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Coprime subdegrees for primitive permutation groups and completely reducible linear groups

In this paper we answer a question of Gabriel Navarro about orbit sizes of a finite linear group H acting completely reducibly on a vector space V: if the orbits containing the vectors a and b have coprime lengths m and n, we prove that the orbit containing a+b has length mn. Such groups H are always reducible if n and m are greater than 1. In fact, if H is an irreducible linear group, we show that, for every pair of non-zero vectors, their orbit lengths have a non-trivial common factor. In the more general context of finite primitive permutation groups G, we show that coprime non-identity subdegrees are possible if and only if G is of O'Nan-Scott type AS, PA or TW. In a forthcoming paper we will show that, for a finite primitive permutation group, a set of pairwise coprime subdegrees has size at most 2. Finally, as an application of our results, we prove that a field has at most 2 finite extensions of pairwise coprime indices with the same normal closure.

1. Introduction 1.1. Completely reducible linear groups. In this paper we are concerned with the orbit lengths of a completely reducible linear group and with the subdegrees of a primitive permutation group. Given a field k, a kH-module V is said to be completely reducible if V is a direct sum of irreducible kH-modules. Furthermore, the set of subdegrees of a finite transitive permutation group G is the set of orbit lengths of the stabilizer G ω of a point ω.
Our first main result is a positive answer to a question of Gabriel Navarro [24] about actions of a finite linear group. Indeed, Navarro asked whether a finite completely reducible kH-module V with two H-orbits of relatively prime lengths m and n has an orbit of size mn. Theorem 1.1. Let k be a field, let H be a finite group, let V be a completely reducible kH-module and let a and b be elements of V . If the H-orbits a H and b H have sizes m and n, and m, n are relatively prime, then C H (a+b) = C H (a)∩C H (b) and the H-orbit (a + b) H has size mn.
We note that Theorem 1.1 explicitly exhibits an H-orbit of size mn, namely the orbit containing a + b. Furthermore, Example 3.1 shows that the "completely reducible" hypothesis in Theorem 1.1 is essential. Martin Isaacs [16,Theorem] has proved a similar result under stronger arithmetical conditions on m, n and the characteristic of k.
Our second main result is a somehow remarkable theorem (in our opinion) on irreducible linear groups. Theorem 1.2 shows that the groups arising in Theorem 1.1 are always reducible if n, m > 1.
Theorem 1.2. Let k be a field, let H be a finite group, let V be a non-trivial irreducible kH-module and let a and b be in V \ {0}. Then the sizes of the H-orbits a H and b H have a non-trivial common factor.
The strategy for proving Theorem 1.2 is to reduce the problem inductively to the case where H is a non-abelian simple group which admits a maximal factorization H = AB with |H : A| relatively prime to |H : B|. Table 1 contains all such triples (H, A, B). In the case where H is a sporadic simple group, we use the information in Table 1 for the proof of Theorem 1.2. Furthermore, we observe that as a consequence of Theorem 1.2, if (H, A, B) is one of the triples in Table 1, then there are no irreducible representations of H with A and B vector stabilizers.
A direct application of Theorem 1.2 gives the following corollary.
Corollary 1.3. Let k be a field, let H be a finite group, let V be a non-trivial finite dimensional kH-module and let a and b be elements of V . If both H-orbits a H and b H span V , then |a H | and |b H | have a non-trivial common factor.
In the same direction as Corollary 1.3, in the case of p-soluble groups, we prove the following theorem. Here, by abuse of notation, "0-soluble finite group" means "finite group" and a "Sylow 0-subgroup" is the "identity subgroup". In the proof of Theorem 1.4 we do not make use of the Classification of the Finite Simple Groups. Moreover, since in an irreducible kH-module V every non-trivial H-orbit spans V , we see that Theorem 1.4 (b) generalizes (for the class of p-soluble groups) Theorem 1.2 and, in particular, offers an independent and more elementary proof. Note that if p does not divide the order of H (including the case p = 0), then H is p-soluble, the Sylow p-subgroups of H are trivial and, in particular, Theorem 1.4 applies in this situation.
1.2. Coprime subdegrees in primitive permutation groups. In the more general context of finite primitive permutation groups G, we investigate coprime subdegrees according to the O'Nan-Scott type of G. (We say that a subdegree d of G is non-trivial if d = 1.) In particular, Theorem 1.2 yields that the primitive permutation group G = V ⋊H acting on V has no pair of non-trivial coprime subdegrees. One of the most important modern methods for analyzing a finite primitive permutation group G is to study the socle N of G, that is, the subgroup generated by the minimal normal subgroups of G. The socle of an arbitrary finite group is isomorphic to a direct product of simple groups, and, for finite primitive groups these simple groups are pairwise isomorphic. The O'Nan-Scott theorem describes in detail the embedding of N in G and collects some useful information on the action of N . In [26] eight types of primitive groups are defined (depending on the structure and on the action of the socle), namely HA (Holomorphic Abelian), AS (Almost Simple), SD (Simple Diagonal ), CD (Compound Diagonal ), HS (Holomorphic Simple), HC (Holomorphic Compound ), TW (Twisted wreath), PA (Product Action), and it is shown in [18] that every primitive group belongs to exactly one of these types. Theorem 1.5. Let G be a finite primitive permutation group. If G has two nontrivial coprime subdegrees, then G is of AS, PA or TW type. Moreover, for each of the O'Nan-Scott types AS, PA and TW, there exists a primitive group of this type with two non-trivial coprime subdegrees.
