Finding a simple submodule of $\mathbb C[Q_8]$ that is isomorphic to a certain irrep Given I know a certain $2$-dimensional irreducible representation $V$ of $\mathbb C[Q_8]$, is there a standard process to find a simple submodule of $\mathbb C[Q_8]$ that is isomorphic to that irreducible representation?
There's a method described here to find the four-dimensional subalgebra of $\mathbb C[Q_8]$ that is isomorphic to $V^{\oplus 2}$ but I'm looking for the simple submodule of $\mathbb C[Q_8]$ that is isomorphic to $V$, not $V^{\oplus 2}$.
 A: Your 4D subalgebra $A\subset\mathbb{C}[Q_8]$ corresponding to the 2D irrep $V$ is generated by the isotypical projector
$$ e=\frac{\dim V}{|G|}\sum_{g\in G}\chi_V(g^{-1})=\frac{1-\varepsilon}{2} $$
where we call $\varepsilon$ the order two element of $Q_8$. Since $\varepsilon e=-e$, we get
$$ A=\mathrm{span}\{e,\mathbf{i}e,\mathbf{j}e,\mathbf{k}e\}\cong\mathrm{End}(V)\cong M_2\mathbb{C}. $$
Indeed, $\mathbb{H}\otimes\mathbb{C}\to A$ given by $q\otimes z\mapsto qez$ (suitably interpreted!) is an isomorphism. Here, the second factor is considered the center $\mathbb{C}=Z(\mathbb{C}[Q_8])$ of scalars, so $i\in\mathbb{C}$ and $\mathbf{i}\in Q_8$ are different. This isomorphism is actually standard: the elements of $\mathbb{H}$ can be interpreted as linear transformations of $\mathbb{H}$ itself (viewing $\mathbb{H}$ a right $\mathbb{C}$-vector space) by thinking of a quaternion $q$ as a left-multiplication-map $L_q(x):=qx$, and this means by picking a basis for $\mathbb{H}$ (e.g. $\{1,\mathbf{j}\}$) we can write quaternions as $2\times 2$ complex matrices.
The subreps of $A$ correspond to its ideals as an algebra, and two nice complementary ideals of $M_2\mathbb{C}$ correspond to the sets of matrices of the form $[\begin{smallmatrix}\ast&0\\\ast&0\end{smallmatrix}]$ and $[\begin{smallmatrix}0&\ast\\0&\ast\end{smallmatrix}]$. If we write out our aforementioned isomorphism reasonably explicitly, we can figure out basis elements for the corresponding ideals in $A$.
