$D_4$ representation as a $C^*$-algebra Let us consider the natural representation $\pi$  of the $D_4$ group on the vector space with orthonormal basis $\{e_1, e_2, e_3, e_4 \}$. Note that this gives us a $C^*$ algebra $A$ such that $A = \tilde{\pi} (\mathbb{C}(D_4))$, where $\tilde{\pi}: \mathbb{C}(D_4) \rightarrow B(H)$ is  a *- homomorphism of $\mathbb{C}(D_4)$.
My question is: What is an explicit description of $A$ in terms of the definition above?
I know that by above and since $D_4$ has 8 elements, $A$ can be written as a linear combination of 8 matrices.
I tried to compute the image of each element by force, but I'm not quiet sure how to do so.
I have that $$  \mathrm{Rotation} = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}  \text{.}  $$
$$  \mathrm{Reflection} = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}  \text{.}  $$.
But I'm confused about what the 8 elements are.
Any other suggestions will be appreciated too!
Thank you in advance!
 A: Let
$$  \rho = \begin{bmatrix} 0 & 0 & 0 & 1  \\  1 & 0 & 0 & 0  \\  0 & 1 & 0 & 0  \\  0 & 0 & 1 & 0  \end{bmatrix}  $$
and
$$ \sigma = \begin{bmatrix} 0 & 0 & 0 & 1  \\  0 & 0 & 1 & 0  \\  0 & 1 & 0 & 0  \\  1 & 0 & 0 & 0 \end{bmatrix}  $$
The group operation is multiplication and the identity is the identity matrix.
$$  I = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}  $$
Now we proceed by length of word in the free group generated by $\rho$ and $\sigma$.  We know we have a finite group, so we will only generate finitely many distinct matrices and we will find a bunch of relations.
The missing words of length $1$ are $\rho^{-1} = \rho^T$ (the transpose) and $\sigma^{-1} = \sigma$.  This last says $\sigma^2 = I_4$, so we need consider no further powers of $\sigma$ and need not ever consider a word with consecutive $\sigma$s.
$\rho^2$, $\rho \sigma$, and $\rho^{-1} \sigma$ are three new elements of the group.  $\rho^{-2} = \rho^2$, $\sigma \rho = \rho^{-1}\sigma$ and $\sigma \rho^{-1} = \rho\sigma$.  (Notice that we're done with powers of $\rho$ and can consider only these powers of $\rho$ in further words: $\rho$, $\rho^{-1}$, and $\rho^2$.)
Being a little more systematic:  \begin{align*}
\rho^2 \sigma &= \text{(new)}  \\
\rho \sigma \rho &= \sigma  \\
\rho \sigma \rho^{-1} &= \rho^2 \sigma  \\
\rho^{-1} \sigma \rho &= \rho^2 \sigma  \\
\rho^{-1} \sigma \rho^{-1} &= \sigma  \\
&\\
\rho^2 \sigma \rho &= \rho\sigma  \\
\rho^2 \sigma \rho^{-1} &= \sigma \rho^{-1}
\end{align*}
Notice that every long word in the free group has a subword in the list above that is equal to another (typically shorter) subword.  In fact, every word can be reduced to one of the eight words we haven't shown are reducible to a word previously added to the list.
So the group is the set of matrices $\{I, \rho, \rho^{-1}, \sigma, \rho^2, \rho\sigma, \rho^{-1}\sigma, \rho^2\sigma \}$ under the operation of matrix multiplication.
