# Representation theory - determining al irreducible FG-modules of dimension $n$.

I am asked to determine all irreducible $$FG$$-modules of dimension $$n$$ for $$G=C_k$$, the cyclic group of order $$k\in \mathbb N$$ in the following cases:

• $$n=1, F= \mathbb R$$
• $$n=2, F= \mathbb R$$
• $$n=1, F= \mathbb C$$
• $$n=2, F= \mathbb C$$

I honestly do not know where to start. This seems like a classification question, can someone help me to get started?

• $C_k-$cyclic group of order $k$? If so, what do you know about the dimension of irreducible representations of abelian groups? Apr 25, 2020 at 17:06
• math.stackexchange.com/questions/1661134/… Apr 25, 2020 at 17:13
• $\rho$ is an irreducible representation $\iff$ dim$(V)=1$? where $V$ denotes the FG-module.
– user459879
Apr 25, 2020 at 17:14
• No. invertible matrices. So it is $\Bbb R^* = \Bbb R \backslash \{0\}$. Apr 25, 2020 at 17:24
• Indeed, it is not true that an irreducible representation of an abelian group over $\mathbb{R}$ has to be $1$-dimensional. It can be $2$-dimensional. Apr 26, 2020 at 12:45

Let $$V_k$$ be the $$2$$-dimensional $$\mathbb{R}C_k$$-representation that you described. We shall show that $$V_k$$ is irreducible when $$k > 2$$, and reducible when $$k = 1,2$$
Suppose that $$k>2$$ and suppose, for the sake of contradiction, that $$V$$ is reducible, that is it contains some non-zero proper subrepresentation $$U \subseteq V$$. Since $$V$$ is $$2$$-dimensional, that forces $$U$$ to be one-dimensional. In other words any non-zero $$u \in U$$ is simultaneously an eigenvector for the action of $$C_k$$ (or alternatively an eigenvector for its generator since it is cyclic). But this generator has no eigenvectors when $$k > 2$$, and so we have reached our contradiction.
When $$k = 1,2$$ the vector $$\begin{pmatrix} 1 \\ 0\end{pmatrix}$$ is an eigenvector of the generator of $$C_k$$, and so generates a non-trivial one-dimensional submodule of $$V_k$$.
For an extension, why does this argument fail if we replace $$\mathbb{R}$$ with $$\mathbb{C}$$?