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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!

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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.

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  • $\begingroup$ Perhaps I misunderstood your previous concern but I thought you wanted a self-contained way to show that there were only two irreps of degree 1? Namely, without appealing to the degree theorem. If you were to use that and know a priori that there were exactly two irreps of degree one, then it's a simple matter of producing two nonisomorphic ones. $\endgroup$ – Future Dec 27 '15 at 22:25
  • $\begingroup$ Yes you are right. I am not really satisfied with this answer I give (the use of the degree theorem at this stage, you rightly pointed out, is the unsatisfying part) but as I said, this is the only way I can think of that uses the explicited structure of $S3$. I thought it is worth posting, though I will be happier with a proof avoiding an early use of the degree theorem, and I woulld gladly accept it. $\endgroup$ – A.B. Dec 28 '15 at 0:37
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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.

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  • $\begingroup$ Thanks. But he seems to be going the other way round: from the short discussion of the beginning, he derives that there are two degree one irreducible characters, then, using the theorem you mentionned, he knows there remains exactly one other irreducible character whose degree $n_3$ can be calculated since the sum of the squares of the degrees is $|G|=6$ and whose values can be known given that the weighted sum of irreducible characters is the character is the regular representation. $\endgroup$ – A.B. Dec 25 '15 at 20:25
  • $\begingroup$ I see. So you want to know how just using the group relations you can deduce there are exactly two irreducible representations of degree one? $\endgroup$ – Future Dec 25 '15 at 20:33
  • $\begingroup$ And why do you know beforehand that the trivial representation and the sign representation are always there, which he didn't prove before, and would have made his introductory discussion useless (this discussion, namely that $t^2=1$, $c^3=1$ and $tc=c^2 t$, seems to be relevent- he writes "hence" after that)? $\endgroup$ – A.B. Dec 25 '15 at 20:33
  • $\begingroup$ I hadn't seen your comment before posting mine.That's exactly what I want to know! $\endgroup$ – A.B. Dec 25 '15 at 20:35
  • $\begingroup$ So the trivial representation is always there. I am not sure whether he mentioned the sign rep up until this point (I have read Serre, but it was not the first source I used to learn representation theory), I also am not sure why he would need to explicitly write out the relations since they are not necessary to compute the character table. I will read a bit more and see if I can find out why. $\endgroup$ – Future Dec 25 '15 at 20:38

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