It is possible, for a single primitive group to have several different pairs of nontrivial coprime subdegrees. We give a construction of groups of PA type with this property in Example 4.3. However, in the case of primitive groups of TW type, it is not possible to have as many as three pairwise coprime non-trivial subdegrees. Theorem 1.6. For a finite primitive permutation group of TW type, the maximal size of a set of pairwise coprime non-trivial subdegrees is at most 2.
Using the Classification of the Finite Simple Groups, we have proved the following theorem in [8].
Theorem 1.7. Let G be a finite primitive permutation group. The maximal size of a set of pairwise coprime non-trivial subdegrees of G is at most 2. Theorem 1.7 is related to a result on primitive groups first observed by Peter Neumann to be a consequence of a 1935 theorem of Marie Weiss. Its statement [25,Corollary 2,p. 93] is: if a primitive group has k pairwise coprime non-trivial subdegrees, then its rank is at least 2 k . Theorem 1.7 shows that this result can only be applied with k = 1 or k = 2. In light of Theorems 1.5 and 1.6, proving Theorem 1.7 reduces to consideration of primitive permutation groups of AS and PA type. In this paper we show that a proof of Theorem 1.7 reduces to a similar problem for transitive nonabelian simple permutation groups (it is this reduction that is used in [8] to prove Theorem 1.7). Definition 1.8. Let T be a nonabelian simple group and L a subgroup of T . We say that L is pseudo-maximal in T if there exists an almost simple group H with socle T and a maximal subgroup M of H with T M and L = T ∩ M .
We announce here a proof of the following theorem about nonabelian simple groups (which again will be proved in [8]). Theorem 1.9. Let T be a transitive nonabelian simple permutation group and assume that the stabilizer of a point is pseudo-maximal in T . Then the maximal size of a set of pairwise coprime non-trivial subdegrees of T is at most 2.
Since a pseudo-maximal subgroup of T is not necessarily a maximal subgroup, we see that Theorem 1.9 is formally stronger than Theorem 1.7 for the class of nonabelian simple permutation groups. However, we prove in this paper that these two theorems are strongly related. Theorem 1.10. Theorem 1.7 follows from Theorem 1.9.
We conclude with a problem on relatively prime subdegrees in primitive groups. Problem 1.11. Determine the finite primitive permutation groups G having two non-trivial coprime subdegrees m and n for which mn is not a subdegree of G.
This problem is related to another classical result due to Marie Weiss [25, Theorem 3, p. 92]: if m and n are non-trivial coprime subdegrees of a primitive group G and m < n, then G has a subdegree d such that d divides mn and d > n. In Problem 1.11 we suggest that (apart a small list of exceptions) d can be chosen to be mn. Actually, we know only one almost simple group G where d cannot be taken to be nm. Example 1.12. Let G be the sporadic simple group HS in its primitive permutation representation of degree 3850 on the cosets of 2 4 . Sym (6). Using the computational algebra system magma [4], it is easy to check that the subdegrees of G are 1,15,32,90,120,160,192,240,240,360,960,1440. In particular, we see that 15 and 32 are coprime but there is no subdegree of size 15 × 32=480.
We are grateful to Michael Giudici for providing this beutiful example. We now give some examples which demonstrate that Theorem 1.7 is false for transitive groups that are not primitive. These examples show that there is no upper bound on the number of pairwise coprime non-trivial subdegrees for general transitive groups. Example 1.13. Let G be the direct product F 1 ×· · ·×F ℓ of ℓ Frobenius groups. For each i ∈ {1, . . . , ℓ}, let N i be the Frobenius kernel of F i and let K i be a Frobenius complement for N i in F i . Assume that |K i | is coprime to |K j |, for every two distinct elements i and j in {1, . . . , ℓ}. Write N = N 1 × · · · × N ℓ and K = K 1 × · · · × K ℓ . Clearly, the group G acts on N as a holomorphic permutation group, that is, N acts on N by right multiplication and K acts on N by group conjugation. The stabilizer in G of the element 1 of N is K. Now, for each i ∈ {1, . . . , ℓ}, let n i ∈ N i \ {1} and let ω i = (1, . . . , 1, n i , 1, . . . , 1) be the element of N with n i in the i th coordinate and 1 everywhere else. Clearly, |ω K i | = |K : C K (ω i )| = |K i |. Therefore G has a set of at least ℓ pairwise coprime non-trivial subdegrees.
However, if we restrict to faithful subdegrees of a transitive group G, that is, subdegrees d such that there exists an orbit of length d of a stabilizer G α on which G α acts faithfully, then in fact we can show that a conclusion analogous to the statement of Theorem 1.7 does hold. We note that, in particular, every primitive permutation group has a faithful subdegree [12,Theorem 3]. Theorem 1.14. Let G be a finite transitive permutation group of degree n > 1. Assume that G is not regular and let H be the stabilizer of a point. Then a set of faithful subdegrees that are pairwise coprime has size at most 2. Moreover, if the Fitting subgroup of H is non-trivial, then any two faithful subdegrees of G have a non-trivial common factor.
The proof of Theorem 1.14 also yields the following result about field extensions. Theorem 1.15. Let k be a field and let k 1 , . . . , k t be finite extensions of k all with the same normal closure K. Assume that the indices [k i : k] are pairwise coprime. Then t ≤ 2.
