How to derive the character table for $D_4$?

$D_4$ being the group of symmetries on a square. So, I understand that there are 5 conjugacy classes in this group: $\{e\}$, $\{r, r^3\}$, $\{r^2\}$, $\{sr, sr^3\}$, $\{s, sr^2\}$. I need 5 representations, one of them will be the trivial one. Also, I know that the sum of the squares of the irreps will be equal to the order of the group (8), thus my first column will consist of all ones and one two. But how do I get the rest of them?

\begin{bmatrix} &\ \{e\} & \{r, r^3\} & x_{13} & \{r^2\} &\{sr, sr^3\} &\ \{s, sr^2\} \\ triv &\ 1 &\ 1 &\ 1 &\ 1 &\ 1 &\ 1 \\ \sigma_1 &\ 1 &\ ? &\ ? &\ ? &\ ? &\ ? \\ \sigma_2 &\ 1 &\ ? &\ ? &\ ? &\ ? &\ ? \\ \sigma_3&\ 1 &\ ? &\ ? &\ ? &\ ? &\ ? \\ \sigma_4&\ 2 &\ ? &\ ? &\ ? &\ ? &\ ? \end{bmatrix}

How do I figure out my functions $\sigma$ and how do I figure out the trace? I realize this might be a ridiculous question, and sometimes I do feel like I might have stumbled into the wrong classroom.

• You have an extra $x_{13}$ column in your table, I don't think that's meant to be there. Apr 19 '17 at 4:47

The one-dimensional representations $G \to \mathbb{C}^\times$ will factor through the abelianisation $\frac{G}{[G, G]}$, via the universal property of the abelianisation. (This will work for any 1-dimensional representation, since they're all abelian). You can easily check that $[G, G] = \langle r^2 \rangle$, and so the quotient $\frac{G}{[G, G]}$ is of order 4, generated by the order 2 elements $\overline{r}$ and $\overline{s}$. Since each of $\overline{r}$ and $\overline{s}$ must be sent to an element of order 2, they are sent to $1$ or $-1$. There are four possibilities here. If they are both sent to $1$, this lifts back to the trivial representation in $G$. If $\overline{r}$ is sent to $-1$ and $\overline{s}$ to $1$, this lifts to the representation $$\begin{bmatrix} &\ \{e\} & \{r, r^3\} & \{r^2\} &\{sr, sr^3\} &\ \{s, sr^2\} \\ \sigma_1 &\ 1 &\ -1 &\ 1 &\ -1 &\ 1 \end{bmatrix}$$ and so on. You now have all the one-dimensional representations.
As for the two-dimensional representation, just take a guess! There is an obvious representation of $G$ on $\mathbb{R}^2$ acting via the symmetries of a square, where $r$ is a $90^\circ$ rotation clockwise, and $s$ is a reflection over the $y$-axis. (Your exact convention for $r$ and $s$ may vary, but $r$ will be a rotation and $s$ a reflection). So $$r = \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}, \quad s = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$$ You can easily compute traces of these to get the character $$\begin{bmatrix} &\ \{e\} & \{r, r^3\} & \{r^2\} &\{sr, sr^3\} &\ \{s, sr^2\} \\ \chi & 2 &\ 0 &\ -2 &\ 0 &\ 0 \end{bmatrix}$$
And now you can compute the inner product $\langle \chi, \chi \rangle$ and check that it is $1$, to conclude that $\chi$ is an irreducible character, so this real representation on $\mathbb{R}^2$ is actually irreducible over $\mathbb{C}$ as well.