Decomposition of Permutation Representation

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, $\mathrm{dim}\ \mathrm{End}_{\mathbb{C}G} (V_1) = 1$.