Character Table S4 I am trying to understand how to build a character table of S4. I've already read many articles about it but I am stuck at one point.
S4 has 5 conjugacy classes and therefore 5 irreducible representation. One of them is the identity which is one-dimensional. 
The dimensions of all representations is determined by the sum of dimensions squares, which must equal 24, because 4! = 24. From this, the representations have dimensions 1, 1, 2, 3, 3.
And now I am stuck. I know that one row will be filled with 1 (identity, dimension 1), but what about the rest? The other 1-dimensional representation is the signum of permutations, but I honestly don't know why (mostly because signum is not present in the character table of A4).
I'll appreciate any help.
 A: Two 1-dimensional representations for $S_4$ are immediate: as you said, the trivial representation $\rho_\mathrm{trivial}$ and the sign representation $\rho_\mathrm{sign}$. $A_4$, however, do not have a distinct sign representation because it is indeed trivial.
Another immediate representation is the one in $\mathbb{C}^4$ sending each permutation to its associated permutation matrix. This representation is reducible, with an obvious invariant subspace the span of $e_1+e_2+e_3+e_4$. As in the case of $S_3$ we may study its orthogonal complement of this 1-dimensional invariant space, i.e., the subspace spanned by $\{e_2-e_1,e_3-e_1,e_4-e_1\}$, which is again invariant. This sub-representation is the standard representation $\rho_\mathrm{standard}$. Using this particular basis, for example, we may compute the character $\chi_\mathrm{standard}$:

*

*$\chi_\mathrm{standard}(\mathrm{id})=3$ (size 1);

*$\chi_\mathrm{standard}((1 \ 2))=1$ (size 6);

*$\chi_\mathrm{standard}((1 \ 2 \ 3))=0$ (size 8);

*$\chi_\mathrm{standard}((1 \ 2 \ 3 \ 4))=-1$ (size 6);

*$\chi_\mathrm{standard}((1 \ 2) (3 \ 4))=-1$ (size 3).

Since $\langle \chi_\mathrm{standard}, \chi_\mathrm{standard} \rangle = 1$, it follows that $\rho_\mathrm{standard}$ is irreducible.
To cook up the other 3-dimensional irreducible representation, we may simply multiply $\rho_\mathrm{sign}$ and $\rho_\mathrm{standard}$. We may simply compute the character and verify that it has norm 1 to show that it is irreducible.
Finally we only have 1 row left in the character table, which corresponds to a 2-dimensional representation. Using orthogonality and the three other known irreducible representations, (say start with the class function that is 2 on identity element and 0 otherwise, and subtract the orthogonal projections, then normalize). It turns out the last character of irreducible representation $\chi$ is

*

*$\chi(\mathrm{id})=2$ (size 1);

*$\chi((1 \ 2))=0$ (size 6);

*$\chi((1 \ 2 \ 3))=-1$ (size 8);

*$\chi((1 \ 2 \ 3 \ 4))=0$ (size 6);

*$\chi((1 \ 2) (3 \ 4))=2$ (size 3).

In case you want to know, the corresponding representation is the composition of the canonical surjection $S_4 \to S_4/V \cong S_3$, where $V$ is the subgroup generated by the double transpositions, and the standard 2-dimensional representation $\rho_\mathrm{standard} : S_3 \to \mathrm{GL}(n,\mathbb{C})$.
