Identifying $\mathbb{C} D_8$ as a product of matrix rings Let $G = D_8$ be the dihedral group of order $8$, i.e.,
\begin{align*}
G = \langle a,b \mid a^4 = 1 = b^2, ab = ba^{-1} \rangle.
\end{align*}
By standard results from ring/represention theory, the group algebra $\mathbb{C} G$ decomposes as a product of matrix rings as follows:
\begin{align*}
\mathbb{C}G \cong \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times M_2(\mathbb{C}).
\end{align*}
I would like to explicitly find elements of $\mathbb{C} G$ which behave like the elements $(1,0,0,0,0), (0,1,0,0,0), (0,0,1,0,0), (0,0,0,1,0), (0,0,e_{ij})$, where $e_ij$ is the matrix with a $1$ in the $(i,j)$th position, and zeros elsewhere.
I've found the following pairwise orthogonal idempotents which hopefully correspond to $1$ in each copy of $\mathbb{C}$:
\begin{align*}
e_1 = \tfrac{1}{8}(1+a+a^2+a^3+b+ab+a^2b+a^3b)\\
e_2 = \tfrac{1}{8}(1+a+a^2+a^3-b-ab-a^2b-a^3b)\\
e_3 = \tfrac{1}{8}(1-a+a^2-a^3+b-ab+a^2b-a^3b)\\
e_4 = \tfrac{1}{8}(1-a+a^2-a^3-b+ab-a^2b+a^3b)
\end{align*}
The element
\begin{align*}
e_{11} = \tfrac{1}{4}(1+ia-a^2-ia^3+b+iab-a^2b-ia^3b)\\
\end{align*}
is also an idempotent and is orthogonal to the $e_i$, and so it makes sense to set $e_{22} = 1-e_1-e_2-e_3-e_4-e_{11}$. However, I can't seem to find suitable elements for $e_{12}$ and $e_{21}$. Am I going in the right direction at the moment?
(Also, my method has been largely ad hoc; is there a uniform method for approaching this sort of problem for an arbitrary finite group $G$?)
 A: I managed to work this out, so I'll provide my solution, which carries on from my work in the question.
The elements
\begin{align*}
e_{11} = \tfrac{1}{4}(1+ia-a^2-ia^3) \\
e_{22} = \tfrac{1}{4}(1-ia-a^2+ia^3)
\end{align*}
are orthogonal idempotents which are also orthogonal to each of the $e_i$. Observe that $e_{11}b = be_{22}$ and $be_{11} = e_{22}b$. These identities ensure that if we set
\begin{align*}
e_{12} = e_{11}b = \tfrac{1}{4}(b+iab-a^2b-ia^3b) \\
e_{21} = e_{22}b = \tfrac{1}{4}(b-iab-a^2b+ia^3b),
\end{align*}
then the correct relations hold between these matrices. Additionally, one can check that $e_1+e_2+e_3+e_4+e_{11}+e_{22} = 1$. Finally, the matrix
\begin{align*}
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 & i & -i & 0 & 0 \\
1 & 1 & 1 & 1 & -1 & -1 & 0 & 0 \\
1 & 1 & -1 & -1 & -i & i & 0 & 0 \\
1 & -1 & 1 & -1 & 0 & 0 & 1 & 1 \\
1 & -1 & -1 & 1 & 0 & 0 & i & -i \\
1 & -1 & 1 & -1 & 0 & 0 & -1 & -1 \\
1 & -1 & -1 & 1 & 0 & 0 & -i & i
\end{pmatrix}
\end{align*}
has nonzero determinant, so the eight elements we have found are linearly independent. It follows that the map
\begin{gather*}
\phi: \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times M_2(\mathbb{C}) \to \mathbb{C}D_8 \\
e_1 \mapsto \tfrac{1}{8}(1+a+a^2+a^3+b+ab+a^2b+a^3b)\\
e_2 \mapsto \tfrac{1}{8}(1+a+a^2+a^3-b-ab-a^2b-a^3b)\\
e_3 \mapsto \tfrac{1}{8}(1-a+a^2-a^3+b-ab+a^2b-a^3b)\\
e_4 \mapsto \tfrac{1}{8}(1-a+a^2-a^3-b+ab-a^2b+a^3b)\\
e_{11} \mapsto \tfrac{1}{4}(1+ia-a^2-ia^3) \\
e_{22} \mapsto \tfrac{1}{4}(1-ia-a^2+ia^3) \\
e_{12} \mapsto \tfrac{1}{4}(b+iab-a^2b-ia^3b) \\
e_{21} \mapsto \tfrac{1}{4}(b-iab-a^2b+ia^3b)
\end{gather*}
is a surjective ring homomorphism between $8$-dimensional spaces, and hence an isomorphism.
