Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 4 / 2

We prove the primitive three-cycle criterion through supports. First, suppose a subgroup contains $A_U$, the alternating group on a set $U$ of at least three letters, and a three-cycle with support $E$ meeting $U$.

If $E$ adds one letter, it has two letters in $U$. The group $A_U$ is transitive on unordered pairs of $U$: for three letters its cyclic action permutes the three pairs, and for at least four letters one can adjust the parity of a permutation sending one pair to another while preserving its target pair. Conjugating the new cycle, and using inverses, supplies every three-cycle on two old letters and the new letter. Together with $A_U$, these generate $A_{U\cup E}$.

If $E$ adds two letters, write its cycle as $\tau =(abc)$ with $a\in U$. Choose $a^{\prime }\in U\setminus \{a\}$ and conjugate within $A_U$ to obtain ${\tau }^{\prime }=(a^{\prime }bc)$. ThenThe preceding case first adds $b$; applying it again to $\tau$ adds $c$. This proves connected triple supports generate an alternating group, by successively merging the intersecting triples of a connected support hypergraph. We used that three-cycles generate alternating groups: pairs of transpositions with a common letter are three-cycles, while a pair of disjoint transpositions is a product of two three-cycles.

Now take all conjugates in $G$ of the given three-cycle. Their support hypergraph is $G$-invariant, and its connected components form a block system. Transitivity makes their union the whole set. There is an edge, so a component has more than one vertex. A primitive group action forces a single component. The preceding merging argument proves

For the first family of counterexamples, letIts elements are Möbius transformations $x\mapsto (ax+b)/(cx+d)$, modulo nonzero scalar matrices. Any ordered triple of distinct projective points is the image of $(\infty ,0,1)$ under exactly one such transformation: representatives of the first two target lines form a basis, and scaling the two columns sends their sum to the third target line. Thus the action is sharply three-transitive, hence two-transitive and primitive. Its degree is $p+1$ and order $p(p^2-1)$. It is divisible by $p$, butso it cannot contain $A_{p+1}$.

For the second example takeThe same projective-line construction over ${\mathbb{F}}_8$ is sharply three-transitive. Its order is divisible by 7 and is less than $|A_9|=181440$, so it does not contain $A_9$. For a concrete field one can use ${\mathbb{F}}_8={\mathbb{F}}_2[u]/(u^3+u+1)$.

Use the displayed formula as a right group action: applying $(a,b)$ and then $(c,d)$ gives $(ac,bd)$. It is transitive since $(1,t)$ sends the identity to $t$. An element fixing every $t$ must have $a=b$ by evaluating at the identity, and then $a^{-1}ta=t$ for every $t$. Its kernel is therefore $\{(z,z):z\in Z(T)\}$, which is trivial by the center of a group hypothesis.

The identity stabilizer subgroup is $\Delta =\{(a,a):a\in T\}$. For every subgroup $H$ containing $\Delta$, setConjugation by diagonal elements makes $N$ normal in $T$. Multiplying $(a,b)\in H$ by $(a,a{)}^{-1}$ shows $(1,a^{-1}b)\in H$, soConversely each normal subgroup $N$ gives such an $H$. Thus the subgroups strictly between $\Delta$ and $T\times T$ correspond exactly to proper nontrivial normal subgroups of $T$.

For a transitive action, blocks containing a chosen point correspond to subgroups containing its stabilizer subgroup: a block is the orbit of that point under its setwise stabilizer subgroup, and conversely the orbit under an intermediate subgroup gives a block. Therefore the diagonal action on a centerless group is primitive exactly when $T$ is simple. A nontrivial centerless simple group cannot be abelian, because an abelian group's center of a group is the whole group. HenceThis uses the usual nontrivial degree convention. If one allows a one-point action to count as primitive, $T=1$ is the additional degenerate exception.

For the requested order example use $T=A_5$. Its simplicity follows from its conjugacy-class sizes $1,15,20,12,12$: a normal subgroup is a union of classes containing the identity, and no proper nontrivial such sum divides 60. Set $p=59$, a prime. The primitive diagonal action has