A character-free proof that a permutation group is doubly transitive iff the associated permutation module over $\mathbb C$ has irreducible augmentation submodule? Let $G$ be a group of permutations of a finite set $X$.  By the augmentation submodule of $\mathbb CX$, I mean the set of vectors whose coefficients sum to $0$.  It is easy to show via character theory that the augmentation submodule is irreducible iff $G$ is doubly transitive.  Does anybody know a proof using only the definitions of doubly transitive and irreducible representation and avoiding character theory and the orthogonality relations?  It is known (I learned this from Peter Cameron, but don't know a good reference) that replacing $\mathbb C$ by $\mathbb R$ characterizes double homogeneity.  So the field being algebraically closed is somehow important.
The motivation for this question comes from trying to understand the relationship between double transitivity for transformation monoids and irreducibility of the augmentation submodule.  Character theory for monoids is harder to apply and so an answer to my question may provide some insight.
 A: I am not sure where you want to draw the line about use of character theory.
This equivalence is a statement about the endomorphism ring of the permutation module $\mathbb{C}X$,
and doesn't really require use of characters or orthogonality relations. It does require Schur's lemma, which crucially uses algebraic closure. Using Schur's Lemma, the endomorphism ring of the permutation module is $2$-dimensional if and only if the action of $G$ on $X$ is transitive and the augmentation submodule is irreducible. On the other hand, if the acton of $G$ on $X$ is transitive and $H$ is a point stabilizer, it is just a question of looking at matrices to see that the dimension of the endmorphism ring is the number of $(H,H)$-double cosets. (These things can be proved by Frobenius reciprocity, but they can be seen directly in this case).
A: Maybe this is the sort of direct proof you're looking for?
Suppose the action of $G$ is not $2$-transitive. Then $G$ preserves some nontrivial directed graph $\mathcal{G}$ with vertex set $X$ (the $G$-orbit of your favourite edge). Therefore $G$ preserves the eigenspaces of this graph, i.e., the eigenspaces of the adjacency operator defined by
$$\mathcal{A} f (x) = \sum_{y:(x,y)\in\mathcal{G}} f(y).$$
If the whole augmentation submodule were an eigenspace then $\mathcal{A}$ would act as some scalar $\lambda$, and then it's easy to see that $\lambda$ must be $-1$ and all edges must be in $\mathcal{G}$, a contradiction. Therefore each eigenspace is a proper invariant subspace.
(To be fair, I feel sure that on inspection this proof can be little more than the usual proof, but for dummies. If we insist that $\mathcal{G}$ is the orbit of a single directed edge then $\mathcal{A}$ is closely related to the averaging operator often seen in proofs of complete reducibility, we're using the fact that $G$ and $\mathcal{A}$ commute, and then we're essentially copying the proof of Schur's lemma.)
