Consider the $D_4$ group. Note that the $D_4$ groups is generated by rotation and reflection, where $ rot: 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 $ and $ref:1 \rightarrow 4, 2 \rightarrow 3, 3 \rightarrow 2, 4 \rightarrow 1$, i.e. elements are defined as permutations of vertices as in Way 2 of How are the elements of a dihedral group usually defined?.
Let $\pi$ be the natural representation of $D_4$ on the vector space with orthonormal basis $\{e_1, \cdots, e_4 \}$.
I know that the set of permutations of $\{e_1, \cdots, e_4 \}$ will be a $C^*$-algebra $A$ of $4 \times 4$ matrices. So $A$ is a linear combination of matrices.
My question is: what matrices will give us $A$, and what matrices will give us the commutant $A'$ of $A$?
I'm trying to proceed by writing out the $4 \times 4$ matrices for the rotation and reflection.
Below is the matrix I got for the rotation and reflection (please correct me if I'm wrong, as I'm a bit rusty on this)
$$Rot = \left[\begin{matrix}1& 0 & 0 & 0 \\\ 0& 0 &1& 0 \\\ 0& -1 &0& 0 \\\ 0 & 0 & 0 & 1 \end{matrix}\right].$$
$$Ref = \left[\begin{matrix}1& 0 & 0 & 0 \\\ 0& -1 &0& 0 \\\ 0& 0 &1& 0 \\\ 0 & 0 & 0 & 1 \end{matrix}\right].$$
From here, I'm not sure how to continue.
Thank you for your help!
"rot: $1 \rightarrow 2 \rightarrow 3 \rightarrow 4$" means \begin{align*} \mathrm{Rot}(e_1) &= e_2 \text{, } \\ \mathrm{Rot}(e_2) &= e_3 \text{, } \\ \mathrm{Rot}(e_3) &= e_4 \text{, and} \\ \mathrm{Rot}(e_4) &= e_1 \text{.} \end{align*}
So the first column of $\mathrm{Rot}$ is $e_2$, the second column is $e_3$, the third column is $e_4$, and the fourth column is $e_1$. Think carefully how the product $\mathrm{Rot}\cdot e_i$ picks out the $i^\text{th}$ column of $\mathrm{Rot}$. Therefore, $$ \mathrm{Rot} = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \text{.} $$
Hopefully, you see that your $\mathrm{Ref}$ matrix should also be some permutation of the $e_i$ as columns. In particular, there should be no negative numbers in your matrix.
It should, perhaps, not be a surprise that there is a degree two real representation and a degree one complex representation of this group.
$\mathrm{Rot}$ has order $4$, so $3$ nontrivial powers. $\mathrm{Ref}$ has order $2$, so $1$ nontrivial power. So there are at most $(3+1)^2$ commutators (actually fewer need be explicitly calculated, since many of these can be immediately recognized to give zero).