1.3. Structure of the paper. The structure of this paper is straightforward: we prove Theorem 1.2, Corollary 1.3 and Theorem 1.4 in Section 2; we prove Theorem 1.1 in Section 3; we prove Theorem 1.5 in Section 4; we prove Theorems 1.6 and 1.10 in Section 5; we prove Theorems 1.14 and 1.15 in Section 6; and we give Table 1 in Section 7.
2. Proofs of Theorems 1.2 and 1.4 and Corollary 1.3 We say that a factorization H = AB is coprime if |H : A| is relatively prime to |H : B| and both A, B are proper subgroups of H. Also H = AB is maximal if A and B are maximal subgroups of H. We start by proving a preliminary theorem on finite classical groups. We let τ denote the transpose inverse map of GL n (q), that is, x τ = (x tr ) −1 where x tr is the transpose matrix of x. We denote by CSp(2n, q) the conformal symplectic group, that is, the elements of GL 2n (q) preserving a given symplectic form up to a scalar multiple.
(a): Every element of GL n (q) is conjugate to its inverse in GL n (q) τ . (b): Every element of GU n (q) is conjugate to its inverse in GU n (q) τ .
Proof. We prove (a) and (b) first. Let X = GL n (k) be the algebraic group obtained by taking the algebraic closure k of the finite field F q . Let F : X → X be the Lang-Steinberg map obtained by raising each entry of a matrix x of X to the qth power, and G : X → X the Lang-Steinberg map F • τ . As usual, we denote by X F and by X G the fixed points of F and of G. In our case, we have X F = GL n (q) and X G = GU n (q). Let x be in X F . Then x tr and x are clearly conjugate in the algebraic group X and hence also x τ = (x tr ) −1 and x −1 are conjugate in X. Since the centralizer of any element of X is connected, it follows by the Lang-Steinberg theorem that x τ and x −1 are conjugate in X F . Therefore x and x −1 are conjugate in GL n (q) τ and (a) is proved. Now, let x be in X G . As we have noted in the proof of (a), the elements x τ and x −1 are conjugate in the algebraic group X. It follows by the Lang-Steinberg theorem that x τ and x −1 are conjugate in X G . Therefore x and x −1 are conjugate in GU n (q) τ and (b) is proved.
(c) and (d), when q is even, are the main theorem of [11]. Finally, (c) and (d), when q is odd, are proved in [28].
Given a field k and a kH-module V , we let V * = Hom k (V, k) denote the dual kHmodule of V . Furthermore, we denote by V A the restriction of V to the subgroup A of H. Finally, if A is a subgroup of H and if V is a kA-module, then we denote by V H A = V ⊗ kA kH the module induced by V from A to H. Lemma 2.2. Suppose that H = AB is a factorization. If V is a non-trivial irreducible kH-module, then either A fixes no element of V \ {0} or B fixes no element of V * \ {0}.
Proof. We argue by contradiction and we assume that A fixes a ∈ V \ {0} and that B fixes b ∈ V * \ {0}.
Let Ω (respectively ∆) be the set of right cosets of A (respectively B) in H. Clearly, H acts transitively on Ω and ∆, and as H = AB, the group B is transitive on ∆ and A is transitive on Ω. Let k H A (respectively k H B ) be the permutation module for the action of H on Ω (respectively ∆). Since A is transitive on ∆, the multiplicity of the trivial kA-module k in (k H B ) A is 1, that is, dim Hom kA (k, (k H B ) A ) = 1. From Frobenius reciprocity, it follows that dim Hom kH (k H A , k H B ) = dim Hom kA (k, (k H B ) A ) = 1. Therefore, the only H-homomorphism of k H A to k H B is the homomorphism ϕ with Ker ϕ of codimension 1 in k H A and with Im ϕ the trivial submodule of k H B . Since A fixes the non-zero vector a of V , we have 0 = Hom kA (k, B has a homomorphic image isomorphic to V * . Using duality and the fact that Proof. We argue by contradiction and we assume that both A and B fix some nonzero vector of V . Let r be the characteristic of the field k. Since |H : A| is relatively prime to |H : B| and A ∩ B = 1, we have that either r does not divide |A| or r does not divide |B|. Replacing A with B if necessary, we may assume that r does not divide |B|. Since the characteristic of k is coprime to the order of B, the module V B is a completely reducible kB-module. Therefore, V B = W 1 ⊕ · · · ⊕ W s where W i is an irreducible kB-module, for each i ∈ {1, . . . , s}. Since B fixes a non-zero vector of V , we have that, for some i ∈ {1, . . . , s}, W i is a trivial kB-module. Now, V * B = W * 1 ⊕ · · · ⊕ W * s and hence W * i is a trivial submodule of V * B . This shows that B fixes a non-zero vector of V * , but this contradicts Lemma 2.2.
The following lemma is Lemma 5.1 in [13].
Lemma 2.4. Suppose that every element of H is conjugate to its inverse via an element of Aut(H). If V is an irreducible kH-module, then V * ∼ = V x for some x ∈ Aut(H).
