I would like to talk about an useful class of groups: that of primitive groups. I want to show why this class is important in group theory. I will assume that you know what actions of groups on sets are.
Let \( \displaystyle A,B \) be two sets, \( \displaystyle n \) a positive integer and let \( \displaystyle f: A^n \to B \) be a function (here \( \displaystyle A^n = A \times A \times ... \times A \) is the \( \displaystyle n \) -folded cartesian power of \( \displaystyle A \) ). A naive question is the following (cf. Cameron, "Permutation Groups", section 4.10):
(1) How many distinct values can \( \displaystyle f \) take?
Let us re-formulate the problem in order to make it clearer. Fix \( \displaystyle n \) elements \( \displaystyle a_1,...,a_n \in A \) . How many values do we get by applying \( \displaystyle f \) to the \( \displaystyle n \) -tuples obtained by permuting \( \displaystyle a_1,...,a_n \) in all possible ways? In other words, what is the cardinality of the set \( \displaystyle D := \{f(a_{\sigma(1)},...,a_{\sigma(n)})\ |\ \sigma \in S_n\} \) ? Or better, what is the object intrinsically dependent on the function \( \displaystyle f \) whose understanding would help us to solve this problem? This object turns out to be the subgroup \( \displaystyle H \) of \( \displaystyle S_n \) which consists of the permutations \( \displaystyle h \) with the following property: \( \displaystyle f(a_{h(\sigma(1))},...,a_{h(\sigma(n))}) = f(a_{\sigma(1)},...,a_{\sigma(n)}) \) for every \( \displaystyle \sigma \in S_n \) . Indeed, the elements of \( \displaystyle D \) correspond to the right cosets of \( \displaystyle H \) in \( \displaystyle S_n \) , thus the cardinality of \( \displaystyle D \) is just the index of \( \displaystyle H \) in \( \displaystyle S_n \) .
This means that problem (1) can be re-stated as follows:
(2) What are the possible indices of subgroups of \( \displaystyle S_n \) ?
A permutation group is a group \( \displaystyle G \) endowed with a faithful action (on the right) on a set \( \displaystyle \Omega \) (denote this action by \( \displaystyle \Omega \times G \ni (\omega,g) \mapsto \omega^g \in \Omega \) ). Equivalently, a permutation group on a set \( \displaystyle \Omega \) is a group \( \displaystyle G \) endowed with an injective homomorphism \( \displaystyle G \to \text{Sym}(\Omega) \) (cf. the basic theory on actions). The "degree" of the permutation group \( \displaystyle G \) is the size of \( \displaystyle \Omega \) . If the permutation group \( \displaystyle G \) is transitive (i. e. for every \( \displaystyle \omega_1,\omega_2 \in \Omega \) there exists \( \displaystyle g \in G \) such that \( \displaystyle \omega_1^g = \omega_2 \) ) then its action is equivalent to the action of right multiplication on the set of right cosets of a point stabilizer. In other words if \( \displaystyle \omega \in \Omega \) and \( \displaystyle G_{\omega} := \{g \in G\ |\ \omega^g = \omega\} \) then \( \displaystyle G_{\omega} g = G_{\omega^g} \) for every \( \displaystyle g \in G \) , and this means that we can think of the action of \( \displaystyle G \) on \( \displaystyle \Omega \) as the action of right multiplication on \( \displaystyle \{G_{\omega}g\ |\ g \in G\} \) . This implies that the degree of a transitive group is equal to the index of a point stabilizer, i.e. \( \displaystyle |\Omega| = |G:G_{\omega}| \) .
Therefore (2) can be re-stated as follows:
(3) What are the degrees of the faithful transitive actions of \( \displaystyle S_n \) ?
Indeed, since if \( \displaystyle n \geq 5 \) the only non-trivial normal subgroup of \( \displaystyle S_n \) is \( \displaystyle A_n \) , the non-trivial actions of \( \displaystyle S_n \) of degree at least \( \displaystyle 3 \) are faithful if \( \displaystyle n \geq 5 \) . Therefore assuming faithfulness is little loss of generality.
