Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 2 / 3

Since the two-transitive group action is primitive, the orbits of any nontrivial normal subgroup form only the single block $\Omega$. Thus $N$ is transitive. If $N$ is regular, identify $\Omega$ with $N$ by $x\mapsto x\alpha$. Then $G=N{G}_{\alpha }$, $N\cap G_{\alpha }=1$, and conjugation by $G_{\alpha }$ gives automorphisms of $N$. The two-transitive group action makes these automorphisms transitive on $N\setminus \{1\}$. All nonidentity elements therefore have the same order, which must be a prime $p$: a suitable power of an element of composite order would have a smaller nontrivial order. Cauchy's theorem for finite groups now shows $|N|=p^d$.

The center of a group of prime-power order is nontrivial and is a characteristic subgroup. The transitive conjugation action on nonidentity elements therefore forces $Z(N)=N$. Hence $N$ is an elementary abelian group, which we identify with the additive vector space ${\mathbb{F}}_p^d$. Conjugation by $G_{\alpha }$ is linear on this vector space. It is faithful: an element fixing $\alpha$ and centralizing $N$ fixes every $x\alpha$, hence is the identity permutation. ConsequentlyHere the general affine group acts by translations followed by invertible linear maps.

The orbits of $N_{\alpha }⊴G_{\alpha }$ on $\Omega \setminus \{\alpha \}$ have a common size $r$, because $G_{\alpha }$ is transitive there. Suppose $N$ has a block system of block size $b$, $1<b<n$. Since $r\mid n-1$ and $b\mid n$, $\gcd (r,b)=1$. Let $B$ be the block of $\alpha$ and $C\ne B$ another block. The union $N_{\alpha }C$ is a union of blocks and of r-element point-orbits, so its size is divisible by $br$; it is at most $br$. Equality means each of the b point-orbits meets $C$ once. Thus $N_{\alpha \beta }$ fixes $C$ pointwise for $\beta \in C$. Interchanging $\alpha ,\beta$ also makes it fix $B$. All two-point stabilizers have order $|N_{\alpha }|/r$. Therefore $N_{\alpha \beta }=N_{\alpha \gamma }$ for every $\gamma \in B\setminus \{\alpha \}$. Varying $\beta$ outside $B$ shows this same subgroup fixes every point, so it is trivial. We have proved uniform subdegrees force an imprimitive action to be Frobenius: no nonidentity element of $N$ fixes two points.

Write $|N_{\alpha }|=h$. The distinct point stabilizers meet only at $1$, and $|N|=nh$. Thus the set $D$ of fixed-point-free elements of $N$ has sizeCount triples $(\alpha ,\beta ,x)$ with $x\in D$ and $x\alpha =\beta$: there are $n(n-1)$, exactly the number of ordered distinct pairs. Conjugation by the two-transitive group makes every pair occur, hence each pair has a unique such $x$. It follows that $G$ conjugates all elements of $D$ transitively.

Also $h\mid n-1$, since $N_{\alpha }$ acts freely on the remaining points. For each prime $p\mid n$, Cauchy's theorem for finite groups supplies an order-p element of $N$; it cannot fix a point, since $p\nmid h$. Conjugacy of $D$ forces all these primes to coincide, so $n=p^d$ and every element of $D$ has order $p$. A Sylow subgroup $P$ of $N$ has order $n$, because $p\nmid h$. Every nonidentity element of $P$ is fixed-point-free, so $P=\{1\}\cup D$. This uniquely described subgroup is normal in $G$. Minimal normality of $N$ gives $P=N$, and $|N|=n$. Therefore $N$ is regular. The proof establishes closure through a Sylow subgroup; it does not assume that a set of derangements is automatically a subgroup.

A finite minimal normal subgroup is a characteristically simple group, and has the direct-product structure of a finite minimal normal subgroupwith isomorphic simple groups $T_i$. For completeness, choose a minimal nontrivial normal subgroup $T$ of $N$. Its distinct $G$-conjugates are minimal normal subgroups of $N$, so they centralize one another. Their product is normal in $G$, hence equals $N$. If $T$ is abelian the product is abelian and minimal primary characteristic subgroups make $N$ elementary abelian. Otherwise each $T$ has trivial center: its center is a characteristic subgroup of $T$, hence normal in $N$, and minimality would make $T$ abelian. Each new conjugate then meets the product of previous conjugates trivially, because it centralizes that product and is centerless. The product is direct. Every normal subgroup of a direct factor is normal in $N$, so minimality of that factor makes it simple.

Assume now that $N$ acts primitively and nonregularly. Any nontrivial normal subgroup of this faithful primitive action is transitive. Thus each $T_i$ is transitive. If $k\ge 2$, the transitive commuting groups $T_1$ and $T_2$ are regular: an element of one fixing $\alpha$ commutes with the other and hence fixes its entire orbit. But the centralizer of a regular group has order $n$, as its elements are determined by the image of $\alpha$. Therefore all remaining factors lie in the order-n centralizer of $T_1$, and the transitive group $T_2$ already fills it. There can be only two factors, with $|N|=n^2$. Now $|N_{\alpha }|=n$, whereas its common nontrivial subdegree $r$ divides both $n$ and $n-1$. Hence $r=1$, making $N_{\alpha }$ trivial, a contradiction. Thus $k=1$, and $N$ is simple. An abelian transitive group is regular, so $N$ is nonabelian.

Finally $C_G(N)$ is normal in the primitive group $G$. If nontrivial it would be transitive, and commuting with transitive $N$ would force $N$ regular. Thus $C_G(N)=1$, and conjugation embeds $G$ in $\mathrm{Aut}(N)$. The centerless simple group $N$ is identified with its inner automorphism group. Therefore

By 3(a), identify $\Omega$ with $V={\mathbb{F}}_p^d$ and let $H=G_0\le G{L}_d(p)$. Four-transitivity means $H$ is transitive on ordered triples of distinct nonzero vectors, and in particular on ordered pairs of distinct nonzero vectors.

If $p>2$ and $d\ge 2$, the pairs $(u,au)$ with $a\ne 0,1$ and $(u,v)$ with $u,v$ linearly independent cannot be interchanged by a linear map. Thus $d=1$ if $p>2$. Then $|H|\le p-1$, whereas pair-transitivity on the $p-1$ nonzero points requires $(p-1)(p-2)\le |H|$. This forces $p\le 3$, incompatible with four-transitivity, whose degree is at least four.

Hence $p=2$. If $d\ge 3$, the triples $(u,v,u+v)$ and $(u,v,w)$ for linearly independent $u,v,w$ cannot be interchanged: the first is linearly dependent and the second is not. Thus $d=2$. The general affine group $AG{L}_2(2)$ has order $4(4-1)(4-2)=24$. Four-transitivity on four points requires $24\mid |G|$, and $G\le AG{L}_2(2)\le S_4$. Consequently