Cliques in derangement graphs for innately transitive groups

Theorem 1.1

There exists a function f 1 : N → N such that, if 𝐺 is innately transitive of degree 𝑛 and the derangement graph of 𝐺 has no clique of size 𝑘, then n ≤ f 1 ⁢ ( k ) .

Theorem 1.2

If 𝐺 is innately transitive of degree 𝑛 and the derangement graph of 𝐺 has no clique of size 4, then n ≤ 3 .

Theorem 1.3

There exists a function f 2 : N → N such that, if 𝐺 is innately transitive of degree 𝑛 and 𝐺 has no semiregular subgroup of order at least 𝑘, then either n ≤ f 2 ⁢ ( k ) , or 𝐺 is primitive of degree 12 κ and G = M 11 wr A , for some positive integer 𝜅 and for some transitive subgroup 𝐴 of Sym ⁡ ( κ ) , where M 11 is the Mathieu group.

Moreover, if 𝐺 has no semiregular subgroup of order at least 4, then one of the following holds:

1. n ≤ 3 ;

2. n = 6 , 𝐺 is primitive and G ≅ Alt ⁢ ( 5 ) ;

3. n = 6 , 𝐺 is primitive and G ≅ Alt ⁢ ( 6 ) ;

4. n = 36 , 𝐺 is primitive and PSU 3 ⁢ ( 3 ) ≤ G ≤ P ⁢ Γ ⁢ U 3 ⁢ ( 3 ) ;

5. n = 12 κ , 𝐺 is primitive and G = M 11 wr A for some positive integer 𝜅 and for some transitive subgroup 𝐴 of Sym ⁡ ( κ ) .

Proof of Theorem 1.1

Let f 2 be the function from Theorem 1.3 and let

We show that Theorem 1.1 holds true with this choice of f 1 .

As semiregular subgroups are cliques in the derangement graph, Theorem 1.1 follows immediately from Theorem 1.3 using f 1 , except when 𝐺 is primitive of degree 12 κ and G = M 11 wr A for some positive integer 𝜅 and some transitive subgroup 𝐴 of Sym ⁡ ( κ ) . Therefore, it suffices to deal with this case.

Let Ω be the domain of 𝐺. Then Ω admits a Cartesian decomposition Δ κ , where | Δ | = 12 and 𝐺 acts on Δ κ via its natural primitive product action. In particular, we identify Ω with Δ κ and we denote the elements of 𝐺 as

with h 1 , … , h κ ∈ M 11 and a ∈ A . Moreover, given ( δ 1 , … , δ κ ) ∈ Ω , we have

From Jordan’s theorem, M 11 has a derangement ℎ in its action on Δ. Now, the set

has cardinality 2 κ and it is a clique in Γ G . In particular, if 2 κ ≥ k , then Γ G has a clique of size at least 𝑘. Otherwise, 2 κ < k , and hence κ < log 2 ⁡ ( k ) . Therefore,

Thus Theorem 1.1 follows also in this case. ∎

Proof of Theorem 1.2

Let 𝐺 be innately transitive and assume that Γ G has no clique of size 4. Arguing as in the previous proof, we may assume that 𝐺 is one of the groups appearing in parts 2–5 of Theorem 1.3. Moreover, in part 5, we may assume that κ = 1 . We have checked with a computer, using the computer algebra system magma [2], that, in the derangement graph of each these permutation groups, there is a clique of size 4. ∎

Lemma 3.1

Let 𝐺 be an innately transitive group on Ω, let 𝑁 be a minimal normal subgroup of 𝐺 transitive on Ω, let Σ be a system of imprimitivity, let π : G → Sym ⁡ ( Σ ) be the natural homomorphism given by the action of 𝐺 on Σ and let G Σ be the image of 𝜋. If | Σ | > 1 and X ̄ is a semiregular subgroup of G Σ , then π − 1 ⁢ ( X ̄ ) is a semiregular subgroup of 𝐺.

Proof

Let K = Ker ⁡ ( π ) . Since 𝑁 is a minimal normal subgroup of 𝐺, we have N ≤ K or K ∩ N = 1 . If N ≤ K , then 𝐾 is transitive because so is 𝑁. Since 𝐾 acts trivially on Σ, 𝐾 fixes every element of Σ setwise and, since | Σ | > 1 , we deduce that 𝐾 is intransitive. This contradiction yields K ∩ N = 1 . Hence 𝑁 centralizes 𝐾. Since 𝑁 is transitive on Ω, we deduce from [8, Theorem 4.2A] that 𝐾 is semiregular on Ω.

