Find character table for symmetric group $S_3$

Question

This group contains all permutations of 3 elements, so it has order 3!=6.

Its three congruency classes are {1}, {(1,2),(1,3),(2,3)}, {(123),(132)}. As we know that the number of congruency classes equals the number of irreducible characters, we know that there are 3 irreducible characters (of degrees, say, $a,b,c$, such that $a^2+b^2+c^2=6$).

The two most obvious representations are the trivial one and the sign one, both of degree 1. Hence $a=b=1$, and therefore $c=2$.

How can I find $c$? My book says to use the orthogonality relations (i.e. irreducible characters form an orthonormal system), but I don't really see how.

Answer 1

You have already find two representations and the dimension of the last one, so you have the table : $$\begin{array}{|c|c|c|c|} \hline & \text{Id} & (i,j) & (i,j,k) \\ \hline 1\!\!1 & 1 & 1 & 1 \\ \hline \epsilon & 1 & -1 & 1 \\ \hline \chi & 2 & \alpha & \beta \\ \hline \end{array}$$

Do you know why the first entry of the last row $\chi(\text{Id})=2$ ? Now, to find $\alpha$ and $\beta$, you can use the orthogonality of the columns. For example, we know that $(1,1,2) \cdot (1,-1,\alpha)=0$ so $2\alpha=0$.

Answer 2

Not what you asked, but first here's the "natural" way to construct the third irreducible representation: $S_3$ has a natural 3-dimensional representation as permutation matrices. An obvious invariant subspace is $\{(u,v,w)\ | \ u+v+w=0\}$. Then the quotient is two-dimensional and irreducible.

Now to actually answer your question, yes, you can do this with the orthogonality relations. Let $x,y$ be the unknown character values (as you've noted, the character must send $1$ to $2$). Call your character $\chi$. Then $\langle\chi,1\rangle=0$ and $\langle\chi,sign\rangle=0$. This will give you two linear equations in the two unknowns $x$ and $y$, so you'll be able to find a unique solution.