\text{order}\\\hline
\{I\}&1\\
\{I,-I\}&2\\
\langle\mathbf i\rangle=\{I,-I,\mathbf i,-\mathbf i\}&4\\
\langle\mathbf j\rangle=\{I,-I,\mathbf j,-\mathbf j\}&4\\
\langle\mathbf k\rangle=\{I,-I,\mathbf k,-\mathbf k\}&4\\
Q&8
\end{array}}
$$
To see completeness, <Lagrange's theorem> allows only orders $1,2,4,8$. The only element of order two is $-I$, which fixes the order-two subgroup. Every noncentral element has order four and generates the corresponding listed <cyclic subgroup>. Any order-four subgroup contains such an element, because there are only two central elements; it must equal the <cyclic subgroup> generated by that element. This accounts for all possibilities.
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