Clearly the action of a group \( \displaystyle G \) on a set \( \displaystyle \Omega \) is determined by its action on the orbits (the orbit of \( \displaystyle \omega \in \Omega \) is the set \( \displaystyle O_{\omega} := \{\omega^g\ |\ g \in G\} \) - clearly \( \displaystyle G \) acts transitively on \( \displaystyle O_{\omega} \) ), and this means that understanding actions amounts to understanding transitive actions.
Let \( \displaystyle G \) be a permutation group on a set \( \displaystyle \Omega \) . A "block" of this action is a subset \( \displaystyle B \) of \( \displaystyle \Omega \) with the following property: if \( \displaystyle g \in G \) and \( \displaystyle B^g \cap B \neq \emptyset \) then \( \displaystyle B^g=B \) . Clearly \( \displaystyle \Omega \) and \( \displaystyle \{\omega\} \) are blocks of \( \displaystyle G \) for every \( \displaystyle \omega \in \Omega \) . They are called the "trivial" blocks. Note that if \( \displaystyle B \) is a block then \( \displaystyle B^g \) is a block, and \( \displaystyle |B^g|=|B| \) , for every \( \displaystyle g \in G \) .
Let \( \displaystyle G \) be transitive, and call \( \displaystyle G_{\omega} \) the point stabilizer of \( \displaystyle \omega \in \Omega \) . Let \( \displaystyle B \) be a non-trivial block, and let \( \displaystyle G_B := \{g \in G\ |\ B^g=B\} \) be the stabilizer of \( \displaystyle B \) . Observe that:
(a) \( \displaystyle G \) acts transitively on the set \( \displaystyle \{B^x\ |\ x \in G\} \) by the rule \( \displaystyle (B^x,g) \mapsto B^{xg} \) .
(b) If \( \displaystyle \omega \in B \) then \( \displaystyle G_{\omega} \subseteq G_B \) .
(c) The action of \( \displaystyle G_B \) on \( \displaystyle B \) induced by \( \displaystyle G \) is transitive.
The union of the elements of a block of the action in (a) gives a block for \( \displaystyle G \) . Assume that \( \displaystyle B \) is a block of maximal size. Then the action in (a) has no non-trivial blocks. Such an action is called "primitive":
Definition. A permutation group \( \displaystyle G \) is called "primitive" if it does not admit non-trivial blocks.
Observe that the orbits of a permutation group are blocks, and this implies that every primitive permutation group of degree at least \( \displaystyle 3 \) is transitive.
We have just shown that every transitive permutation group \( \displaystyle G \) admits a block \( \displaystyle B \) such that the action of \( \displaystyle G \) on \( \displaystyle \{B^g\ |\ g \in G\} \) is primitive. Arguing in the exact same way for the (transitive) action of \( \displaystyle G_B \) on \( \displaystyle B \) , and iterating this process, we end up with a sequence of primitive permutation groups which we call "primitive components" of \( \displaystyle G \) . These components are not unique: they depend on the particular blocks we have chosen.
Summarizing, every transitive group can be understood by looking at the so-called "primitive components", permutation groups given by considering first the action of the group on the set of translates of a block, and then the action of the stabilizer of a block on the block. This allows to re-state (3) as follows:
(4) What are the degrees of the faithful primitive actions of \( \displaystyle S_n \) ?
As we have noticed, if \( \displaystyle B \) is a block for the transitive permutation group \( \displaystyle G \) and \( \displaystyle \omega \in B \) then \( \displaystyle G_{\omega} \subseteq G_B \) . This means that if the point stabilizer is a maximal subgroup then the action is primitive. Conversely, if the action is primitive and the point stabilizer \( \displaystyle H=G_{\omega} \) is contained in a proper subgroup \( \displaystyle K \) of \( \displaystyle G \) then the set \( \displaystyle \{Hk\ |\ k \in K\} \) is a block for \( \displaystyle G \) , hence it is trivial, i.e. \( \displaystyle Hk = H \) for every \( \displaystyle k \in K \) , i.e. \( \displaystyle H=K \) . This implies that \( \displaystyle H \) is a maximal subgroup of \( \displaystyle G \) . See also here. This means that the primitive actions of a group correspond to its maximal subgroups. This allows us to re-state (4) as follows:
(5) What are the possible indices of the maximal subgroups of \( \displaystyle S_n \) ?