Let X = π − 1 ⁢ ( X ̄ ) . We prove that 𝑋 is semiregular on Ω. Indeed, let ω ∈ Ω and let σ ∈ Σ with ω ∈ σ . Clearly, X ω ≤ X σ because each permutation of 𝐺 fixing 𝜔 must fix the block of the system of imprimitivity Σ containing 𝜔. As X ̄ is semiregular on Σ, we have X ̄ σ = 1 , that is, X σ fixes every element of Σ setwise. Therefore, π ⁢ ( X σ ) = 1 and X ω ≤ X σ ≤ Ker ⁡ ( π ) = K . As 𝐾 is semiregular on Ω, we obtain X ω ≤ K ω = 1 . This shows that 𝑋 is semiregular on Ω. ∎

Lemma 3.2

Let 𝐺 be an innately transitive group on Ω, let Σ be a system of imprimitivity such that the permutation group G Σ induced by 𝐺 on Σ is isomorphic to M 11 wr A with its natural primitive product action on 12 κ points, for some positive integer 𝜅 and some transitive subgroup 𝐴 of Sym ⁡ ( κ ) , and let π : G → Sym ⁡ ( Σ ) be the natural homomorphism given by the action of 𝐺 on Σ. Then either 𝐺 in its action on Ω has semiregular subgroups of order at least min ⁡ ( | Ker ⁡ ( π ) | ⋅ 11 κ , 660 κ ) , or Σ = Ω and the action of G = M 11 wr A on Ω is the natural primitive product action on 12 κ points.

Proof

Let 𝐾 be the kernel of 𝜋. By definition, G Σ is the image of 𝜋. Let 𝑁 be a minimal normal subgroup of 𝐺 with 𝑁 transitive on Ω: the existence of 𝑁 is guaranteed by the fact that 𝐺 is innately transitive on Ω.

We first assume K = 1 . As G ≅ G Σ = M 11 wr A , we deduce 𝑁 is the unique minimal normal subgroup of 𝐺, 𝐺 is quasiprimitive on Ω and N = M 11 κ . Let σ ∈ Σ and let ω ∈ σ . Since G Σ is endowed with its natural primitive product action of degree 12 κ , we get

If Σ is the trivial system of imprimitivity { { ω } ∣ ω ∈ Ω } , then the action of G = M 11 wr A on Ω is the natural primitive product action on 12 κ points, and the lemma is satisfied. Therefore, for the rest of the proof, we assume that Σ is not the trivial system of imprimitivity. Therefore, 𝐺 is imprimitive on Ω and G ω < G σ . In particular, there exists a maximal subgroup 𝑅 of G σ with G ω ≤ R . Since there is a one to one order-reversing correspondence between the lattice of subgroups of 𝐺 containing G ω and the systems of imprimitivity for 𝐺 acting on Ω, 𝑅 corresponds to the stabilizer of a block 𝜆 in a system of imprimitivity, Λ say. As G λ = R ≤ G σ , Λ is a refinement of the system of imprimitivity Σ. See Figure 1.

Set H = G σ and Λ σ = { μ ∈ Λ ∣ μ ⊆ σ } . We claim that 𝐻 acts primitively and faithfully on Λ σ . The fact that 𝐻 acts primitively on Λ σ follows from the fact that, by definition, 𝑅 is a maximal subgroup of G σ = H and from the fact that R = G λ is the stabilizer of the part λ ∈ Λ σ in the system of imprimitivity Λ. Let

Observe that PSL 2 ⁢ ( 11 ) κ is the unique minimal normal subgroup of 𝐻. Therefore, if 𝐻 were not faithful on Λ σ , that is L ≠ 1 , then 𝐿 would contain the socle PSL 2 ⁢ ( 11 ) κ = N σ of 𝐻. Now, since 𝑁 is transitive on Ω, we have G = G ω ⁢ N . Intersecting both sides of this equality with G σ and using the modular law, we deduce G σ = G ω ⁢ N σ . Therefore, N σ seen as a permutation group on Ω is transitive on the points contained in the block 𝜎. As L ≥ N σ , we deduce that 𝐿 is transitive on the points contained in the block 𝜎, which is a contradiction because L ≤ R = G λ fixes the subset 𝜆 of Ω setwise and λ ⊊ σ .

