Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 7 / 2

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

Suppose an extension $H$ existed. Its degree would be $p$, and its order would be $2p(p-1)$ by the orbit-stabilizer theorem. A Sylow $p$-subgroup $P$ consequently has order $p$, and the Sylow theorems imply that its number $n_p$ divides $2(p-1)$ and is congruent to one modulo $p$. Since $2(p-1)<2p$, the only possibilities are $1$ and $p+1$. The latter would require $p+1\mid 2(p-1)$ and hence $p+1\mid 4$, impossible for $p\ge 5$. Thus $P$ is a normal subgroup of $H$.

A nonidentity element of $P$ is a $p$-cycle, so $P$ acts regularly on the $p$ points. Label them by ${\mathbb{F}}_p$ so that $P$ consists of translations. A permutation normalizing $P$ must have the form $x\mapsto ax+b$: its conjugation sends translation by one to translation by some $a\ne 0$, and its value at zero determines $b$. Therefore the normalizer of $P$ in $S_p$ has order $p(p-1)$. Normality would place $H$ inside this normalizer, contradicting $|H|=2p(p-1)$. Hence the specified dihedral permutation group has no one-point extension.

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