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It suffices just to consider a linear transformation $f$ such that $f^n=id$ and require $V$ to have no proper subspace invariant under $f$. But I still don't have a picture of what's going on.

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    Hint: rotations.2012-10-23
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    Hi @nik do you mean $V$ must be less than 2 dimensional?2012-10-23
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    It depends on $n$. If $n = 1,2$, then you can easily find a representation of degree 1; otherwise, if you want to find all the representations, for some of them you will need $V$ to be of dimension at least 2 (and in fact 2 is enough).2012-10-23
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    You mean if the dimension is greater than 2 I can always decompose into direct sums of 1 or 2 dimensional subspaces? Why2012-10-23
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    How much do you know about linear algebra and representations? One possible answer is that if $f \in \mathrm{GL}(V)$ is such that $f^n = id$, then $f$ is diagonalizable over $\mathbb{C}$; its eigenvalues are roots of unity, and it's then easy to show that the eigenvalues are either real or come in pairs of conjugate numbers. Then you're left with a decomposition of $f$ as a direct sum of $\pm 1 \in \mathrm{GL}(\mathbb{R})$ and of rotations $\in \mathrm{GL}(\mathbb{R}^2)$.2012-10-23
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    @NajibIdrissi Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2015-11-07
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    @JulianKuelshammer Done.2015-11-08

1 Answers 1

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Let $\rho : C_n \to \mathrm{GL}(V)$ be a representation, equivalently let $f : V \to V$ be an automorphism such that $f^n = \operatorname{id}_V$. Then $f$ is a root of the polynomial $X^n - 1$, hence it is diagonalizable over $\mathbb{C}$ and its complex eigenvalues are $n$th roots of unity. These roots are either real or come in pair of complex conjugate numbers: let $A \in M_d(\mathbb{R})$ be the matrix associated to $f$; then if $\lambda \in \mathbb{C} \setminus \mathbb{R}$ is an eigenvalue of $A$, i.e. $Av = \lambda v$ for nonzero $v$, then $$\overline{Av} = \overline{\lambda v} \implies A \bar{v} = \bar{\lambda} \bar{v}$$ hence $\bar{\lambda}$ is also an eigenvalue of $A$.

Now you can pair up $\lambda = e^{2ik\pi/n}$ and $\bar{\lambda} = e^{-2ik\pi/n}$. Let $\theta = 2k\pi/n$, such that $n \theta \equiv 0 \pmod{2\pi}$. The two following matrices are similar: $$\begin{pmatrix} e^{i \theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix} \sim \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

Conclusion: The irreducible real representations of $C_n$ are either

  • 1-dimensional, with matrix a real $n$th root of unity;
  • 2-dimensional, with matrix $\left(\begin{smallmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{smallmatrix}\right)$ where $n\theta \equiv 0 \pmod{2\pi}$ but $\sin\theta \neq 0$.