We apply the O’Nan–Scott theorem to the primitive permutation group 𝐻 in its action on Λ σ . Since 𝐻, as an abstract group, is isomorphic to PSL 2 ⁢ ( 11 ) wr A , 𝐻 in its primitive action on Λ σ is of type

• AS (when κ = 1 ), or PA (when κ > 1 ), or

• SD, or CD, or

• TW.

We deal with each of these cases in turn.

Assume that 𝐻 in its action on Λ σ has type AS or PA. Thus we have

for some maximal subgroup 𝐵 of PSL 2 ⁢ ( 11 ) . This shows that 𝐺 in its action on Λ has stabilizer the wreath product B wr A . Therefore, Λ admits a 𝐺-invariant Cartesian decomposition Λ ′ ⁣ κ , where Λ ′ is the set of right cosets of 𝐵 in M 11 and has cardinality | M 11 : B | . Now, PSL 2 ⁢ ( 11 ) has four conjugacy classes of maximal subgroups: isomorphic to 11 : 5 , 6 : 2 , and two conjugacy classes isomorphic to Alt ⁢ ( 5 ) . Therefore, 𝐵 is M 11 -conjugate to one of these four subgroups. With the help of a computer, we have computed these four permutation representations and their semiregular subgroups: in the action of M 11 on the cosets of 11 : 5 , there are semiregular subgroups of order 144; in the action of M 11 on the cosets of 6 : 2 , there are semiregular subgroups of order 55; in the action of M 11 on the cosets of Alt ⁢ ( 5 ) (for each of the two choices of M 11 -conjugacy classes), there are semiregular subgroups of order 11. In particular, M 11 in its action on Λ ′ has semiregular subgroups of order at least 11. Therefore, G = M 11 wr A in its action on Λ = Λ ′ ⁣ κ has semiregular subgroups of order at least 11 κ . Applying Lemma 3.1 (with Σ = Λ ), we deduce that 𝐺 in its action on Ω has semiregular subgroups of order at least 11 κ .

Assume 𝐻 in its action on Λ σ has type SD or CD. Recall that N σ = PSL 2 ⁢ ( 11 ) κ is the socle of 𝐻. Let T 1 , … , T κ be the 𝜅 simple direct factors of N σ . Then ( N σ ) λ = N λ is isomorphic to the direct product of 𝑎 diagonal subgroups; indeed, up to relabeling the indexed set, there exists a divisor^[11] a > 1 of 𝜅 such that

Now, if we let

then we see that X ∩ N λ = 1 . Therefore, 𝑋 acts semiregularly on Λ σ . What is more, from the definition of 𝑋, we deduce N λ n ∩ X = N λ n ∩ X = 1 for all n ∈ N . As 𝑁 is transitive on Λ, we get that 𝑋 is semiregular on Λ. Applying Lemma 3.1 (with Σ = Λ ), we deduce that 𝐺 in its action on Ω has semiregular subgroups of order at least | X | = | PSL 2 ⁢ ( 11 ) | κ − κ / a = 660 κ − κ / a ≥ 660 κ / 2 ≥ 11 κ .

Assume that 𝐻 in its action on Λ σ has type TW. Then the socle N σ of H = G σ acts regularly on Λ σ , that is, N λ = 1 . As N ω ≤ N λ = 1 , we deduce N ω = 1 , and hence 𝑁 is regular on Ω. So 𝐺 in its action on Ω has semiregular subgroups of order at least | N | = | M 11 | κ ≥ 11 κ .

It remains to consider the case that 𝐺 does not act faithfully on Σ, that is, K ≠ 1 ; we pivot on the previous part of the proof. Let Λ be the system of imprimitivity consisting of the 𝐾-orbits, that is, Λ = { ω K ∣ ω ∈ Ω } . Let ω ∈ Ω , let σ ∈ Σ with ω ∈ σ and let λ = ω K ∈ Λ . Observe that the stabilizer of the block 𝜆 in 𝐺 is G λ = K ⁢ G ω . As 𝐾 and G ω are both subgroups of G σ , we get G λ ≤ G σ . Therefore, Λ is a refinement of the system of imprimitivity Σ. See again Figure 1: here the system of imprimitivity Λ is formed by the 𝐾-orbits.