Furthermore, since every element of H is conjugate to its inverse via an element of G, we obtain that the Brauer character of M is real valued. Now, from [23, Theorem 1.19 and Lemma 2.2], we see that completely reducible modules with real Brauer characters are self dual and hence M is self dual, that is, Lemma 2.5. Suppose that every element of H is conjugate to its inverse via an element of Aut(H). If H = AB is a coprime factorization and V is a non-trivial irreducible kH-module, then either A fixes no element of V \ {0} or B fixes no element of V \ {0}.
Proof. From Lemma 2.4, V * ∼ = V x for some x ∈ Aut(H). As H = AB is a coprime factorization, we obtain H = AB x .
We argue by contradiction and we assume that A fixes the non-zero vector a of V and that B fixes the non-zero vector b of V . So B x fixes the vector b x of V x and, as V x ∼ = V * , the group B x fixes some non-zero vector of V * . This contradicts Lemma 2.2 applied to G = AB x , and so the lemma is proven.
In the following proposition we prove Theorem 1.2 in the case that the group H is a non-abelian simple group. If H is a classical group, we see from Theorem 2.1 that every element of H is conjugate to its inverse via an element of Aut(H). Clearly, the same result holds true if H is an alternating group. Therefore, if H is a classical group or an alternating group, we obtain a contradiction from Lemma 2.5 (applied with A = C H (a) and B = C H (b)). This shows that H is either an exceptional group of Lie type or a sporadic simple group. From Table 1, we see that exceptional groups of Lie type do not admit coprime factorizations. Therefore, H is a sporadic simple group. Again, using Table 1, we see that the only sporadic simple groups admitting a coprime factorization are M 11 , M 23 and M 24 . In the rest of this proof we consider separately each of these groups. Note that Table 1 Table 1 are exact and hence the result follows again from Lemma 2.3. (6), |H : A| = 24, |H : B| = 1771 = 7 · 11 · 23 and |A ∩ B| = 5760 = 2 7 · 3 2 · 5.
Since |H : C H (a)| is divisible by 2 and 3, and |H : C H (b)| is relatively prime to |H : C H (a)|, we have that C H (b) contains a Sylow 2-subgroup and a Sylow 3-subgroup of H. Thus |C H (b)| is divisible by 2 10 · 3 3 and |B : C H (b)| ≤ 5. Since B has no subgroup of index 5, we obtain B = C H (b). With a similar argument applied to C H (a), we get that 7 · 11 · 23 divides |C H (a)|. From [6], we see that M 23 has no proper subgroup of order divisible by 7, 11 and 23. Therefore Let M be the permutation module k H A . (Thus M is the permutation module of the 2-transitive action of H on a set Ω of size 24. In particular, M is one of the modules investigated by Mortimer in [22].) Since A fixes the non-zero vector a of V , we have Hom kA (k, V A ) = 0 and so, from Frobenius reciprocity, we obtain Hom kH (M, V ) = 0. Hence the kH-module V is isomorphic to M/W , for some maximal kH-submodule W of M . Let (e ω ) ω∈Ω be the canonical basis of M and let p be the characteristic of k.
Let e = ω∈Ω e ω , C = e and C ⊥ = { ω∈Ω c ω e ω | ω∈Ω c ω = 0}. Clearly, C and C ⊥ are submodules of M . Assume that p = 2, 3. From [22, Table 1], we see that the module M is completely reducible, M = C ⊕ C ⊥ and C ⊥ is an irreducible kH-module. Since V is a non-trivial kH-module, we obtain V ∼ = C ⊥ . Since M is self dual, we obtain that M * ∼ = M and hence V * ∼ = V . Therefore, since B fixes the non-zero vector b of V , it also fixes a non-zero vector of V * , but this contradicts Lemma 2.2. Now assume p = 3. From [22, Lemma 2], we have that C ⊆ C ⊥ and C ⊥ is the unique maximal submodule of M . Therefore W = C ⊥ and V ∼ = M/C ⊥ ∼ = C is a trivial kH-module, a contradiction. Therefore it remains to consider the case p = 2.
Now we are ready to prove Theorems 1.2, 1.4 and Corollary 1.3.
Proof of Theorem 1.2. We argue by contradiction and we let H be a minimal (with respect to the group order) counterexample. Let a, b ∈ V \ {0} with |a H | relatively prime to |b H |.
If H is a cyclic group of prime order p, then every H-orbit on V \ {0} has size p, a contradiction. Similarly, from Proposition 2.6 we see that the group H is not a non-abelian simple group. Thus H has a non-identity proper normal subgroup N . From the Clifford correspondence, V N = W 1 ⊕ · · · ⊕ W k with W i a homogeneous kN -module, for each i ∈ {1, . . . , k}, and with H acting transitively on the set of direct summands In particular, replacing a and b by a h and b k if necessary, we may assume that i = j = 1. Since N is a normal subgroup of H, we get that |a N | divides |a H | (respectively |b N | divides |b H |). Furthermore, since N acts trivially on {W 1 , . . . , W k }, we obtain that C N (a) ⊆ C N (a 1 ) and In particular, |a N 1 | and |b N 1 | are coprime. As W 1 is a homogeneous kN -module, there exists an irreducible kN -module U such that W 1 = U 1 ⊕ · · · ⊕ U r with U i ∼ = U , for each i ∈ {1, . . . , r}. Assume that U is the trivial kN -module. So, N acts trivially on W 1 . Since H permutes transitively the set of direct summands {W 1 , . . . , W k } and since N is normal in H, we obtain that N acts trivially on V and N = 1, a contradiction. Therefore U is a non-trivial irreducible kN -module and, in particular, In particular, for each i ∈ {1, . . . , r}, we have that |x N i | divides |a N 1 | and |y N i | divides |b N 1 |. Since a 1 , b 1 = 0, there exist i, j ∈ {1, . . . , r} with x i = 0 and y j = 0. Fix ϕ i : U i → U and ϕ j : U j → U two N -isomorphisms and write x = ϕ i (x i ) and y = ϕ j (y j ). In particular, x and y are non-zero elements of the non-trivial irreducible kN -module U . Furthermore, since ϕ i and ϕ j are N -isomorphisms, we obtain C N (x i ) = C N (x) and C N (y j ) = C N (y) and thus |x N | and |y N | are coprime. This contradicts the minimality of H and hence the theorem is proved.
