irreducible representations of $\mathbb{Z}_n$ I'm trying to find all the irreducible representations (which are not isomorphic) of $(\mathbb{Z}_n,+)$ in the field of real numbers. I saw that we can realize by mapping a group generator $g$ to $1$ and to $−1$, that there are two one-dimensional irreducibles. The others are all 2-dimensional and consist of the sum of a complex irreducible and its dual, which we can have by mapping $g$ to a $2\times 2$ orthogonal matrix representing a rotation through $\frac{2k\pi}{n}$ for some $k$. Why is this right? How can I write the full answer in detail?
I'll be grateful for having answers.
 A: Here is a sketch:

*

*The group $(\mathbb{Z}_n,+)$ is generated by a single element ("$1$"), which is of order $n$. So any representation $\rho:(\mathbb{Z},+)\mapsto \mathbb{R}^{d\times d}$ is fully defined by the value of $\rho(1)$, which must be of order $\le n$. More precisely, the order of $\rho(1)$ must be a divisor of $n$.


*All finite-order elements of $\mathbb{R}^{d\times d}$ are orthogonal matrices. Orthogonal matrices can always be written as a rotation around an axis or as rotation + reflection.


*In an appropriate basis, a pure rotation matrix can always be written as
$$\begin{pmatrix}
\cos(\phi)&-\sin(\phi)  \\
\sin(\phi)&\cos(\phi)  \\
&&1 \\
&&&1\\
&&&&...
\end{pmatrix}$$
So the irreducible represenations are $2$-dimensional, which can also be written as complex representations with $\rho(1)=e^{2k\pi/n}$ for some $k$. (i.e. $\phi=2k\pi/n$).


*The reflection case leads to the one-dimensional "$\pm1$" represenation (only possible with even $n$). But I think you can figure out the details yourself ;)