We have

Therefore, 𝐾 is also the kernel of the action of 𝐺 on Λ. In particular, applying the first part of the proof with the group 𝐺 replaced by G / K and with the set Ω replaced by Λ, we deduce that either

• G / K in its action on Λ has a semiregular subgroup of order at least 11 κ , or

• Λ = Σ and the action of G / K on Λ is the natural primitive product action on 12 κ points.

In the first case, Lemma 3.1 implies that 𝐺 in its action on Ω has a semiregular subgroup of order at least

This concludes the analysis of the first case.^[12]

We deal with the second case. Observe that, in this case, Figure 1 is somehow misleading because, in this case, we have Λ = Σ . Let M = N ⁢ K = N × K . We have

where the first equality follows from the fact that 𝑁 is transitive on Ω and the second equality follows from the fact that 𝐾 fixes the 𝐾-orbit λ = ω K setwise. Since M ω ≤ M λ = N λ × K , we may write each element of m ∈ M ω as an ordered pair a ⁢ b for a unique a ∈ N λ and a unique b ∈ K . Let π N λ : M ω → N λ and π K : M ω → K be the natural projections. From (3.1), π K is surjective. Let m ∈ Ker ⁡ ( π N λ ) . Then m ∈ K ∩ M ω = K ω = 1 because 𝐾 is semiregular on Ω. Thus π N λ is injective. Moreover, since M λ is transitive on the points contained in the block 𝜆 and since λ = ω K , from (3.2), we deduce

This yields | N λ | = | M ω | . As π N λ : M ω → N λ is injective, we obtain that π N λ is surjective, and hence it is a bijection. This shows that

is a surjective group homomorphism. Furthermore, from the definitions of π N λ and π K , we have

Recall that, in the case under consideration, Λ = Σ . As N λ = N σ = PSL 2 ⁢ ( 11 ) κ and as ψ : N λ → K is a surjective group homomorphism, we get K ≅ PSL 2 ⁢ ( 11 ) ℓ for some 1 ≤ ℓ ≤ κ , and Ker ⁡ ( ψ ) = PSL 2 ⁢ ( 11 ) κ − ℓ .

Now, if a ∈ N ω , then a ∈ M ω , and hence we have a ∈ Ker ⁡ ( ψ ) . Conversely, if a ∈ Ker ⁡ ( ψ ) , then a = a ⁢ a ψ ∈ M ω ∩ N = N ω . This yields

Since 𝑁 is transitive on Ω, we have G = N ⁢ G ω . This implies that G ω acts transitively by conjugation on the 𝜅 simple direct factors of N = M 11 κ . Thus G ω acts transitively by conjugation on the 𝜅 simple direct factors of N λ = PSL 2 ⁢ ( 11 ) κ . As N ⊴ G , we have N ω ⊴ G ω . Putting together the fact that G ω acts transitively on the simple direct factors of N λ = PSL 2 ⁢ ( 11 ) κ and the fact that

we deduce ℓ = κ . Therefore, | K | = | PSL 2 ⁢ ( 11 ) κ | , and hence 𝐺 in its action on Ω has a semiregular subgroup of order at least

Proposition 3.3

Suppose that Theorem 1.3 holds true for primitive permutation groups. Then Theorem 1.3 holds true.

Proof

Let g : N → N be a function witnessing that Theorem 1.3 holds true for primitive permutation groups. This means that, if 𝐺 is primitive of degree 𝑛 and 𝐺 has no semiregular subgroup of order at least 𝑘, then either n ≤ g ⁢ ( k ) , or 𝐺 has degree 12 κ and G = M 11 wr A for some positive integer 𝜅 and for some transitive subgroup 𝐴 of Sym ⁡ ( κ ) . Moreover, if 𝐺 has no semiregular subgroup of order at least 4, then one of parts 1–5 holds.

Let f : N → N be the function defined by

We show that the first part of Theorem 1.3 holds true using this function 𝑓. Let 𝐺 be an innately transitive group of degree 𝑛 and suppose that 𝐺 has no semiregular subgroup of order at least 𝑘. If 𝐺 is primitive, then we have nothing to prove because we are assuming the veracity of Theorem 1.3 for primitive groups. Therefore, we may suppose that 𝐺 is imprimitive.

