\(\def\C{\mathbf{C}}\)
Cyclic Groups
 The cyclic group \(\C_n\) is of order \(n\). We use additive notation, so the identity element is 0.
 As the group is Abelian, no two distinct elements are conjugate, so there are \(n\) conjugacy classes each containing 1 element
 This is close to the theory of Fourier series.
C_{2}
 0
 1

A_{0}
 1
 1

A_{1}
 1
 1

C_{3}
 0
 1
 2

A_{0}
 1
 1
 1

E
 2
 1
 1

 The representation E is irreducible but not absolutely irredicible. Indeed, over \(\mathbb{C}\) it splits as \(E = E_+ \oplus E_\), where \(E_+\) has character \((1,\,\omega,\,\bar\omega)\) and \(E_\) has character \((1,\,\bar\omega,\,\omega)\), where \(\omega\) is a cube root of unity.
C_{4}
 0
 1
 2
 3

A_{0}
 1
 1
 1
 1

A_{1}
 1
 1
 1
 1

E
 2
 0
 2
 0

C_{5}
 0
 1
 2
 3
 4

A_{0}
 1
 1
 1
 1
 1

E_{1}
 2
 \(\gamma\)
 \(\bar\gamma\)
 \(\bar\gamma\)
 \(\gamma\)

E_{2}
 2
 \(\bar\gamma\)
 \(\gamma\)
 \(\gamma\)
 \(\bar\gamma\)

 \(\gamma = \textstyle\frac12(\sqrt51)\) and \(\bar\gamma = \textstyle\frac12(\sqrt5+1)\)
C_{6}
 0
 1
 2
 3
 4
 5

A_{0}
 1
 1
 1
 1
 1
 1

A_{1}
 1
 1
 1
 1
 1
 1

E_{1}
 2
 1
 1
 2
 1
 1

E_{2}
 2
 1
 1
 2
 1
 1

And so the pattern goes on ...
 n even: \(\C_n\) has two 1dimensional representations and ½(n2) 2dimensional representations.
 n odd: \(\C_n\) has one 1dimensional representation and ½(n1) 2dimensional representations.
 Denote by \(E_r\) the 2d rep where the generator of \(\C_n\) has character \(2\cos(2\pi r/n)\), so acts on the plane by rotation through an angle of \(2\pi r/n\) (for \(1 \leq r \leq n/2\)). These 2dimensional reps are irreducible but not absolutely irreducible, and their complexification splits as a sum of two 1d reps.
 Indeed, the complexification \(E_r^{\mathbb{C}} = U_r \oplus U_{nr}\), where \(U_r\) is the 1d complex rep with the generator of \(\C_n\) acting as multiplication by \(\exp(2\pi ir/n)\).
More on Permutation Reps