Let $G$ be a finite group acting on a finite set, 2-transitively, and $(\rho,V)$ be the corresponding permutation representation of $G$ over $\mathbb{C}$. Then it is known that $V$ is direct sum of a trivial representation and an irreducible representation. The known proofs of this fact use character theory.
I tried to prove by "character free method" as follows:
Let $\{ e_1,\cdots, e_n\}$ be a basis of $V$ on which $G$ is acting 2-transitively. Then $V$ is direct sum of two $G$-invariant subspaces:
$V_0=\langle e_1+e_2+\cdots + e_n\rangle$, and $V_1=\langle e_1-e_2,e_2-e_3,\cdots, e_{n-1}-e_n\rangle$, and $G$ acts trivially on $V_0$.
To show that $V_1$ is irreducible, I proceed as follows:
Let $W\subseteq V_1$ be $G$-invariant subspace.
Case $1$: If $e_i-e_{i+1}\in W$ for some $i$, then by 2-transitivity of $G$, $\exists g\in G$ such that $g.e_i=e_j$ and $g.e_{i+1}=e_{j+1}$, hence $g.(e_i-e_{i+1})=e_j-e_{j+1}$, hence $W$ contains all basis vectors of $V_1$, hence $W=V_1$.
Question How can we proceed in Case $2$ for the proof?
Best Answer
To prove that $V_1$ is irreducible, it is sufficient to prove that $\mathrm{End}_{\mathbb{C}G}(V_1)$ is one-dimensional. (Subproof: if $V_1$ has a non-zero proper subrepresentation $W$ then $V_1 = W \oplus C$ for some complementary representation $C$. The projection maps onto $W$ and onto $C$ corresponding to this decomposition are linearly independent.)
Let $\theta \in \mathrm{End}_{\mathbb{C}G}(V_1)$. Extend $\theta$ to an endomorphism of $V = V_0 \oplus V_1$ by setting $\theta(V_0) = 0$.
By definition $G$ acts on $\{1,2,\ldots, n\}$. Let $H = \mathrm{Stab}_{G}(1)$. Suppose that
$$\theta(e_1) = a e_1 + \sum_{i=2}^n b_i e_i$$
where $a \in \mathbb{C}$ and $b_i \in \mathbb{C}$ for each $i \in \{2,\ldots, n\}$. Since $h \cdot 1 = 1$ for each $h \in H$, and $\theta$ is a $\mathbb{C}G$-homomorphism, the vector
$$\sum_{i=2}^n b_i e_i \in V_1 $$
is $H$-invariant. But $H$ has a single orbit on $\{2,\ldots,n\}$ because $G$ is $2$-transitive. Therefore $b_i$ is constant for $i \in \{2,\ldots,n\}$. Let $b$ be the common value.
To complete the proof, pick $g_1, \ldots, g_n \in G$ such that $g_i \cdot 1 = i$ for each $i \in \{1,2,\ldots, n\}$. We have
$$ 0 = \theta\bigl( \sum_{i=1}^n e_i \bigr) = \theta\bigl( \sum_{i=1}^n g_i \cdot e_1 \bigr) = \sum_{i=1}^n g_i \cdot \theta(e_1). $$
Comparing the coefficient of $e_1$ on both sides gives $a+(n-1)b = 0$. So $\theta$ is determined by $a$ and, as claimed, $\mathrm{dim}\ \mathrm{End}_{\mathbb{C}G} (V_1) = 1$.