Let Ω be the domain of 𝐺 and let 𝑁 be a minimal normal subgroup witnessing that 𝐺 is innately transitive, that is, 𝑁 is transitive on Ω. Let Σ be a system of imprimitivity for the action of 𝐺 on Ω with the property that 𝐺 acts primitively on Σ.^[13] Let 𝐾 be the kernel of the action of 𝐺 on Σ and let G Σ ≅ G / K be the permutation group induced by 𝐺 on Σ. We denote by π : G → Sym ⁡ ( Σ ) the natural homomorphism; by definition, G Σ is the image of 𝜋.

Let X ̄ be a semiregular subgroup of G Σ and let X = π − 1 ⁢ ( X ̄ ) be the preimage of X ̄ via 𝜋. By Lemma 3.1, 𝑋 is semiregular on Ω. As 𝐺 has no semiregular subgroups of order at least 𝑘, we get

As | K | ≥ 1 , (3.3) shows that the primitive group G Σ has no semiregular subgroups of order at least 1 + ( k − 1 ) / | K | ≤ k . Since we are assuming the veracity of Theorem 1.3 for primitive permutation groups, we deduce that either

• | Σ | ≤ g ⁢ ( k ) , or

• G Σ is primitive of degree 12 κ and G Σ = M 11 wr A for some positive integer 𝜅 and for some transitive subgroup 𝐴 of Sym ⁡ ( κ ) .

Assume first that | Σ | ≤ g ⁢ ( k ) . In particular,

Since n ≤ | G | , we deduce

where we are using (3.3) in the second inequality.

Assume that the second possibility above holds. Our auxiliary Lemma 3.2 implies that either 𝐺 in its action on Ω has a semiregular subgroup of order at least min ⁡ ( | K | ⋅ 11 κ , 660 κ ) ≥ 11 κ , or Σ = Ω and the action of G = M 11 wr A on Ω is the natural primitive product action on 12 κ points. The second case is impossible in our situation because we are assuming that 𝐺 is imprimitive on Ω. Moreover, if k ≤ 11 κ , then 𝐺 does have a semiregular subgroup of order at least 𝑘. Assume then that 11 κ < k . Observe that G Σ = M 11 wr A has a faithful permutation representation of degree 11 κ , and hence | G Σ | ≤ ( 11 κ ) ! ≤ k ! . Therefore,

where as above we are using (3.3) in the second inequality.

It remains to prove the second part of the statement of Theorem 1.3 for innately transitive groups. Therefore, let 𝐺 be innately transitive with no semiregular subgroups having order at least 4. We use the notation above (with k = 4 ). In particular, we may assume that 𝐺 is not primitive because we are assuming the veracity of Theorem 1.3 for primitive groups. Recall that G Σ is primitive and either

• each semiregular subgroup of G Σ has order at most ( k − 1 ) / | K | = 3 / | K | , see (3.3) with k = 4 , or

• G Σ has degree 12 κ and G Σ = M 11 wr A for some positive integer 𝜅 and for some transitive subgroup 𝐴 of Sym ⁡ ( κ ) .

Assume that the second possibility above holds. Our auxiliary Lemma 3.2 implies that either 𝐺 in its action on Ω has a semiregular subgroup of order at least 11 κ , or Σ = Ω and the action of G = M 11 wr A on Ω is the natural primitive product action on 12 κ points. In the first case, we have a semiregular subgroup of order at least 11 ≥ 4 , and in the second case, we obtain that part 5 holds. This concludes the proof for this case.

Assume now that the first possibility above holds. In particular, as 3 / | K | < 4 and as G Σ is primitive, by hypothesis, one of parts 1–5 holds for G Σ . Observe that part 5 is exactly the second possibility, which we have already dealt with; therefore, we may disregard this part from further consideration. Before dealing with each of the remaining four possibilities, we make some preliminary observations.

Recall that | K | ≤ 3 because 𝐾 is semiregular on Ω and 𝐺 has no semiregular subgroups of order at least 4. Assume | K | ∈ { 2 , 3 } . Then 3 / | K | < 2 , and hence G Σ has no non-trivial semiregular subgroups. A direct inspection on parts 1–4 shows that | Σ | = 1 and G Σ = 1 . This is impossible because Σ is a non-trivial system of imprimitivity of Ω, and hence | Σ | > 1 .