The maximal subgroups of \( \displaystyle S_n \) are known in a satisfactory way: the O'Nan-Scott theorem (see here) provides a classification of them, thus it can be considered as a solution of (1).
C1 - The intrinsic point of view.
Given any group \( \displaystyle G \) and a subgroup \( \displaystyle H \leq G \) , consider its action \( \displaystyle \phi \) of right multiplication on the set of right cosets of \( \displaystyle H \) in \( \displaystyle G \) . The kernel of this action is \( \displaystyle H_G := \bigcap_{g \in G} g^{-1}Hg \) . \( \displaystyle H_G \) is called the "normal core" of \( \displaystyle H \) in \( \displaystyle G \) , and it coincides with the largest normal subgroup of \( \displaystyle G \) contained in \( \displaystyle H \) . In the following sense: every normal subgroup of \( \displaystyle G \) contained in \( \displaystyle H \) is contained in \( \displaystyle H_G \) . The action \( \displaystyle \phi \) is faithful if and only if \( \displaystyle H \) is "core-free" (i.e. \( \displaystyle H_G = \{1\} \) ), it is primitive if and only if \( \displaystyle H \) is maximal. Therefore, \( \displaystyle \phi \) makes \( \displaystyle G \) a primitive permutation group if and only if \( \displaystyle H \) is a core-free maximal subgroup of \( \displaystyle G \) , and the primitivity degrees of \( \displaystyle G \) are just the indices of its core-free maximal subgroups.
For example, the primitivity degrees of \( \displaystyle S_5 \) are \( \displaystyle 5 \) (point stabilizers), \( \displaystyle 6 \) (normalizers of \( \displaystyle 5 \) -Sylow subgroups), \( \displaystyle 10 \) (intransitive subgroups of type \( \displaystyle S_3 \times S_2 \) ).
For example, the primitivity degrees of \( \displaystyle S_6 \) are \( \displaystyle 6 \) (point stabilizers and subgroups isomorphic to \( \displaystyle S_5 \) acting on their six \( \displaystyle 5 \) -Sylow subgroups), \( \displaystyle 15 \) (intransitive subgroups of type \( \displaystyle S_4 \times S_2 \) and imprimitive subgroups of type \( \displaystyle S_2 \wr S_3 \) ), \( \displaystyle 20 \) (intransitive subgroups of type \( \displaystyle S_3 \times S_3 \) ), \( \displaystyle 10 \) (imprimitive subgroups of type \( \displaystyle S_3 \wr S_2 \) ).
Remark. The existence of a subgroup \( \displaystyle H \) of \( \displaystyle S_6 \) of index \( \displaystyle 6 \) which is not a point-stabilizer provides an (outer!) automorphism of \( \displaystyle S_6 \) given by the action of right multiplication of \( \displaystyle S_6 \) on the six right cosets of \( \displaystyle H \) in \( \displaystyle S_6 \) : cf. here.
For example, any solvable primitive permutation group has prime power degree (cf. here).
C2 - Examples.
Now I'll list some examples of primitive groups, which you may like to work out:
1. The cyclic group \( \displaystyle \langle (1 \ldots n) \rangle \cong C_n \) acting in the natural way on \( \displaystyle \{1, \ldots, n\} \) is primitive if and only if \( \displaystyle n \) is a prime.
2. The symmetric group \( \displaystyle S_n \) and the alternating group \( \displaystyle A_n \) acting in the natural way on \( \displaystyle \{1, \ldots ,n\} \) .
