Irreducible representations (over $\mathbb{C}$) of dihedral groups 
Find  number of complex irreps of the group $D_n$. Find dimension of the irreps.

I know that 


*

*The number of complex irreps of a finite group is equal to the number of conjugacy classes of the group.

*All the reflections are conjugate to each other in case $n$ is odd, but

*they fall into two conjugacy classes if $n$ is even. But there are $n$ rotations in $D_n$. I can't find the number of conjugacy classes for these.

*The number of 1-dimensonal complex irreps is equal to $[D_n:D_n^\prime]$. I don't know anything about the commutator subgroup of $D_n$.

*The matrices for elements of $D_n$ have the following form:
$$R_k=\pmatrix{\cos{\frac{2 \pi k}{n}}&-\sin{\frac{2 \pi k}{n}}\\
               \sin{\frac{2 \pi k}{n}}& \cos{\frac{2 \pi k}{n}}}$$
$$S_k=\pmatrix{\cos{\frac{2 \pi k}{n}}& \sin{\frac{2 \pi k}{n}}\\
               \sin{\frac{2 \pi k}{n}}& -\cos{\frac{2 \pi k}{n}}}$$


$R_k$ is a rotation matrix, expressing a counterclockwise rotation through an angle of $2\pi k/n$. $S_k$ is a reflection across a line that makes an angle of $2\pi k/n$ with the $x$-axis.
This helps me to find all $2$-dimensional real irreps, but what about the complex ones?
Answer in my book:
"If $n=2k$, then there are $4$ 1-dimensional  complex irreps and $(k-1)$ 2-dimensional complex irreps;
If $n=2k+1$, then there are $2$ 1-dimensional  complex irreps and $(k+1)$ 2-dimensional complex irreps."
 A: Here is one way to get the conjugacy classes of $D_n$ and irreducible representations over $\mathbf{C}$.
Setup
First I will fix some notation. A presentation of $D_n$ is $\langle r, s\mid r^n = s^2 = 1, sr=r^{-1}s\rangle$, which means we can pin down $D_n$ as a group of rotations $\{ 1, r, r^2\ldots r^{n-1}\}$ together with a bunch of reflections $\{ s, sr, sr^2\ldots sr^{n-1}\}$. (In what follows I will frequently be sloppy and implicitly use the identifications $r^{-i} = (r^i)^{-1} = r^{n-i}$.)
Analysis of conjugacy classes of $D_n$
To figure out the conjugacy classes, we can just use brute force. Every element is $r^i$ or $sr^i$ for $0\leq i < n$, so it's not too hard just to write down every possible conjugation:
\begin{array}{rlclcl}
\text{Conjugate} &r^i &\text{by}&r^j &:&(r^j) r^i(r^{-j})=r^i\\
                 &r^i &\text{by}&sr^j&:&(sr^j)r^i(r^{-j}s)= sr^is=r^{-i}\\
                 &sr^i&\text{by}&r^j &:&(r^j) sr^i(r^{-j}) = sr^{-j}r^ir^{-j} = sr^{i-2j}\\
                 &sr^i&\text{by}&sr^j&:&(sr^j)sr^i(r^{-j}s) = r^{i-2j}s=sr^{2j-i}\\
\end{array}
Conjugacy classes of rotations
The first two rows tell us that the set of rotations decomposes into inverse pairs $r^i$ and $(r^{i})^{-1}$, i.e. the classes $\{1\}, \{r, r^{n-1}\}, \{r^2, r^{n-2}\},\ldots$
Counting these, there are $\frac{n}{2}+1$ when $n$ is even (note that $r^{n/2}$ is its own inverse) and $\frac{n+1}{2}$ when $n$ is odd.
Conjugacy classes of reflections
Now observe from the third and fourth lines of the table that $sr$ is conjugate to $sr^3, sr^5, \ldots$ while $s$ is conjugate to $sr^2, sr^4,\ldots$ and these two sets are disjoint if $n$ is even. However, $sr$ is conjugate to $sr^{n-1}$ (via $r$) so if $n$ is odd, all the nontrivial reflections are in one conjugacy class. (You said you already knew this but I'm putting it here for completeness.)
Together, this brings us to the total number of conjugacy classes of $D_n$:
\begin{array}{rl}
\left(\frac{n}{2}+1\right)+2 = \color{#090}{\frac{n}{2}+3}&\text{for }n\text{ even}\\
\left(\frac{n+1}{2}\right)+1 = \color{#090}{\frac{n+3}{2}}&\text{for }n\text{ odd.}
\end{array}
Analysis of irreducible representations of $D_n$
One dimensional irreducibles
The commutators of $D_n$ look like $[r^i, sr^j]$ or the inverse of such, and
$$[r^i, sr^j] = r^{-i}(sr^j)r^i(sr^j) = sr^{2i+j}sr^j = (r^i)^2$$ so the commutators generate the subgroup of squares of rotations. This means $G/[G,G]$ has order 2 if $n$ is odd (since all rotations are squares) or order 4 if $n$ is even (since only half the rotations are squares). Now you can use your fact #4, which tells us that we have precisely 2 ($n$ odd) or 4 ($n$ even) irreps of dim. 1 obtained from pulling back those from $G/[G,G]$.
Other irreducibles
This is related to your item #5. We can define some 2-dim'l representations over $\mathbf{R}$, namely
\begin{array}{ccc}
r&\mapsto&\pmatrix{\cos(2\pi k/n)&-\sin(2\pi k/n)\\
                 \sin(2\pi k/n)&\cos(2\pi k/n)} \\
s&\mapsto&\pmatrix{0&1\\1&0}
\end{array}
for $0\leq k \leq \lfloor \frac{n}{2}\rfloor$. We would like to know if these rep'ns are irreducible if we consider them as matrices over $\mathbf{C}$.
They are reducible if $k=0$ or $k = n/2$ (can you decompose them?). If $k$ is different from $0$ or $n/2$, a quick computation shows that the matrix for $r$ has distinct complex eigenvalues $\pm e^{2\pi ki/n}$ with corresponding eigenvectors $\pmatrix{1\\-i}$ and $\pmatrix{1\\i}$. The spans of each e-vector are the only candidates for invariant subspaces, but the matrix for $s$ interchanges the two eigenspaces, so there are no invariant subspaces and thus these repn's are irreducible.
Final count
We have 2-dim'l irreps for each integer $1\leq k < \frac{n}{2}$, specifically we have $\frac{n}{2}-1$ for $n$ even and $\frac{n-1}{2}$ for $n$ odd. If we count these with the 1-dim'l irreps, we have
\begin{array}{rl}
\left(\frac{n}{2}-1\right)+4 = \color{#090}{\frac{n}{2}+3}&\text{for }n\text{ even}\\
\left(\frac{n-1}{2}\right)+2 = \color{#090}{\frac{n+3}{2}}&\text{for }n\text{ odd.}
\end{array}
which matches up with the number of conjugacy classes, so we must be done by your fact #1. (Also, we can verify that the sum of squares of the dimensions of the irreps is $2n$ in both cases.)