Assume | K | = 1 . In particular, G ≅ G Σ as abstract groups. We have constructed with a computer the abstract group 𝐺 (for each of the cases arising in parts 1–4), we have determined all the imprimitive innately transitive faithful actions of these groups and we have verified that, in each action, 𝐺 admits a semiregular subgroup of order at least 4. ∎

Lemma 3.4

Let 𝐺 be a primitive group of degree 𝑛 of type HA, HS, HC, TW, SD or CD. Then 𝐺 has a semiregular subgroup of order 𝑛. In particular, Theorem 1.3 holds true in these cases.

Proof

Let 𝑁 be the socle of 𝐺 and let Ω be the domain of 𝐺.

When 𝐺 is of type HA or TW, 𝑁 acts regularly on Ω. In particular, 𝑁 is a clique in Γ G of cardinality | N | = | Ω | = n .

When 𝐺 is of type HS or HC, 𝑁 is the direct product of two minimal normal subgroups of 𝐺, say M 1 and M 2 . From the description of the primitive groups of type HS or HC, we see that M 1 and M 2 act regularly on Ω and they form a clique in Γ G of cardinality | M i | = | Ω | = n .

Suppose 𝐺 is of type SD. Then N = T 1 × ⋯ × T ℓ + 1 , where T 1 , … , T ℓ + 1 are pairwise isomorphic non-abelian simple groups. From the description of the primitive groups of type SD, we see that | Ω | = | T 1 | ℓ and that T 1 × ⋯ × T ℓ acts regularly on Ω. Therefore, as above, T 1 × ⋯ × T ℓ forms a clique in Γ G of cardinality | Ω | = n .

Suppose that 𝐺 is of type CD. Then Ω admits a non-trivial Cartesian decomposition, that is, Ω = Δ κ for some finite set Δ and for some positive integer κ ≥ 2 , and we have an embedding G ≤ H wr Sym ⁡ ( κ ) , where the wreath product H wr Sym ⁡ ( κ ) acts on Δ κ primitively, H ≤ Sym ⁡ ( Δ ) and 𝐻 is of type SD in its action on Δ. Now, if the socle of 𝐻 is isomorphic to T ℓ + 1 for some non-abelian simple group 𝑇 and for some positive integer ℓ ≥ 1 , then the socle of 𝐺 is isomorphic to T κ ⁢ ( ℓ + 1 ) . In particular, the socle of 𝐺 contains a subgroup isomorphic to T κ ⁢ ℓ acting regularly on Ω, and we may argue as above. ∎

Proposition 3.5

Suppose that Theorem 1.3 holds true for primitive simple groups. Then Theorem 1.3 holds true.

Proof

In view of Proposition 3.3, we may suppose that 𝐺 is primitive. Moreover, in view of Lemma 3.4, we may suppose that 𝐺 is of AS or PA type.

Let Ω be the domain of 𝐺. Then Ω admits a Cartesian decomposition Δ κ , for some κ ≥ 1 ,^[14] and 𝐺 embeds into the wreath product H wr Sym ⁡ ( κ ) endowed with the primitive product action. Replacing Sym ⁡ ( κ ) by a suitable transitive subgroup 𝐴, we may suppose that 𝐺 embeds into the wreath product H wr A and 𝐺 projects surjectively to 𝐴. Moreover, 𝐻 is of type AS. Let 𝑇 be the socle of 𝐻. Then the socle of 𝐺 is T κ . When T = M 11 and | Δ | = 12 , as we have mentioned in the introduction, Giudici [11] has shown that 𝐺 has no non-identity semiregular element, and hence, for the rest of the argument, we may suppose that 𝑇 is not M 11 in its degree 12 action.

Observe that 𝑇 acts transitively on Δ because 𝐻 is primitive on Δ, but 𝑇 is not necessarily primitive on Δ. Let Σ be a non-trivial system of imprimitivity for the action of 𝑇 on Δ; by choosing the blocks of Σ as large as possible, we may assume that 𝑇 acts primitively on Σ.