Proof of Corollary 1.3. We argue by contradiction and we assume that V is a nontrivial finite dimensional kH-module and that a and b are elements of V with V = a h | h ∈ H = b h | h ∈ H and with gcd(|a H |, |b H |) = 1. Now we argue by induction on dim k V . If V is irreducible, then the result follows from Theorem 1.2. So, we assume that this is not the case. Let W be a minimal submodule of V and suppose that V /W is non-trivial. Clearly, (a + W ) H and (b + W ) H span V /W and hence, by induction, the lengths of the orbits of (a + W ) H and (b + W ) H are not coprime. As |(a + W ) H | divides |a H | and |(b + W ) H | divides |b H |, we have that |a H | and |b H | are not coprime.
Suppose now that V /W is the trivial kH-module. We claim that in this case V splits over W , that is, V = v ⊕ W for some element v of V fixed by H. If the characteristic of V is zero, then V is semisimple and our claim is immediate. Suppose that V has characteristic p > 0. Replacing a by b if necessary, we may assume that p ∤ |a H | and hence C H (a) contains a Sylow p-subgroup P of H. We claim that V ∼ = k ⊕ W , that is, V splits over W . The module V corresponds to an element δ of Ext 1 G (k, W ) ∼ = H 1 (G, W * ) (see [3, Section (III) 2] for the last isomorphism). On the other hand, V splits over W as a kP -module because P ⊆ C H (a) and a / ∈ W . Thus δ = 0 in H 1 (P, W * ). However, from [3, Theorem 10.3], we see that the restriction map res : H 1 (G, W * ) → H 1 (P, W * ) is injective. So δ = 0 is H 1 (G, W * ) and V splits over W . In particular, H fixes a vector v ∈ V \ W and We claim that C V (L) = ψ L (V ). For v ∈ V and y ∈ L, we have is a multiple of the sum of the elements of a L = a H . We deduce that Now we are ready to prove (a). Since a H = a L and a H spans V , the vector space V is generated by a as a kL-module. Thus, the map π : kL → V , given by π( x∈L α x x) = x∈L α x a x , defines a kL-homomorphism of kL onto V . Since p is coprime to |L|, by Maschke's theorem the kL-module V is isomorphic to a direct summand of the group-algebra kL. Therefore dim We now prove (b). Let b ∈ V with gcd(|a H |, |b H |) = 1. Write B = C V (b) and observe that H = AB. Since a is fixed by a Sylow p-subgroup of H, we see that p does not divide |H : A| = |B : (A∩B)| and so A∩B contains a Sylow p-subgroup of B. As H is p-soluble, we get that B is p-soluble and that B contains a p-complement L, say. So, B = (A ∩ B)L and H = AB = AL. In particular, we are in the position to apply the first part of the proof to L. Thus b ∈ C V (B) ⊆ C V (L) = C V (H).

Proof of Theorem 1.1
In this section we use Theorem 1.2 to prove Theorem 1.1. We start by showing that the hypothesis "completely reducible" is essential.
Example 3.1. Let p be an odd prime, V be the 2-dimensional vector space of row vectors over a field F p of size p, λ be a generator of the multiplicative group F p \ {0} and The group H has order p(p − 1) and has p + 1 orbits on V . Namely, for each Proof of Theorem 1.1. As V is completely reducible, we have V = C V (H) ⊕ W for some direct summand W of V . Clearly, replacing V by W if necessary, we may assume that C V (H) = 0, that is, H fixes no non-zero vector of V . There is also no loss in assuming that V is generated by a and b as a kH-module. Let V (a) and V (b) denote the kH-submodules generated by a and b, respectively. We claim that Suppose not. Let S be a simple kH-submodule of V (a)∩V (b). Since C H (V ) = 0, H does not act trivially on S. Since V is completely reducible, V = S ⊕ T as kHmodules. Let π denote the projection of V onto S with kernel T . Clearly, |a H | is a multiple of |π(a) H | and similarly for b. Since S ≤ V (a), π(a) = 0 (and similarly for b). Thus, the lengths of the H-orbits in S of π(a) and π(b) are coprime contradicting Theorem 1.2.
We point out that from Theorem 1.1 we can easily deduce the following wellknown result of Yuster (see [29] or [17, 3.34]).

Corollary 3.2.
Let H and A be finite groups with |H| relatively prime to |A| and with H acting as a group of automorphisms on A. If, for a, b ∈ A, the H-orbits a H and b H have relatively prime size, then H has an orbit of size |a H ||b H |.