3. The group of isometries of \( \displaystyle \mathbb{R}^n \) acting in the natural way on \( \displaystyle \mathbb{R}^n \) .
For the following exercises recall the "counting principle": if a group \( \displaystyle G \) acts on a set \( \displaystyle \Omega \) and \( \displaystyle \omega \in \Omega \) then the size of the orbit of \( \displaystyle \omega \) equals the index of its stabilizer: \( \displaystyle |O_G(\omega)| = |G:\text{Stab}_G(\omega)| \) .
C3 - Exercise set 1.
E1-1. Let \( \displaystyle G \) be a primitive group, and let \( \displaystyle N \) be a non-trivial normal subgroup of \( \displaystyle G \) . Show that \( \displaystyle N \) is transitive.
Testo nascosto, fai click qui per vederlo
Prove that the \( \displaystyle N \) -orbits are blocks for \( \displaystyle G \) .
E1-2. Let \( \displaystyle G \) be any permutation group, and let \( \displaystyle H \) be a transitive subgroup of \( \displaystyle G \) . Prove that \( \displaystyle C_G(H) := \{g \in G\ |\ gh=hg\ \forall h \in H\} \) is semi-regular.[Recall that a permutation group \( \displaystyle X \) on a set \( \displaystyle \Omega \) is called semi-regular if \( \displaystyle \text{Stab}_X(\omega) = \{1\} \) for every \( \displaystyle \omega \in \Omega \) .]
E1-3. Let \( \displaystyle G \) be a primitive group.
(a) If \( \displaystyle G \) admits two non-trivial normal subgroups \( \displaystyle N_1,N_2 \) such that \( \displaystyle N_1 \cap N_2 = \{1\} \) then every non-trivial normal subgroup of \( \displaystyle G \) contains either \( \displaystyle N_1 \) or \( \displaystyle N_2 \) .
(b) If \( \displaystyle G \) admits a non-trivial abelian normal subgroup then the intersection of the non-trivial normal subgroups of \( \displaystyle G \) is non-trivial.
(c) If you like, prove that (b) holds even replacing "abelian" with "solvable".
Testo nascosto, fai click qui per vederlo
Use the previous exercises. For the solvable case try to reduce to the abelian case.
E1-4. Prove that 3(b) is false in general if \( \displaystyle G \) has no non-trivial solvable normal subgroups.Testo nascosto, fai click qui per vederlo
Take a non-abelian simple group \( \displaystyle S \) and consider \( \displaystyle G := S \times S \) .
E1-5. Let \( \displaystyle G \) be a primitive perfect group, and assume that its point stabilizers are solvable. Then \( \displaystyle G \) is simple.[Recall that a group \( \displaystyle X \) is called "perfect" if \( \displaystyle X'=X \) , where \( \displaystyle X' \) is the derived subgroup - equivalently, a group is called perfect if it has no non-trivial abelian quotients.]
Testo nascosto, fai click qui per vederlo
Consider a non-trivial normal subgroup \( \displaystyle N \) of \( \displaystyle G \) and prove that \( \displaystyle G/N \) is solvable. Prove that this contradicts perfectness.
E1-6. Show that if a primitive group \( \displaystyle G \) of degree \( \displaystyle n \) contains a transposition then \( \displaystyle G=S_n \) .E1-7. A permutation group \( \displaystyle G \) on a set \( \displaystyle \Omega \) is called \( \displaystyle 2 \) -transitive if its action on the set \( \displaystyle \{(\alpha,\beta)\ |\ \alpha,\beta \in \Omega,\ \alpha \neq \beta\} \) given by \( \displaystyle (\alpha,\beta)^g := (\alpha^g,\beta^g) \) is transitive. Prove that every \( \displaystyle 2 \) -transitive permutation group is primitive.
C4 - More on primitive groups.
Let \( \displaystyle G \) be a transitive permutation group on a set \( \displaystyle \Omega \) . The "suborbits" of \( \displaystyle G \) are the orbits of a point stabilizer \( \displaystyle G_{\omega} \) acting on \( \displaystyle \Omega-\{\omega\} \) . The "subdegrees" of \( \displaystyle G \) are the sizes of its suborbits.