We now prove the first part of the statement of Theorem 1.3. Let g : N → N be a function witnessing that Theorem 1.3 holds for primitive simple groups. Without loss of generality, we may suppose that g ⁢ ( 1 ) = 1 . Define f : N → N by

Observe that 𝑓 is well-defined because, when ℓ ≥ k , we have ⌊ k 1 / ℓ ⌋ = 1 and hence g ⁢ ( ⌊ k 1 / ℓ ⌋ ) ! ℓ = 1 . In particular,

Let k ∈ N . By hypothesis, either 𝑇 in its action on Σ has a semiregular subgroup 𝑋 of order at least k 1 / κ , or | Σ | ≤ g ⁢ ( ⌊ k 1 / κ ⌋ ) . Observe that 𝑋 is also semiregular for the action of 𝑇 on Δ. Therefore, in the first case, X κ is a semiregular subgroup of 𝐺 of order at least ( k 1 / κ ) κ = k . Assume then | Σ | ≤ g ⁢ ( ⌊ k 1 / κ ⌋ ) . Thus

We now prove the second part of the statement of Theorem 1.3. Therefore, we suppose that 𝐺 has no semiregular subgroup of order at least 4. We use the notation established above. Assume first that κ = 1 , i.e., 𝐺 is almost simple. Since we are assuming that Theorem 1.3 holds for simple primitive groups, we deduce that 𝑇 is isomorphic to Alt ⁢ ( 5 ) , Alt ⁢ ( 6 ) , M 11 or PSU 3 ⁢ ( 3 ) . We have constructed with a computer the abstract group 𝐺 having socle 𝑇, we have determined all the primitive actions of these groups and we have established the veracity of Theorem 1.3. Assume next that κ ≥ 2 . If 𝑇 has a semiregular subgroup of order at least 2, then T κ ≤ G has a semiregular subgroup of order at least 4. From an inspection of the cases arising in parts 2–5, we see that the only group not having a semiregular subgroup of order at least 2 is T = M 11 in its primitive action of degree 12, which we have already dealt with above. ∎

Theorem 4.1

Let ℓ be a positive integer. Then the product of ℓ consecutive integers greater than ℓ is divisible by a prime 𝑝 greater than ℓ.

Lemma 4.2

Let 𝑚 be a positive integer with m ≥ 5 and let ℓ ∈ { 1 , … , m − 1 } . Suppose that, for every prime p ≥ 5 , ℓ p ≤ m p . Then either

1. ℓ ∈ { 1 , m − 1 } and m = 2 a ⋅ 3 b for some a , b ∈ N , or

2. m = 9 and ℓ ∈ { 2 , 7 } .

Proof

Suppose that 𝑚 and ℓ satisfy the property

Now, consider ℓ ′ = m − ℓ and let p ≥ 5 be a prime number. By hypothesis, we have ℓ p ≤ m p , and hence m p − ℓ p is the remainder of ℓ ′ = m − ℓ in the division by 𝑝, that is, ℓ p ′ = ( m − ℓ ) p = m p − ℓ p ≤ m p . This shows that, if the pair ( m , ℓ ) satisfies ( † ), then so does ( m , ℓ ′ ) = ( m , m − ℓ ) . Therefore, without loss of generality, replacing ℓ by m − ℓ if necessary, we may suppose that ℓ ≤ m / 2 .

Now, consider the ℓ consecutive numbers

As m ≥ 2 ⁢ ℓ , these numbers are greater than ℓ, and hence, by Sylvester’s theorem, there exists a prime

dividing m − i for some i ∈ { 0 , … , ℓ − 1 } . As i ≤ ℓ − 1 and p ∣ m − i , we have m p ≤ i ≤ ℓ − 1 . However, as p > ℓ , we have ℓ p = ℓ , and hence m p < ℓ p . Since ( m , ℓ ) satisfies ( † ), we have p < 5 , that is, p ∈ { 2 , 3 } .

Assume ℓ = 1 . If 𝑚 is divisible by a prime p ≥ 5 , then ℓ p = 1 > m p = 0 , and hence ( m , ℓ ) does not satisfy ( † ). Therefore, m = 2 a ⋅ 3 b for some a , b ∈ N , and we obtain part 1.

