How many irreducible representations are there in a cyclic group? $$
  \begin{array}{c|c|c|}
      \backslash
    & e
    & p
    \\
    \hline
      e
    & e
    & p
    \\
    \hline
      p
    & p
    & e
    \\
    \hline
  \end{array}
$$
(Original picture of the table here.)
First I want to look at the cyclic group $\mathbb{Z}_2$. In this group, the symmetric functions transform under the trivial representation ($D(e) = D(p) = 1$, so all group actions do nothing to the function), and the antisymmetric functions transform under the sign representation ($D(e) = 1, D(p) = -1$, so we need to act $D(p)$ twice to get the orignal function, acting once won't do). There are two irreducible representarion, and they all create their own subspace. All nice and dandy.
$$
  \begin{array}{c|c|c|c|}
      \backslash
    & e
    & a
    & b
    \\
    \hline
      e
    & e
    & a
    & b
    \\
    \hline
      a
    & a
    & b
    & e
    \\
    \hline
      b
    & b
    & e
    & a
    \\
    \hline
  \end{array}
$$
(Original picture of the table here.)
But what I don't understand is the group $\mathbb{Z}_3$, because it seems to me that they only create two irreducible representation also, whereas it should be three. Nothing wrong with the trivial representation subspace, but the two remaining representation:


*

*$D(a) = e^{2\pi i/3}$, $D(b) = e^{4\pi i/3}$

*$D(a) = e^{4\pi i/3}$, $D(b) = e^{2\pi i/3}$


seems indistinguishable. An "antisymmetric" function need $D(a)$ or $D(b)$ to act thrice to return to the original function, in both representation; and $D(a)D(b)$ or $D(b)D(a)$ also return the original function, in both representation. 
So what did I misunderstand here? Are there only two irreducible representations? Thank you!
 A: The point is that you are right if you are talking about real representations (indeed there are only two real irreducible representation of $\mathbb{Z}_3$. But the non trivial one is the $\frac{2 \cdot \pi}{3}$ rotation (which is isomorphic to the $\frac{4 \cdot \pi}{3}$ rotation) in $\mathbb{R}^2$, so its dimension is $2$).
But, since you are using complex numbers, I think you mean complex representations. 
In this case the two representation (let me call them $\rho_1$ and $\rho_2$ with $\rho_i : \mathbb{Z}_3 \to GL_1(\mathbb{C}) \cong \mathbb{C}$ ) are not isomorphic because there is no $\mathbb{C}-$linear invertible function $f: \mathbb{C} \to \mathbb{C}$ (maybe your confusion is here, you seem to be looking at $\mathbb{R}-$linear functions) such that $f( \rho_1(a)z) = \rho_2(a)( f(z))$ for any $a \in \mathbb{Z}_3$ and $z \in \mathbb{C}$.
This fact is indeed true because since $f: \mathbb{C} \to \mathbb{C}$ is $\mathbb{C}-$linear and invertible, there exist $\lambda \neq 0$ such that $f(z) = \lambda z$ for all $z \in \mathbb{C}$. 
Let $\omega = e^{\frac{2 \pi i}{3}}$, then for $a = [1]$ we have $ \rho_1(a)(z) = \omega z$ and $\rho_2(a)(z) = \omega^2 z$.
Now, the equation implies $\lambda \omega z = \lambda \omega^2 z$ for any $z \in \mathbb{C}$, which is absurd (because $\lambda \neq 0$).
Hope this is helpful.