Exercise: the transitive permutation group \( \displaystyle G \) is 2-transitive if and only if \( \displaystyle G_{\omega} \) acts transitively on \( \displaystyle \Omega-\{\omega\} \) .
Let \( \displaystyle G \) be a transitive permutation group on a set \( \displaystyle \Omega \) . \( \displaystyle G \) acts on the cartesian product \( \displaystyle \Omega \times \Omega \) as follows: \( \displaystyle (\alpha,\beta)^g := (\alpha^g,\beta^g) \) . The orbits of this action are called "orbitals" of \( \displaystyle G \) , we denote them by \( \displaystyle \Gamma_0,...,\Gamma_{r-1} \) , where \( \displaystyle \Gamma_0 \) is the trivial orbital \( \displaystyle \{(\alpha,\alpha)\ |\ \alpha \in \Omega\} \) .
Fix \( \displaystyle \alpha \in \Omega \) . There is a canonical bijective correspondence between the set of orbitals of \( \displaystyle G \) and the set of suborbits: the orbital \( \displaystyle \Gamma_i \) corresponds to the suborbit \( \displaystyle \Gamma_i(\alpha):=\{\beta \in \Omega\ |\ (\alpha,\beta) \in \Gamma_i\} \) ; the suborbit \( \displaystyle \Delta \) corresponds to the orbital \( \displaystyle \{(\alpha^g,\delta^g)\ |\ \delta \in \Delta,\ g \in G\} \) . Let \( \displaystyle n_i:=|\Gamma_i(\alpha)| \) for \( \displaystyle i=0,...,r-1 \) . Let us order the \( \displaystyle n_i \) 's in such a way that \( \displaystyle 1=n_0 \leq n_1 \leq ... \leq n_{r-1} \) . The number \( \displaystyle r \) is the rank of \( \displaystyle G \) .
For example, a transitive permutation group is 2-transitive if and only if it has rank 2.
Some results follow.
Theorem 1. If \( \displaystyle G \) is primitive then \( \displaystyle n_1 n_{i-1} \geq n_i \) for every \( \displaystyle 2 \leq i \leq r-1 \) .
Theorem 2 (Marie Weiss). Let \( \displaystyle G \) be a primitive group of degree \( \displaystyle n \) , different from \( \displaystyle C_p \) of degree \( \displaystyle p \) prime. If \( \displaystyle 1 \leq i,j \leq r-1 \) are such that \( \displaystyle (n_i,n_j)=1 \) then there exists \( \displaystyle k \in \{1,...,r-1\} \) such that \( \displaystyle n_k > n_i \) , \( \displaystyle n_k > n_j \) and \( \displaystyle n_k \) divides \( \displaystyle n_in_j \) .
Corollary. Let \( \displaystyle G \) be a primitive groups different from \( \displaystyle C_p \) of degree \( \displaystyle p \) prime. Then \( \displaystyle (n_{r-1},n_i) \neq 1 \) for every \( \displaystyle 1 \leq i \leq r-1 \) .
With such results one is able to prove things of the following flavour.
Example: every primitive group of degree \( \displaystyle 8 \) has rank \( \displaystyle 2 \) .
Proof. The subdegrees must be \( \displaystyle n_0=1,n_1,...,n_{r-1} \) with \( \displaystyle \sum_{i=1}^{r-1}n_i=8-1=7 \) and \( \displaystyle (n_i,n_{r-1}) \neq 1 \) for every \( \displaystyle i=1,...,r-1 \) . The only possibility is \( \displaystyle r=2 \) and \( \displaystyle n_1=7 \) .
C5 - The O'Nan-Scott Theorem: some details.