Proof. As |H| is relatively prime to |A|, from [14, Lemma 2.6.2] we see that we may assume that A is a direct product of elementary abelian groups. In particular, from Maschke's theorem, A is a completely reducible H-module, possibly of mixed characteristic. Now the result follows from Theorem 1.1.

Proof of Theorem 1.5
The main ingredient in the proof of Theorem 1.5 is Theorem 1.2 and the positive solution of Fisman and Arad [9] of Szep's conjecture.
Theorem 4.1. [9] Let G = AB be a finite group such that A and B are both subgroups of G with non-trivial centres. Then G is not a non-abelian simple group.

We start by considering some examples.
Example 4.2. Primitive groups of AS type. From [21], we see that the sporadic simple group G = J 1 has a primitive permutation representation of rank 5 on a set ∆ of size 266. The subdegrees of G are 1, 11, 12, 110 and 132. In particular, G has two coprime subdegrees. No primitive group of smaller rank has this property: the proof of this assertion requires the classification of the finite simple groups [5,Remark,p. 33]. Now we give an infinite family of examples. Let p be a prime with p ≡ ±1 mod 5 and with p ≡ ±1 mod 16, and let G = PSL (2, p). From [27, Chapter 3, Section 6], we see that G contains a maximal subgroup H with H ∼ = Alt (5). Consider G as a primitive permutation group acting on the set ∆ of right cosets of H in G. Let K be a maximal subgroup of H with K ∼ = Alt(4). As 8 divides |G|, we see from [27,Chapter 3,Section 6] that N G (K) ∼ = Sym(4). Let g ∈ N G (K) \ H.
Then K = H ∩ H g , |H : H ∩ H g | = 5 and so G has a suborbit of size 5 on ∆. Similarly, let now K be a Sylow 5-subgroup of H. Using the generators of H given in [27, Chapter 3, Section 6], we see, with a direct computation, that there exists g ∈ N G (K) \ H with K = H ∩ H g . Therefore |H : H ∩ H g | = 12 and so G has a suborbit of size 12. Furthermore, another explicit computation with the generators of H shows that there exists g ∈ G with H ∩ H g = 1. So G has a suborbit of size 60 = 5 · 12.
Example 4.3. Primitive groups of PA type. Let G be a primitive group of AS type on ∆ with non-trivial coprime subdegrees a and b. Let δ, δ 1 and δ 2 be in ∆ with a = |δ G δ 1 |, b = |δ G δ 2 |. For each n ≥ 2, the wreath product W = G wr Sym(n) endowed with its natural product action on Ω = ∆ n is a primitive group of PA type. Consider the elements α = (δ, . . . , δ), β = (δ 1 , . . . , δ 1 ) and γ = (δ 2 , . . . , δ 2 ) of Ω. We have |β Wα | = a n and |γ Wα | = b n and so a n and b n are two coprime subdegrees of W . In many cases there are several pairs of coprime non-trivial subdegrees of W . For example, if n ≥ 3 and n is coprime to b, then the point β ′ = (δ 1 , δ, . . . , δ) lies in a W α -orbit of size na and we have also na and b n as coprime non-trivial subdegrees.
In particular, this construction can be applied with G and ∆ as in Example 4.2.
Example 4.4. Primitive groups of TW type. In this example we construct a primitive group of TW type with two non-trivial coprime subdegrees. We start by recalling the structure and the action of a primitive group of twisted wreath type. We follow [7,Section 4.7].
The group H is a primitive group of SD type in its action on the right cosets of L. Let T be PSL(2, 7) and let ϕ : L → Aut(T ) be the mapping sending (x, x)ι i to the inner automorphism of T given by x, for each x ∈ PSL(2, 7) and i ∈ {0, 1}. Clearly, ϕ is a homomorphism whose image contains the inner automorphisms of T . Let G be the twisted wreath product determined by H and ϕ and let N be its socle as described above. Since H has no homomorphic image isomorphic to T , we see that G is a primitive permutation group on Ω = N . The group H is the stabilizer G f of the function f : H → T mapping every element of H to the identity of T .
Fix z in H. Distinguishing the cases z ∈ CL and z / ∈ CL, it is easy to verify that g(zl) = g(z) ϕ(l) for each l ∈ L and z ∈ H, and hence g ∈ Ω. For each c ∈ C and z ∈ H, we have g c (z) = g(cz) = g(z), and hence C ⊆ C H (g). We claim that C = C H (g). Let h = (h 1 , h 2 )ι j be in C H (g). Suppose that h / ∈ CL. As g(1) = γ = 1 and g h (1) = g(h) = 1, we obtain g = g h . Thus h ∈ CL and C H (g) ⊆ CL. As C ⊆ C H (g), replacing h by ch for a suitable element c ∈ C, we may assume that h ∈ L, that is, h = (x, x)ι i for some x ∈ T and i ∈ {0, 1}. Now . Hence x ∈ C T (γ) = γ , h ∈ C ∩ L and our claim is proved. Thus the H-orbit containing g has size |H : C| = 24 2 = 576, and 576 is a subdegree of G. Define We claim that the function h is well-defined. In fact, if z = d 1 tl 1 = d 2 tl 2 for d 1 , d 2 ∈ D and l 1 , l 2 ∈ L, then l 2 l −1 In particular, since u centralizes (η, η), we obtain η ϕ(l1) = η ϕ(u)ϕ(l1) = η ϕ(ul1) = η ϕ(l2) and so the image η ϕ(l) is independent of the representation z = d i tl i of z.
