The irreducible representation of $S_n$ corresponding to the partition $n = (n − 1) + 1$

Question

Let a finite group $G$ act doubly transitively on a finite set $X$, i.e., given $x,y,z,w\in X$ such that $x\neq y$ and $z\neq w$, there is a $g\in G$ such that $g.x=z$ and $g.y=w$. So, we can write $$\chi = \mathbf{1}_G + \theta$$ where $\chi$ is the trivial representation and $\theta$ is a character in which $\mathbf{1}_G$ does not appear. Show that $\theta$ is irreducible.

This was an exercise that I have solved with a little help from here. The next exercise is

Show that the irreducible representation of $S_n$ corresponding to the partition $$n = (n − 1) + 1$$ is the representation with character $\theta$ from the previous exercise.

I have proved that $n=n$ corresponds to the trivial representation and $n=1+\dots +1$ corresponds to the sign representation by showing how they act on a basis element and extending that to the other elements. But, I can't figure the given question out. In fact, I can't even understand how to construct the column stabilizer to start with.

Answer 1

Maybe you know the irreducible representation of $S_n$ corresponding to partition $n=(n-1)+1$ is the Specht module $S^{(n-1,1)}$. Actually, $S^{(n-1,1)}$ is isomorphic to the $(n-1)$-dimensional irreducible submodule in $\mathbb{C}S_n\{\mathbf{1,2,3,\dots,n}\}$, and we denote the character of $S^{(n-1,1)}$ as $\psi$, the character of $\mathbb{C}S_n\{\mathbf{1,2,3,\dots,n}\}$ as $\chi$. By Maschke's Theorem, we can see $\chi=1+\psi$ (it is easy to prove). Back to the question, you should prove that the action of $S_n$ on $\{\mathbf{1,2,3,\dots,n}\}$ is 2-transitively. It is obviously by 3-cycles. Then, by the front exercise, $\chi = 1+\theta$ which means $\theta = \psi$.