Degree one irreducible representations

Question

In section 2.5 of his Linear representations of finite groups (I have the french copy), Serre gives an example of determination of the character table of a group $G$. The group $G$ is taken to be $S_3$. He points out that the order of $G$ is $6$, that there are three conjugacy classes (the identity $1$, the three transpositions and the two cyclic permutations, which imply there are exactly three irreducbile characters), and that if $t$ is a transposition and $c$ a cyclic permutation, then $t^2=1$, $c^3=1$ and $tc=c^2t$. No problem until here. But then he says

Hence we have two irreducible characters of degree one: the unit character (character of the unitary representation) and the character giving the signature of a permutation.

What I would like to understand is how the above imply the quoted part, since the argument seems to come up again in the example of the following section. Note that he doesnt directly define the character as a homomorphism, which would have made the answer almost obvious to me, but as a trace (if $\rho$ is a reperesenation with character $\chi$, then $\chi (s)=Tr(\rho_s)$ ).

Thanks!

Answer 1

The next line (in translated version) says

Theorem 7 shows that there exists one other irreducible character [of degree 2]$

This shows that the irreducible representations are two degree one representations and a degree two representation.

The signature representation is just the sign representation, which one always has and is of degree 1 and one always has the trivial representation, also of degree one. This shows exactly what the two irreducible representations of degree one are.

Answer 2

Here is the closest I came to the argument of Serre, in case someone might be interested.

By prop.7, the number of irreducible characters is 3. If we denote by $n_1$, $n_2$ and $n_3$ their degrees, we know that $n_1^2 + n_2^2 + n_3^2=|G|=6$. So the only possibility is that two of them are of degree one. Lets denote by $\chi$ one of those two and $\rho$ the corresponding representation. Then $\rho_t$ and $\rho_c$ are scalar functions of ratio $\lambda$ and $\mu$ respectively, and we have $\chi (t)=\lambda$, $\chi (t^2)=\lambda^2=1$, $\chi (c)=\mu$,$\chi (c^3)=\mu^3=1$ and $\chi (tc)=\lambda\times\mu=\mu^2\times\lambda$. So we are left we exactly two possibilities for $\chi$: the unit character and the sign character.