# Abstract Nonsense

## Character Table of S_3 Without Finding Irreducible Characters

Point of post: In this post we construct the character table of $S_3$ without having to actually find the irreducible characters.

Motivation

As was stated in our last post this post shall serve to show how nice the theory we’ve devoloped  can make the construction of character tables. In particular, it is often very unapparent percisely how to construct all the irreducible characters. It is just an odd coincidence that for $S_3$ the irreps are so obvious. So, we shall show that in this post except for the trivial irrep which requires no thought to construct we don’t even need to construct either of the other two characters.

Character Table of $S_3$ Without Finding the Irreducible Characters

You’ll recall from our last post that we were able to ascertain that the character table for $S_3$ is of the form

$\begin{array}{c|ccc} & e & (1,2) & (1,2,3)\\ \hline \chi^{\text{triv}} & 1 & 1 & 1\\ \chi^{(\alpha_2)} & 1 & a & b\\ \chi^{(\alpha_3)} & 2 & c & d\end{array}$

Moreover, since $1+\#\left(\mathcal{C}_{(1,2)}\right)+\#\left(\mathcal{C}_{(1,2,3)}\right)=6$ and $\#\left(\mathcal{C}_{(1,2)}\right),\#\left(\mathcal{C}_{(1,2,3)}\right)\mid 6$ it clearly follows that one of the conjugacy classes of $(1,2)$ and $(1,2,3)$ contain three elements and the other two. But, it’s clear by inspection that $(2,3)$ and $(1,3)$ are conjugate to $(1,2)$ (not only by inspection, but a common theorem I will discuss later which says that two elements of a symmetric group are conjugate if and only if they have the same ‘cycle type’) and so $\mathcal{C}_{(1,2)}=\left\{(1,2),(1,3),(2,3)\right\}$ and more importantly $\#\left(\mathcal{C}_{(1,2)}\right)=3$ and so $\#\left(\mathcal{C}_{(1,2,3)}\right)=2$. But, from CT 3 we know that

$\displaystyle 1+|a|^2+|c|^2=\frac{6}{3}=2\quad\mathbf{(2)}$

Note next that $|a|=1$ by CT 9. Thus, by $\mathbf{(2)}$ we may conclude that $c=0$. But, by CT 2 we have (considering the first and third rows)

$0=1\cdot(1\cdot 2)+3\cdot(1\cdot c)+2\cdot(1\cdot d)=2+2d$

so that $d=-1$. Moreover, by considering CT 2 again between rows two and three we see that

$0=1\cdot(2\cdot 1)+3\cdot(a\cdot c)+2\cdot(b\cdot d)=2-2b$

and so $b=1$. Finally, considering CT 2 one last time this time considering row two and row one we see that

$0=1\cdot(2\cdot 1)+3\cdot(1\cdot a)+2\cdot(1\cdot b)=3+3a$

and thus we may conclude that $a=-1$. Thus, we may finally conclude that the character table for $S_3$ is

$\begin{array}{c|ccc} & e & (1,2) & (1,2,3)\\ \hline \chi^{\text{triv}} & 1 & 1 & 1\\ \chi^{(\alpha_2)} & 1 & -1 & 1\\ \chi^{(\alpha_3)} & 2 & 0 & -1\end{array}$

Consequences of the Character Table

Using CT 5 and CT 6 which are really just statements about the kernel and center of a character respectively we may conclude that the set of all normal subgroups of $S_3$ is equal to $\bigg\{\{e\},\left\{e,(1,2,3),(1,3,2)\right\},S_3\bigg\}$ and the center is equal to $\{e\}$.

References:

1. Isaacs, I. Martin. Character Theory of Finite Groups. New York: Academic, 1976. Print.

2. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.