The O'Nan-Scott theorem (cf. here) provides a classification of the maximal subgroups of the symmetric group \( \displaystyle \text{Sym}(n) \) . Roughly, a maximal subgroup \( \displaystyle G \) of \( \displaystyle \text{Sym}(n) \) can be of three types:
1) Intransitive, i.e. admitting a non-trivial orbit. In this case there exist \( \displaystyle I_1,I_2 \) proper non-empty subsets of \( \displaystyle \{1,\ldots,n\} \) such that \( \displaystyle G \) is the direct product of their pointwise stabilizers: \( \displaystyle G = \text{Stab}(I_1) \times \text{Stab}(I_2) \cong \text{Sym}(a) \times \text{Sym}(b) \) , where \( \displaystyle |I_1|=b \) and \( \displaystyle |I_2|=a \) . Clearly \( \displaystyle a+b=n \) .
Remark. Note that the index of a maximal intransitive subgroup \( \displaystyle \text{Sym}(a) \times \text{Sym}(b) \) of \( \displaystyle \text{Sym}(n) \) is just the binomial coefficient \( \displaystyle \binom{n}{a} \) .
2) Imprimitive, i.e. transitive but not primitive. In this case there exists a non-trivial block \( \displaystyle B \subset \{1,\ldots,n\} \) , let \( \displaystyle a \) be its cardinality. Call \( \displaystyle H := \{g \in G\ |\ g(B)=B\} \) the stabilizer of the block \( \displaystyle B \) , and call \( \displaystyle B_1=B,\ldots,B_b \) the \( \displaystyle G \) -translates of \( \displaystyle B \) , \( \displaystyle \pi:G \to \text{Sym}(b) \) the homomorphism induced by the action of \( \displaystyle G \) on \( \displaystyle \{B_1,\ldots,B_b\} \) (where \( \displaystyle B_i \) is identified with \( \displaystyle i \) ), and let \( \displaystyle K := \pi(G) \leq \text{Sym}(b) \) . It is easy to see that \( \displaystyle G \) embeds in the wreath product \( \displaystyle H \wr K \) and \( \displaystyle ab=n \) . Modulo some details, this proves that a maximal imprimitive subgroup of \( \displaystyle \text{Sym}(n) \) is of the form \( \displaystyle \text{Sym}(a) \wr \text{Sym}(b) \) where \( \displaystyle ab=n \) , \( \displaystyle a \) is the size of a block with \( \displaystyle b \) translates.
Remark. Note that the index of a maximal imprimitive subgroup \( \displaystyle \text{Sym}(a) \wr \text{Sym}(b) \) is \( \displaystyle n!/(a!^b b!) \) . In particular, this implies that if \( \displaystyle a,b \) are positive integers then \( \displaystyle a!^b b! \) divides \( \displaystyle (ab)! \) .
3) Primitive.
This means that what the O'Nan-Scott Theorem really is, is a classification of the primitive permutation groups.
C5-1 - About the classification of primitive permutation groups.
Recall that a "minimal normal subgroup" of a group is a non-trivial normal subgroup minimal with respect to inclusion. Note that every finite group admits minimal normal subgroups. By Lemmas D and E here, a minimal normal subgroup is of the form \( \displaystyle S \times \ldots \times S = S^n \) where \( \displaystyle S \) is a minimal sub-normal subgroup (in particular, \( \displaystyle S \) is simple). A group is called "monolithic" if it has a unique minimal normal subgroup.
Exercise. Prove that a monolithic group with non-abelian minimal normal subgroup is primitive (i.e. admits a core-free maximal subgroup).
Testo nascosto, fai click qui per vederlo
Use the fact that the Frattini subgroup of a finite group is nilpotent (cf. here).
Let \( \displaystyle G \) be a primitive permutation group, and let \( \displaystyle \text{soc}(G) \) (the socle of \( \displaystyle G \) ) denote the subgroup generated by the minimal normal subgroups of \( \displaystyle G \) . If \( \displaystyle \text{soc}(G) \) is abelian then by E1-3(b) \( \displaystyle G \) is monolithic. If \( \displaystyle \text{soc}(G) \) is non-abelian then by E1-3(a) \( \displaystyle \text{soc}(G) \) is either monolithic or of the form \( \displaystyle N_1 \times N_2 \) where \( \displaystyle N_1,N_2 \) are non-abelian minimal normal subgroups.