Assume ℓ ≥ 2 . From (4.1), we have p > ℓ ≥ 2 . Since p ∈ { 2 , 3 } , we deduce p = 3 and, more importantly, ℓ = 2 . For each prime divisor r ≥ 5 of m − 1 or 𝑚, we have m r ≤ 1 and hence ℓ r ≤ m r ≤ 1 because ( m , ℓ ) satisfies ( † ). As ℓ = 2 , the condition ℓ r ≤ 1 can be satisfied if and only if m − 1 and 𝑚 are only divisible by the primes 2 and 3. Thus

for some a , a ′ , b , b ′ ∈ N . Since

we obtain

• m = 2 a and m − 1 = 3 b ′ , or

• m = 3 b and m − 1 = 2 a ′ .

We deal with each of these two cases in turn. Assume m = 2 a and m − 1 = 3 b ′ . As m ≥ 5 , 3 divides m − 1 = 2 a − 1 , and hence 𝑎 is even. Thus a = 2 ⁢ α for some integer 𝛼. This gives 2 a − 1 = 4 α − 1 = 3 b ′ . In particular, since 3 divides 4 1 − 1 , 4 α − 1 has no primitive prime divisors. Using the theorem of Zsigmondy [38], we deduce that this case is impossible unless m = 4 . However, this contradicts our hypothesis m ≥ 5 . Assume m = 3 b and m − 1 = 2 a ′ . Thus 3 b − 1 = 2 a ′ . In particular, since 2 divides 3 1 − 1 , 3 b − 1 has no primitive prime divisors. Using the theorem of Zsigmondy [38], we deduce that b ∈ { 1 , 2 } . When b = 1 , m = 3 and we contradict our hypothesis m ≥ 5 . When b = 2 , m = 9 and we obtain the exceptional case in 2. ∎

Lemma 4.3

Let 𝑚 be a positive integer. If m ≥ 8 , then there exists a prime 𝑝 with m / 2 < p ≤ m − 3 .

Proof

Bertrand’s postulate [16, page 498] says that, when n ≥ 4 , there is a prime 𝑝 satisfying n < p < 2 ⁢ n − 2 .

In particular, when 𝑚 is even, the proof follows by applying Bertrand’s postulate with n = m / 2 . Whereas, when 𝑚 is odd, the proof follows by applying Bertrand’s postulate with n = ( m − 1 ) / 2 . ∎

Lemma 4.4

Let 𝑚 be a positive integer. Then ( m / 2 ) m ≥ m ! / 2 .

Proof

This follows from an inductive argument on 𝑚. ∎

Theorem 4.5

There exists a positive constant 𝜅 such that, if 𝑎, 𝑏 and 𝑐 are coprime positive integers with c = a + b , then c ≤ exp ⁡ ( κ ⋅ rad ⁢ ( a ⁢ b ⁢ c ) 15 ) .

Lemma 4.6

Given n ∈ N with n ≥ 3 and q ∈ N with q ≥ 2 , there exist two positive constants 𝑐 and c ′ depending on 𝑛 only such that q n − 1 admits a primitive prime divisor p ≥ c ⁢ log ⁡ log ⁡ q for every q ≥ c ′ .

Proof

This follows from a remarkable result of Siegel [33, Satz 7]. Let f ∈ Z ⁢ [ x ] be a polynomial with integer coefficients and at least 2 distinct roots. Then there exist two positive constants c f and c f ′ depending on 𝑓 only such that

As n ≥ 3 , it follows that Φ n ⁢ ( x ) has φ ⁢ ( n ) ≥ 2 distinct roots, and hence we may apply Siegel’s theorem with f ⁢ ( x ) = Φ n ⁢ ( x ) . In particular, there exist two positive constants 𝑐 and c ′ depending on 𝑛 only such that P ⁢ [ Φ n ⁢ ( q ) ] ≥ c ⁢ log ⁡ log ⁡ q for every integer 𝑞 with q ≥ c ′ . Replacing c ′ by a larger constant, we may also suppose that P ⁢ [ Φ n ⁢ ( q ) ] ≥ n + 1 for every integer q ≥ c ′ . Let q ∈ N with q ≥ c ′ and let p = P ⁢ [ Φ n ⁢ ( q ) ] .

Following [12, Definition 1], we let Φ n ∗ ⁢ ( q ) denote the largest divisor of Φ n ⁢ ( q ) relatively prime to

Let 𝑟 be the largest prime divisor of 𝑛. From [12, Lemma 3.1], we have

Since p > n ≥ r , we deduce that 𝑝 divides Φ n ∗ ⁢ ( q ) , and hence, by definition, 𝑝 is a primitive prime divisor of q n − 1 . ∎