Doubly transitive group has irreducible augmentation module follow up

This is a follow-up question from Doubly transitive group has irreducible augmentation module?

If $\mathbb{k}$ is of characteristic zero, it is well known that $W$ is irreducible. How can one prove it?

Solution 1:

Let $H$ be a point stabilizer and $\pi=1_H\uparrow^G$ be the permutation character for the natural permutation action. Then (Frobenius reciprocity) $$ (\pi,\pi)_G=(\pi,1_H\uparrow^G)_G=(\pi|H,1_H)=2 $$ since $H$ has two orbits. But then $\pi=1+\chi$ for an irreducible character $\chi$. OTOH $\chi=\pi-1$ is the action on $W$.

Solution 2:

There is also a character-free proof showing that the assertion holds even in positive characteristics under certain conditions (this was the objective of my question you linked above).

Let $\mathbb{K}$ be a field and let $G$ be a doubly transitive permutation group on $\{ 1, \dots, n \}$.

Claim: The $\mathbb{K} G$-Module $W = \{ x \in \mathbb{K}^n : \sum_i x_i = 0 \}$ is irreducible provided that neither $n$ nor $|G_{1,2}|$ is divisible by the characteristic of $\mathbb{K}$. ($G_{i,j}$ denotes the stabilizer of both $i$ and $j$ in $G$)

Sketch of a proof: Since $G$ is doubly transitive, it suffices to show that any nonzero $\mathbb{K}G$-submodule $M \leq W$ contains the vector $(1,-1,0,\dots,0)$ (it immediately follows that $M$ contains a $\mathbb{K}$-basis of $W$ then).

Let $0 \neq x \in M$ be arbitrary. Since $G$ is transitive, we may assume $x_1 \neq 0$. We consider the element $y := \sum_{g \in G_1} gx \in M$. It has the form $y_1 = |G_1| \cdot x_1$ and $$ y_2 = y_3 = \dots = y_n = |G_{1,2}| \cdot \sum_{i=2}^n x_i. $$ Since $x_1 \neq 0$ by assumption and since $x \in W$, we also have $\sum_{i=2}^n x_i \neq 0$. Since $|G_{1,2}|$ is nonzero in $\mathbb{K}$ by assumption, it follows $y_i \neq 0$ for $i \geq 2$. Hence $y \in M$ is an element of the form $y = (a,b,\dots, b)$ with $b \neq 0$ and $a \neq b$. ($a = b$ would imply $n = 0$ in $\mathbb{K}$) Now let $g \in G$ be any element transposing $1$ and $2$. We have $y - gy = (a-b, b-a, 0, \dots, 0) \in M$. Dividing by $(a-b)$ finally implies $(1,-1,0, \dots, 0) \in M$.