Fix z in H. As above, by distinguishing the cases z ∈ DtL and z / ∈ DtL, it is easy to verify that h(zl) = h(z) ϕ(l) for each l ∈ L, and hence h ∈ Ω. From the definition of h, we see that for each d ∈ D and z ∈ H, we have h d (z) = h(dz) = h(z). Hence D ⊆ C H (h). Since D is a maximal subgroup of H, we obtain D = C H (h). Thus the H-orbit containing h has size |H : D| = 7 2 = 49, and 49 is a subdegree of G.
This shows that G has two coprime subdegrees, namely 576 and 49. Now, using the computational algebra system GAP [10], it is easy to show that C H (f g) = C H (f ) ∩ C H (g). In particular, the suborbit of G containing f g has size 576 · 49.
Proof of Theorem 1.5. From Examples 4.2, 4.3 and 4.4 and Theorem 1.2, it suffices to show that if G is a primitive group of type HS, HC, SD or CD, then G has no non-trivial coprime subdegrees. We argue by contradiction and we assume that G is a primitive group on Ω of HS, HC, SD or CD type with two non-trivial coprime subdegrees.

Proofs of Theorems 1.6 and 1.10
Before proving Theorem 1.6 and 1.10 we need the following definition and lemmas.
Definition 5.1. If G is a finite group, we let µ(G) denote the maximal size of a set {G i } i∈I of proper subgroups of G with |G : G i | and |G : G j | relatively prime, for each two distinct elements i and j of I. Lemma 5.2. If K is a direct product of isomorphic nonabelian simple groups, then µ(K) ≤ 2.
Proof. We have K = T 1 × · · · × T ℓ with T i ∼ = T , for some nonabelian simple group T and for some ℓ ≥ 1. We argue by contradiction and we assume that µ(K) ≥ 3, that is, K has three proper subgroups A 1 , A 2 and A 3 with |K : A 1 |, |K : A 2 | and |K : A 3 | relatively prime. Write a i = |K : A i | for i ∈ {1, 2, 3}. Replacing A i with a maximal subgroup of K containing A i if necessary, we may assume that A i is maximal in K, for i ∈ {1, 2, 3}. In particular, (up to relabelling the simple direct summands of K) we have that either . In the latter case we have that a 1 = |T | is not relatively prime to a 2 and to a 3 , a contradiction. Therefore, up to relabeling the indices, we may assume that A i = M i × T 2 × · · · × T ℓ with M i a maximal subgroup of T 1 , for i ∈ {1, 2, 3}. This gives that the nonabelian simple group T 1 admits three coprime factorizations Table 1 shows that this is impossible. The same conclusion can be obtained using [2], where the authors determine all multiple factorizations of finite nonabelian simple groups T = M i M j , for i and j distinct elements of {1, 2, 3}. In particular, it is readily checked that in none of the multiple factorizations in [2] are the indices |T : M 1 |, |T : M 2 | and |T : M 3 | pairwise coprime. Proof of Theorem 1.6. Let G be a primitive group of TW type. We use the notation introduced in the first paragraph of Example 4.4: so G = N ⋊ H is the twisted wreath product determined by H and ϕ : L → Aut(T ). Recall that G acts primitively on N , with N acting on itself by right multiplication and with H acting on N by conjugation. In particular, H is the stabilizer of the point 1 ∈ N . Let K be a minimal normal subgroup of H. From [7, Theorem 4.7B (i)], K is a direct product of nonabelian simple groups. Write ℓ = |H : L|. Hence N = T 1 × · · · × T ℓ with T i ∼ = T , for each i ∈ {1, . . . , ℓ}. Furthermore, L = N H (T i ) for some i ∈ {1, . . . , ℓ}. Relabeling the T j if necessary, we may assume that i = 1. From [7, Theorem 4.7B (ii)], the group L is a core-free subgroup of H and hence K L.
We claim that K ∩ L acting by conjugation on the simple group T 1 induces all the inner automorphisms. If not, then K ∩ L ⊆ C H (T 1 ) because K ∩ L is a normal subgroup of L and T 1 is nonabelian simple. Thus the homomorphism ϕ : L → Aut(T ) can be extended to a homomorphismφ : KL → Aut(T ) of the group KL by settingφ(kl) = ϕ(l), for each l ∈ L and k ∈ K. As L KL, this contradicts the maximality condition of H in G given in [1, Lemma 3.1 (b)], and the claim is proved. In particular, since K is a normal subgroup of H and since H acts transitively on the ℓ simple direct summands {T 1 , . . . , T ℓ }, we obtain that K ∩ N H (T i ) induces by conjugation all the inner automorphisms of T i , for each i ∈ {1, . . . , ℓ}. This gives C N (K) = 1 and so K fixes a unique point of N . Now the proof follows from Lemmas 5.2 and 5.3.
