One-point extensions

Reversing the idea of a point stabilizer Let $(G,\Omega)$ be a transitive permutation group, take a point $\omega \not\in \Omega$ and form $\Omega^+ = \Omega \cup \{\omega\}$, extend the action by $\omega g = \omega$ for all $g \in G$:

Definition A one-point extensionof $(G,\Omega)$ is a transitive permutation group $(G^+,\Omega^+)$ with $(G^+)_\omega = G$

By the stabilizer-orbit theorem $|G_+| = |G|(|\Omega|+1)$. If $G^+$ is $t$-transitive then G is $(t-1)$-transitive.

Example $S_n$ and $A_n$ have one point extensions $S_{n+1}$ and $A_{n+1}$.
Non-example $D_8$ doesn't have one by Sylow theory.

Take $\alpha \in \Omega$, we know the rank $r$ of $G$ equal to the number of double cosets in $G$. For $g_1,\ldots,g_r \in G$ we have a complete representation of the double coset system iff $\alpha g_1,\ldots,\alpha g_r$ is a complete set of representatives of $G_\alpha$-orbits. wlog take $g_1 = 1$, if $(G^+,\Omega^+)$ is a one point extension of $(G,\Omega)$ then it is 2-transitive so it has a primitive action and hence the point stabilizer $G$ is a maximal subgroup of $G^+$: for any $x \in G^+ \setminus G$ we must have $\langle x, G \rangle = G^+$. We can wlog choose $x$ to interchange $\alpha$ and $\omega$.

Theorem Let $(G,\Omega)$ be a transitive permutation group of rank $r$ for $\alpha \in \Omega$, let $g_1=1,g_2\ldots,g_r$ be a complete set of representatives of the double coset system. Take $\omega \not\in \Omega$ and form $\Omega^+ = \Omega \cup \{\omega\}$. Take $x \in S_{(\Omega^+)}$ (so some permutation from the symmetric group) with $\alpha x = \omega$, $\omega x = \alpha$ and set $G^+ = \langle x, G \rangle$ then $(G^+,\Omega^+)$ is a one point extension iff:

1. $x^2 \in G_\alpha$
2. $(G_\alpha)^x = G_\alpha$
3. $g_i^x \in GxG$ for all $i > 1$.

proof: ($\Rightarrow$) (1) definition of $x$ (2) easy (3) As $x \not\in G$ the 2-transitivity of $G^+$ means that $G^+ = G \cup GxG$ (any permutation that can't be done with $g$ can be done with $x$ in the middle), for $2 \le i \le r$ we have $g_i \in G\setminus G_\alpha$ so $\omega g_i^x = \alpha g_i x \not = \alpha x = \omega$ thus $g_i^x \not \in (G^+)_\omega = G$ so $g_i^x \in GxG$. ($\Leftarrow$) Set $H=G \cup GxG$as a set, we will show its a group $H \le S_{\Omega^+}$, to show it's a group we need $HH = H = H^{-1}$. Since $G^{-1}= G$ and $(gxg')^{-1} = g'^{-1} x (x^2)^{-1} g^{-1} \in GxG$ . By (2) $x G_\alpha x= G_\alpha$ so for $i \ge 2$ we have $$x G_\alpha g_i G_\alpha x = G_\alpha (x g_i x) G_\alpha = G_\alpha x^2 g_i^x G_\alpha \subseteq GxG$$ so $$xGx = xG_\alpha x \cup \bigcup_{i=2}^r x G_\alpha g_i G_\alpha x \subseteq G_\alpha \cup GxG \subseteq H.$$ Now $HH = G \cup GxG \cup GxGxG \subseteq H \cup GHG = H$, we have shown it's a group! Now $G^+ = \langle G,x\rangle = H$ and for all $g,g'$ take $\omega (g x g') = \omega x g' = \alpha g' \not = \omega$ so $(G^+)\omega = G$.