Suppose from now on that \( \displaystyle G \) is a monolithic primitive group with non-abelian socle. Write \( \displaystyle \text{soc}(G) = S_1 \times \ldots \times S_n = S \times \ldots S = S^n \) , where \( \displaystyle S \) is a non-abelian simple group. Let \( \displaystyle N := N_G(S_1) \) and \( \displaystyle C := C_G(S_1) \) be respectively the normalizer and the centralizer in \( \displaystyle G \) of the first factor \( \displaystyle S_1 \) . Let \( \displaystyle X := N/C \) . By considering the conjugation action of \( \displaystyle S \) on itself, it is easy to see that \( \displaystyle S \leq X \leq \text{Aut}(S) \) .
Exercise. The normal subgroups of \( \displaystyle S^n \) are of the type \( \displaystyle \prod_{i \in I} S_i \) where \( \displaystyle I \subseteq \{1, \ldots, n\} \) .
It follows that the minimal normal subgroups of \( \displaystyle S_1 \times \ldots \times S_n \) are the \( \displaystyle n \) factors \( \displaystyle S_1, \ldots, S_n \) . This implies that the conjugation action of \( \displaystyle G \) on \( \displaystyle \text{soc}(G) \) induces an action on \( \displaystyle \{1,\ldots,n\} \) given by \( \displaystyle (i,g) \mapsto i^g \) where \( \displaystyle S_i^g = S_{i^g} \) . This gives a homomorphism \( \displaystyle \rho:G \to \text{Sym}(n) \) . Observe that the fact that \( \displaystyle \text{soc}(G) \) is a minimal normal subgroup of \( \displaystyle G \) translates into the fact that \( \displaystyle \rho(G) \) is a transitive subgroup of \( \displaystyle \text{Sym}(n) \) (distinct orbits would correspond to smaller normal subgroups).
Since \( \displaystyle G \) is primitive, it admits a core-free maximal subgroup \( \displaystyle U \) . Since \( \displaystyle U \) is core-free, \( \displaystyle \text{soc}(G) \not \subseteq U \) , so \( \displaystyle U < \text{soc}(G) U \leq G \) , and since \( \displaystyle U \) is maximal \( \displaystyle \text{soc}(G) U = G \) . Note that the primitivity degree of \( \displaystyle G \) associated to \( \displaystyle U \) is \( \displaystyle |G:U| = |\text{soc}(G) U:U| = |\text{soc}(G):U \cap \text{soc}(G)| \) . Moreover since \( \displaystyle G=\text{soc}(G)U \) and \( \displaystyle \text{soc}(G) \) acts trivially on the factors \( \displaystyle S_1,\ldots,S_n \) , \( \displaystyle U \) acts transitively on \( \displaystyle \{1,\ldots,n\} \) , i.e. \( \displaystyle \rho(U) \) is transitive. Call \( \displaystyle \pi_i:S^n \to S_i \) the projection onto the \( \displaystyle i \) -th factor. Let \( \displaystyle i,j \in \{1,\ldots,n\} \) , and let \( \displaystyle u \in U \) send \( \displaystyle i \) to \( \displaystyle j \) . Let \( \displaystyle H := U \cap \text{soc}(G) \) . Then we have an isomorphism
\( \displaystyle \pi_i(H) \cong H/\ker(\pi_i|_H) \to H/\ker(\pi_j|_H) \cong \pi_j(H) \) .
Since \( \displaystyle \rho(U) \) is transitive, we obtain that \( \displaystyle \pi_1(H) \cong \ldots \cong \pi_n(H) \) . There are three possibilities for \( \displaystyle Y := \pi_1(H) \) .
1) Complement type. \( \displaystyle Y = \{1\} \) . Then \( \displaystyle H = \{1\} \) , i.e. \( \displaystyle U \) complements \( \displaystyle \text{soc}(G) \) in \( \displaystyle G \) .