Before concluding this section we show that coprime subdegrees in primitive groups G of AS or PA type determine coprime subdegrees in transitive non-abelian simple groups T , and we give a strong relation between G and T . Let G be a primitive group of AS or PA type. We recall that from the description of the O'Nan-Scott types in [26] the group G is a subgroup of the wreath product H wr Sym(ℓ) endowed with its natural product action on ∆ ℓ , with H an almost simple primitive group on ∆ (we have ℓ = 1 and G = H if G is of AS type, and ℓ > 1 if G is of PA type). Furthermore, if T is the socle of H, then N = T 1 × · · · × T ℓ is the socle of G, where T i ∼ = T for each i ∈ {1, . . . , ℓ}. Write G i = N G (T i ) and denote by π i : G i → H the natural projection. From [26], we see that replacing H by π i (G i ) if necessary, we may assume that π i (G i ) is surjective. In this case, we say that H is the component subgroup of G. In particular, if G is of AS type, the component subgroup of G is G itself.
(We recall that the definition of pseudo-maximal subgroup is in Definition 1.8.) Proposition 5.4. Let G be a primitive permutation group of AS or PA type acting on ∆ ℓ with component subgroup H ⊆ Sym(∆) and let T be the socle of H. For δ ∈ ∆, the stabilizer T δ is a pseudo-maximal subgroup of T . Furthermore, if G has k non-trivial coprime subdegrees, then T acting on ∆ has at least k non-trivial coprime subdegrees.
Proof. As H is a primitive group of AS type on ∆, we have that H δ is a maximal subgroup of the almost simple group H with T H δ , for each δ ∈ ∆. Therefore T δ = T ∩ H δ is a pseudo-maximal subgroup of T and, in particular, N H (T δ ) = H δ . Let Λ be the set of fixed points of T δ on ∆ and let δ 1 , δ 2 ∈ Λ. By transitivity, Therefore h ∈ N H (T δ ) = H δ and Λ is the H δ -orbit containing δ, that is, Λ = {δ} and T δ fixes a unique point of ∆.
Proof of Theorem 1.10. Assume that Theorem 1.9 holds true. Let G be a primitive permutation group on Ω with three non-trivial coprime subdegrees. From Theorems 1.5 and 1.6, G is of AS or PA type. Since Theorem 1.9 holds true, Proposition 5.4 yields a contradiction.
6. Proofs of Theorems 1.14 and 1.15 As usual, we denote by F(G) the Fitting subgroup of the finite group G, that is, the largest normal nilpotent subgroup of G. The proof of Corollary 1.14 and 1.15 will follow from Lemma 6.1 and from the results in Section 5. Proof of Theorem 1.14. Let G be a non-regular finite transitive permutation group on Ω and let α be in Ω. Set H = G α . The result now follows from Lemma 6.1.
Proof of Theorem 1.15. If t = 1, then there is nothing to prove. So we may assume that t ≥ 2. We claim that K/k is a separable extension (and so, as K/k is normal, a Galois extension). This is clear if k has characteristic 0. Suppose then that k has characteristic p > 0. Now, if for some i ∈ {1, . . . , t}, the extension k i /k is separable, then K/k is separable (being the normal closure of a separable extension). Therefore we may assume that k i /k is inseparable, for each i ∈ {1, . . . , t}. This gives that p divides [k i : k], for each i ∈ {1, . . . , t}. As t ≥ 2, we obtain a contradiction and the claim is proved.
Let H be the Galois group Gal(K/k) and set H i = Gal(K/k i ), for i ∈ {1, . . . , t}. Since the normal closure of k i is K, we obtain that H i is core-free in H. Therefore H 1 , . . . , H t is a family of core-free subgroups of H of pairwise coprime index. Now apply Lemma 6.1 to obtain t ≤ 2.

Coprime factorizations of non-abelian simple groups
Liebeck, Praeger and Saxl [19,20] completely classified the maximal factorizations of all finite almost simple groups. Tables 1-6 and Theorem D of [19] determine all the triples (G, A, B) where G is a nonabelian simple group, and A and B are maximal subgroups of G with G = AB. Now, if A ′ and B ′ are subgroups of G with |G : A ′ | relatively prime to |G : B ′ |, then G = A ′ B ′ . In particular, A ′ and B ′ give rise to a coprime factorization of G. Moreover, if A (respectively B) is a maximal subgroup of G with A ′ ⊆ A (respectively B ′ ⊆ B), then G = AB is a maximal coprime factorization. Therefore, the list of all non-abelian simple groups admitting a coprime factorization can be easily obtained with some elementary arithmetic from [19]. Namely, for each triple (G, A, B) in Tables 1-6 and in Theorem D of [19], it suffices to check whether |G : A| is relatively prime to |G : B|. Table 1 in this paper gives all possible maximal coprime factorizations (G, A, B) and, in particular, the list of all nonabelian simple groups admitting a coprime factorization. The notation we use is standard and follows the notation in [ PSp(4, q) P i i ∈ {1, 3}, q odd, q ≡ 1 mod 8 L n (q) PSp(n, q) P i i ∈ {1, n − 1}, n = 2 r , r ≥ 2, q even L n (q) O − 2m (q) P m m odd and q even PSp 4 (3) 2 4 .Alt(5) P i i ∈ {1, 2} Ω 2m+1 (q) N − 1 P m q odd, m ≥ 3 odd Ω + 2m (q) N 1 P i i ∈ {m − 1, m}, m ≥ 5 odd, q even or q ≡ 3 mod 4 Ω 7 (3) Sp 6 (2) P 3 Ω 7 (3) 2 6 . Alt(7) P 3 PΩ + 8 (3) Ω + 8 (2) P i i ∈ {1, 3, 4} Table 1. Maximal coprime factorizations of a finite non-abelian simple group G