Let, $D_{8}$ act on itself via conjugation, find the character of this representation. Where, $D_{8} = <r,s: r^{4} = e,s^{2} = e, sr^{-1} = rs >.$ Let, $g \in D_{8}$, then define $\psi_{g}:\mathbb{C}D_{8} \rightarrow \mathbb{C}D_{8}$ where  $\psi(x) = gxg^{-1}$, and we can extend the above to a linear map. Am I correct in assuming that to obtain the character of this representation that I would have to construct an $8$ by $8$ matrix where each element of $D_{8}$ is encoded into a vector; for example, we can say that $r = (1,0,0,0,0,0,0,0), r^2 = (0,1,0,0,0,0,0,0) $, and so on, and then construct $5$ ( number of conjugacy classes) 8 by 8 matrices that represent our action and take the trace of those $5$ matrices, is this how one would solve the problem?
Thanks!
 A: To expand on my comment, since I have thought about it a bit more and am pretty sure I was not lying.
Note that for each $g,x,y\in D_8$, we have $\psi_g(x) = \psi_g(y)\Leftrightarrow x=y$, so each $\psi_g$ permutes the elements of $D_8$. This is equivalent to permuting the standard basis vectors of $\mathbb{C}D_8$ (which correspond to the elements of $D_8$). Thus each $\psi_g$ can be represented by a permutation matrix with respect to this basis. This means that each row and each column of $\psi_g$ contains exactly one entry of 1, with all other entries being 0. The trace of such a matrix is the number of 1s on the diagonal; in this case, that is the number of basis vectors fixed by $\psi_g$, or, equivalently, the number of $x\in D_8$ such that $\psi_g(x)=x$.
So we can write down the character of this representation without worrying about working explicitly with $8\times 8$ matrices: instead, count how many elements $x\in D_8$ are fixed by each $\psi_g$.
Perhaps this question might also be relevant: Permutation representation of the conjugation action of a finite group