2) Diagonal type. \( \displaystyle Y = S \) . Then \( \displaystyle \pi_1|_H,\ldots,\pi_n|_H \) are surjective. Observe that if \( \displaystyle i,j \in \{1,\ldots,n\} \) then \( \displaystyle \pi_i(\ker(\pi_j|_H)) \unlhd S_i \) . It is not difficult to deduce that there are a number \( \displaystyle \ell \) dividing \( \displaystyle n \) and \( \displaystyle \ell \) diagonals \( \displaystyle \Delta_1,\ldots,\Delta_{\ell} \) such that \( \displaystyle H = \Delta_1 \times \ldots \times \Delta_{\ell} \) .
3) Product type. \( \displaystyle \{1\} < Y < S \) . There exist \( \displaystyle u_1,\ldots,u_n \in U \) such that \( \displaystyle \pi_i(H) = Y^{u_i} \) for \( \displaystyle i=1,\ldots,n \) . Up to conjugate suitably we may assume that \( \displaystyle u_1 = \ldots = u_n = 1 \) . In particular \( \displaystyle Y^n \supseteq H \) . To show that equality holds it suffices to show that \( \displaystyle Y^n U = U \) , i.e. (recall that \( \displaystyle U \) is maximal in \( \displaystyle G \) ) that \( \displaystyle Y^n U \neq G \) . Suppose not. Take \( \displaystyle g \in \text{soc}(G) \) outside \( \displaystyle Y^n \) , and choose \( \displaystyle u \in U, y \in Y^n \) such that \( \displaystyle g = yu \) . Then \( \displaystyle u = y^{-1}g \in \text{soc}(G) \) , so \( \displaystyle u \in H \subseteq Y^n \) , hence \( \displaystyle g \in Y^n \) , a contradiction. Therefore \( \displaystyle H = Y \times Y^{u_2} \times \ldots \times Y^{u_n} \) .
To be continued...
C6 - Exercise set 2.
E2-1. Let \( \displaystyle G \) be a primitive permutation group on a set \( \displaystyle \Omega \) , and let \( \displaystyle \alpha \in \Omega \) . Assume that \( \displaystyle G \) has \( \displaystyle 3 \) as a subdegree, and let \( \displaystyle \{\beta,\gamma,\delta\} \) be a suborbit of size \( \displaystyle 3 \) . Then the stabilizer of \( \displaystyle \beta \) in \( \displaystyle G_{\alpha} \) is a \( \displaystyle 2 \) -group. [cf. here].
E2-2. Let \( \displaystyle G \) a primitive permutation group on a set \( \displaystyle \Omega \) , with \( \displaystyle |\Omega| = n \geq 2 \) . Let \( \displaystyle x \in \Omega \) , and assume \( \displaystyle G_x = \text{Stab}_G(x) \) is abelian. Prove that \( \displaystyle |G_x| \leq n-1 \) .
Suppose now that \( \displaystyle n = p^m \) is a prime power. Prove that the Sylow \( \displaystyle p \) -subgroups of \( \displaystyle G \) are normal in \( \displaystyle G \) , elementary abelian and they supplement \( \displaystyle G_x \) .
C7 - Further results and remarks.
R1. A corollary of a theorem of Jordan [Cameron, "Permutation Groups", Theorem 6.15]. Let \( \displaystyle p \) be a prime, and let \( \displaystyle G \) be a primitive subgroup of \( \displaystyle S_n \) containing a \( \displaystyle p \) -cycle. Then \( \displaystyle G \) is \( \displaystyle (n-p+1) \) -transitive. If \( \displaystyle n>p+2 \) then \( \displaystyle G \supseteq A_n \) .
Note that R1 implies E1-6.
R2. Primitive groups of degree \( \displaystyle n \) not containing \( \displaystyle A_n \) have order at most \( \displaystyle 4^n \) (reference), so they are "small".
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PS. Thank you for pointing out mistakes and misprints.
