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Adjoin a new point $\omega$ fixed by a transitive group $G$. A one-point extension is a transitive group on the enlarged set whose stabilizer of $\omega$ is exactly $G$, acting on the remaining points in the prescribed way. Its order is $(n+1)|G|$ in the finite degree-$n$ case. The double-coset criterion for a one-point extension gives a concrete generator test.

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.

Adjoin $\infty$ to $\Omega =\{1,\dots ,n\}$ and take $G=A_n$ fixing it. Choose three distinct points $a,b,c\in \Omega$ and put $t=(\infty a)(bc)$. This is an even involution interchanging $\infty$ and $a$. Conjugation by it preserves $H=G_a$, because it permutes the remaining $n-1$ points and preserves parity. Thus $tHt=H$.

If $g\in G\setminus H$, the even permutation $k=tgt$ moves $\infty$ into $\Omega$. The alternating group $A_n$ is transitive for $n\ge 3$: a suitable 3-cycle carries any prescribed point to another. Choose $u\in G$ with $u(a)=k(\infty )$. Then $v=t{u}^{-1}k$ is even and fixes $\infty$, so $v\in G$ and $k=utv\in GtG$. All conditions of the double-coset criterion for a one-point extension hold, including when $n=3$ and $H$ is trivial.

The constructed extension lies in $A_{n+1}$ and has order $(n+1)|A_n|=(n+1)!/2$. Therefore

Here is the double-coset criterion for a one-point extension. Let $H=G_{\alpha }$, and suppose $x$ swaps $\omega$ and $\alpha$. Then $⟨G,x⟩$ is a one-point extension if and only ifFor necessity, in an extension $H$ fixes both $\omega$ and $\alpha$, so $x$ normalizes it and $x^2$ lies in it. Since $G$ is transitive on $\Omega$, the extension has exactly two double cosets relative to $G$: $G$ and $GxG$. For $g\notin H$, $xgx$ moves $\omega$ into $\Omega$ and is in the latter double coset.

For sufficiency, the displayed conditions make $G\cup GxG$ closed under multiplication. Products with middle element in $H$ reduce using $xhx=(xh{x}^{-1})x^2\in G$; those with middle element outside $H$ remain in $GxG$. A finite nonempty multiplication-closed set of permutations containing the identity is a group. It contains $G$ and $x$, hence equals $⟨G,x⟩$. Every element in $GxG$ moves $\omega$, while $G$ fixes it, giving the required stabilizer subgroup. The group is transitive because $G$ is transitive on $\Omega$ and $x$ moves the additional point.

Adjoin the label $9=\infty$. Inversion on the projective line, with zero and infinity interchanged, isThe label $2=1\in F$ is fixed. Set $H=G_1=\{M_s:z\mapsto sz\}$, where the subscript one denotes the original point label, namely field zero. Then $x^2=1\in H$ and $x{M}_{s}x=M_{s^{-1}}$, so $x$ normalizes $H$.

For $g(z)=sz+t$ outside $H$, we have $t\ne 0$. On the projective line,The equality uses characteristic two and is valid as an equality of fractional linear transformations, including poles and infinity. ThusAll conditions of the double-coset criterion for a one-point extension hold. Therefore $G^{+}=⟨G,x⟩$ is a one-point extension, with $|G^{+}|=9\cdot 56=504$. Its stabilizer subgroup is sharply two-transitive, so its action on nine points is sharply three-transitive.