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Here is an involution form of the double-coset criterion for a one-point extension. Embed $G$ in $\mathrm{Sym}(\Omega \cup \{\infty \})$ by fixing $\infty$, choose $a\in \Omega$, and put $H=G_a$. The necessary and sufficient condition is the existence of an involution $t$ such thatProducts in this part are compositions of maps, with the rightmost map applied first. The criterion is a finite test on permutations of the enlarged set; it does not presuppose that a larger group has already been found.

For sufficiency, put $K=G\cup GtG$. Multiplying a double coset by an element of $G$ on either side leaves it unchanged. A product of two elements of $GtG$ reduces to $g_{1}tgt{g}_2$. If $g\in H$, then $tgt\in H\le G$; otherwise the last condition puts it in $GtG$. Hence $K$ is closed under multiplication. It contains the identity and is closed under inverses, since $t^{-1}=t$, so it is a group.

Every element $g_{1}t{g}_2$ sends $\infty$ to $g_1(a)\in \Omega$, whereas elements of $G$ fix $\infty$. Thus $K_{\infty }=G$. Moreover the orbit of $\infty$ contains $a$ and then all of $\Omega$, so $K$ is transitive and is the desired one-point extension of a permutation group. For the order check, $G\cap tGt=H$: an element of the intersection fixes both $\infty$ and $a$, and the reverse containment follows from $tHt=H$. Consequently $|GtG|=|G{|}^2/|H|=|\Omega ||G|$, as required.

For necessity, let $K$ be an extension. It is doubly transitive group action, of order $|\Omega |(|\Omega |+1)|G_a|$, which is even. By Cauchy theorem for groups it contains a nontrivial involution, and a pair interchanged by this involution can be carried to $(\infty ,a)$ by double transitivity. Conjugating gives an involution $t$ interchanging these two points. Their common point stabilizer is $H$, so $tHt=H$. Finally $G$ has just two orbits on the enlarged set, namely $\{\infty \}$ and $\Omega$, giving $K=G\cup GtG$. More explicitly, for $k\in K\setminus G$, choose $g\in G$ with $g(a)=k(\infty )$; then $t{g}^{-1}k$ fixes $\infty$ and belongs to $G$. If $g\notin H$, the element $tgt$ moves $\infty$, so it belongs to $GtG$, completing necessity.

Identify $G$ with a subgroup of $\mathrm{Sym}(\Omega )$ and adjoin one new point $\infty$, fixed by every element of $G$. A one-point extension of a permutation group is a transitive subgroup $G^{+}\le \mathrm{Sym}(\Omega \cup \{\infty \})$ whose point stabilizer at $\infty$ is exactly this embedded copy of $G$. Equivalently, one may use an isomorphic permutation representation with a distinguished point whose stabilizer acts on the remaining points as the original group.

Since $G$ is transitive on $\Omega$, the extension is doubly transitive group action. The orbit-stabilizer theorem gives the useful necessary order condition

Extend each element of $G$ to fix $\omega$. A one-point extension of a permutation group is a transitive permutation group $G^{+}\le S_{{\Omega }^{+}}$ such thatwith the induced action on $\Omega$ equal to the prescribed action. The stabilizer subgroup equality is the substantive condition: merely adjoining an element that moves $\omega$ does not suffice. If $|\Omega |=n$, the orbit-stabilizer theorem gives $|G^{+}|=(n+1)|G|$.