The following is from Serre's Linear Representations of Finite Groups.

Take for $G$ the group of permutations of three letters. We have $g = 6$ and there are three classes: the element 1, the three transpositions, and the two cyclic permutations. Let $t$ be a transposition and $c$ a cyclic permutation. We have $t^{2} = 1$, $c^{3} = 1$, $tc = c^{2} t$; whence there are just two characters of degree 1: the unit character $\chi_{1}$ and the character $\chi_{2}$ giving the signature of a permutation.

The last sentence is mysterious to me. How does one conclude from the computations that there are only two degree 1 irreducible representations?


A $1$-dimensional character of $G$ is a function $f : G \to \mathbb{C}^{\times}$ such that $f(gh) = f(g) f(h)$ for all $g, h$. Applying the relations gives $f(t)^2 = 1, f(c)^3 = 1, f(t) f(c) = f(c)^2 f(t)$, which gives $f(c) = 1$ and $f(t) = \pm 1$. The two choices of sign give the trivial and sign representation respectively.

More generally, the $1$-dimensional characters of a finite group $G$ are the same as the $1$-dimensional characters of the abelianization $G/[G, G]$. In this case the relation $tc = c^2 t$ shows that $c$ is a commutator, and in fact the abelianization of $S_3$ is $C_2$, generated by $t$ and with kernel generated by $c$.


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