Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 3 / 4

Choose $\alpha \in \Omega$. Since $G$ acts regularly, $g\mapsto g\alpha$ identifies $\Omega$ with the underlying set of $G$, and the given action becomes left multiplication $L_g(x)=gx$.

A permutation $c$ centralizing all $L_g$ satisfiesConversely every right translation commutes with every left translation. To make the multiplication convention explicit, define $R_g(x)=x{g}^{-1}$. Then $R_g{R}_h(x)=x{h}^{-1}{g}^{-1}=x(gh{)}^{-1}=R_{gh}(x)$, using rightmost-first composition. The map $g\mapsto R_g$ is injective and its image is the entire centralizer, since any possible $c(1)$ can be written $g^{-1}$. ThusThis is centralizer of a regular permutation subgroup. If one instead writes right multiplication as $x\mapsto xg$, the resulting map is an antihomomorphism; including the inverse avoids suppressing that reversal. Choosing a different base point changes the identification but not the isomorphism type.

Retain the identification with the left regular action. A group group automorphism $\phi$ of $G$ satisfies $\phi L_g{\phi }^{-1}=L_{\phi (g)}$, so it normalizes $L(G)$ and fixes the identity point.

Conversely let a normalizing permutation $n$ fix the identity. Its conjugation on $L(G)$ induces a group group automorphism $\phi$ characterized by $n{L}_g{n}^{-1}=L_{\phi (g)}$. Evaluating at the identity gives $n(g)=\phi (g)$, so the whole identity stabilizer in the normalizer is exactly $\mathrm{Aut}(G)$. For arbitrary $n$, put $t=n(1)$. Then $L_t^{-1}n$ fixes the identity and is an group automorphism; thus every element has a unique expression $L_t\phi$.

The translation subgroup is normal in its normalizer, its intersection with the group automorphism subgroup is trivial, and the multiplication rule isConsequentlyThe action in this semidirect product is the natural evaluation of group automorphisms on $G$. This normalizer is the holomorph, realized as all permutations $x\mapsto t\phi (x)$. The decomposition is established from the regular group action itself rather than only from an order comparison.

Choose $\alpha \in \Omega$ and identify $\Omega$ with $N$ by $n\mapsto n\alpha$. Normality means that the point stabilizer $H_{\alpha }$ acts on $N$ by conjugation, andDouble transitivity therefore says that $H_{\alpha }$ is transitive on $N\setminus \{1\}$. All nonidentity elements of $N$ have the same order. By Cauchy's theorem for finite groups, some element has order a prime $p$ dividing $|N|$, so that common order is $p$. Cauchy's theorem excludes every other prime from $|N|$, hence $N$ is a finite p-group.

By the center argument from Q1, $Z(N)\ne 1$. The center is characteristic in $N$ and therefore invariant under the conjugation action of $H_{\alpha }$. Its nonidentity elements form a nonempty invariant subset of the transitive set $N\setminus \{1\}$, so $Z(N)=N$. Thus $N$ is abelian of exponent $p$:This is elementary abelian regular kernels in doubly transitive groups; knowing only that all element orders agree would not by itself prove commutativity.

Every $h\in H$ is uniquely $n{h}_0$ with $n\in N$ and $h_0\in H_{\alpha }$, because regularity supplies the unique $n$ sending $\alpha$ to $h\alpha$. Hence $H=N⋊H_{\alpha }$. The conjugation action of $H_{\alpha }$ on $N$ is faithful: an element centralizing $N$ and fixing $\alpha$ fixes every $n\alpha$, and is the identity permutation. Group automorphisms of an elementary abelian p-group are exactly invertible ${\mathbb{F}}_p$-linear maps, so $H_{\alpha }\le GL(d,p)$. Choosing a vector-space basis therefore givesUnder this embedding, $N$ is the translation subgroup and the point stabilizer is a linear subgroup transitive on nonzero vectors. The action is assumed to have at least two points, as usual for a doubly transitive group.