[proof] Let $H = G_\alpha$ be the stabilizer of some fixed $\alpha \in X$. **Step 1: Group-theoretic setup.** Since $G$ is transitive, all stabilizers are conjugate: $gG_\alpha g^{-1} = G_{g\alpha}$. The assumption that distinct stabilizers meet only in the identity ($G_\alpha \cap G_\beta = \{1\}$ for $\alpha \neq \beta$) and the definition of $K$ give the partition \begin{align*} G = K \cup \bigcup_{\alpha \in X} G_\alpha, \end{align*} with all pieces meeting only at the identity. Counting gives $|G| = |K| + n(|H| - 1)$. By the orbit-stabilizer theorem $|G| = n|H|$, so $|K| = n$. **Step 2: Induced characters from $H$.** One computes directly from the definition of the induced character that for any character $\psi$ of $H$: \begin{align*} \operatorname{Ind}_H^G \psi(g) = \begin{cases} n\psi(1) & g = 1, \\ \psi(g) & g \in H \setminus \{1\}, \\ 0 & g \in K \setminus \{1\}. \end{cases} \end{align*} The key point for $g \in H \setminus \{1\}$ is that $xgx^{-1} \in H$ if and only if $x \in H$ (since $xgx^{-1} \in G_{x\alpha}$, and $g$ fixes exactly one point). The key point for $K \setminus \{1\}$ is that derangements are not conjugate into any stabilizer. **Step 3: The magic characters $\theta_i$.** Let $\psi_1, \ldots, \psi_t$ be the irreducible characters of $H$. For each $i$, define \begin{align*} \theta_i = \psi_i^G - \psi_i(1)\,(1_H)^G + \psi_i(1)\,1_G. \end{align*} By direct substitution using Step 2, $\theta_i$ takes the values $\psi_i(h)$ for $h \in H$ and $\psi_i(1)$ for $k \in K$. In particular $\theta_i$ is constant (equal to $\psi_i(1)$) on all of $K$. To confirm $\theta_i$ is a genuine character (not merely a virtual character), compute \begin{align*} \langle \theta_i, \theta_i \rangle_G = \frac{1}{|G|}\left(n|\psi_i(1)|^2 + n\sum_{h \neq 1, h \in H}|\psi_i(h)|^2\right) = \frac{n}{|G|}\sum_{h \in H}|\psi_i(h)|^2 = \langle \psi_i, \psi_i \rangle_H = 1. \end{align*} So $\pm\theta_i$ is an irreducible character. Since $\theta_i(1) = \psi_i(1) > 0$, it is $\theta_i$ itself. **Step 4: The character $\Theta$ with kernel $K$.** Define \begin{align*} \Theta = \sum_{i=1}^t \theta_i(1)\,\theta_i. \end{align*} Column orthogonality of the irreducible characters $\psi_i$ of $H$ gives, for $1 \neq h \in H$: \begin{align*} \Theta(h) = \sum_{i=1}^t \psi_i(1)\psi_i(h) = 0, \end{align*} and for any $y \in K$: \begin{align*} \Theta(y) = \sum_{i=1}^t \psi_i(1)^2 = |H|. \end{align*} Thus $\Theta(g) = |H|$ for $g \in K$ and $\Theta(g) = 0$ for $g \notin K$ (except that $\Theta(1) = |H|$). The kernel of any representation affording $\Theta$ is exactly $K = \{g \in G : \Theta(g) = \Theta(1)\}$. Since kernels of characters are normal subgroups, $K \lhd G$. [/proof]