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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?

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  • $\begingroup$ $C_k-$cyclic group of order $k$? If so, what do you know about the dimension of irreducible representations of abelian groups? $\endgroup$
    – Want to learn
    Apr 25, 2020 at 17:06
  • $\begingroup$ math.stackexchange.com/questions/1661134/… $\endgroup$
    – Want to learn
    Apr 25, 2020 at 17:13
  • $\begingroup$ $\rho$ is an irreducible representation $\iff $ dim$(V)=1$? where $V$ denotes the FG-module. $\endgroup$
    – user459879
    Apr 25, 2020 at 17:14
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    $\begingroup$ No. invertible matrices. So it is $\Bbb R^* = \Bbb R \backslash \{0\}$. $\endgroup$
    – Want to learn
    Apr 25, 2020 at 17:24
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    $\begingroup$ 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. $\endgroup$
    – Captain Lama
    Apr 26, 2020 at 12:45

1 Answer 1

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This follows on from the discussion in the comments.

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 $\mathbb{C}$?

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