Existence of regular 3-polytopes of order 2𝑛

Dong-Dong Hou; Yan-Quan Feng; Dimitri Leemans

doi:10.1515/jgth-2018-0155

Article Publicly Available

Existence of regular 3-polytopes of order 2^𝑛

Dong-Dong Hou , Yan-Quan Feng and Dimitri Leemans

Published/Copyright: January 30, 2019

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From the journal Journal of Group Theory Volume 22 Issue 4

Abstract

In this paper, we prove that for any positive integers n,s,t such that n≥10, s,t≥2 and n-1≥s+t, there exists a regular polytope with Schläfli type {2s,2t} and its automorphism group is of order 2n. Furthermore, we classify regular polytopes with automorphism groups of order 2n and Schläfli types {4,2n-3},{4,2n-4} and {4,2n-5}, therefore giving a partial answer to a problem proposed by Schulte and Weiss in [Problems on polytopes, their groups, and realizations, Period. Math. Hungar. 53 2006, 1–2, 231–255].

1 Introduction

Classifications of abstract regular polytopes have been a subject of interest for several decades. One path has been to fix (families of) groups of automorphisms and determine the abstract regular polytopes having these groups as full automorphism groups. Some striking results have been obtained, for instance for the symmetric and alternating groups. Fernandes and Leemans classified abstract regular polytopes of rank n-1 and n-2 for Sn (see [9, 12]) and more recently, they extended this classification to rank n-3 and n-4 with Mixer [13]. Cameron, Fernandes, Leemans and Mixer showed that the highest rank of an abstract regular polytope with full automorphism group an alternating group An is ⌊(n-1)/2⌋ when n≥12 (see [3]), and thanks to two previous papers of Fernandes, Leemans and Mixer [11, 10], this bound is known to be sharp. More recently, Gomi, Loyola and De Las Peñas determined the non-degenerate string C-groups of order 1024 in [14].

There exists a well known one-to-one correspondence between abstract regular polytopes and string C-groups. We therefore work with string C-groups as it is more convenient and easier to define them than abstract regular polytopes. In this paper, we study 2-groups acting on regular polytopes. The starting point of our research was the following problem proposed by Schulte and Weiss in [22].

Problem 1.1.

Characterize the groups of orders 2n or 2n⁢p, with n a positive integer and p an odd prime, which are automorphism groups of regular or chiral polytopes.

Conder [5] showed that if 𝒫 is a regular 3-polytope with Schläfli type {k1,k2}, then |Aut⁡(𝒫)|≥2⁢k1⁢k2. If 𝒫 has Schläfli type {2s,2t} and |Aut⁡(𝒫)|=2n, then n-1≥s+t. In this paper, we first show the following theorem.

Theorem 1.2.

For any positive integers n, s and t such that n≥10, s,t≥2 and n-1≥s+t, there exists a string C-group of order 2n with Schläfli type {2s,2t}.

Cunningham and Pellicer [8] classified the regular 3-polytopes 𝒫 for the case when |Aut⁡(𝒫)|=2⁢k1⁢k2. Note that if |Aut⁡(𝒫)|=2n and k1=4 then k2≤2n-3. As a special case, Cunningham and Pellicer [8] obtained the classification of regular 3-polytopes with automorphism groups of order 2n and Schläfli type {4,2n-3}, and this was also given in Loyola [17] by using the classification of 2-groups with a cyclic subgroup of order 2n-3 [19]. We prove the result again independently by using new techniques that are described in this paper, and the techniques work well for more classifications. In particular, we further classify the regular 3-polytopes with automorphism groups of order 2n and Schläfli types {4,2n-4} and {4,2n-5} in this paper.

To state the next result, we need to define some groups:

G1=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-3,(ρ0⁢ρ2)2,[(ρ0ρ1)2,ρ2]〉,

G2=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-3,(ρ0⁢ρ2)2,[(ρ0ρ1)2,ρ2](ρ1ρ2)2n-4〉,

G3=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-4,(ρ0⁢ρ2)2,[(ρ0ρ1)2,(ρ1ρ2)2]〉,

G4=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-4,(ρ0⁢ρ2)2,[(ρ0ρ1)2,(ρ1ρ2)2](ρ1ρ2)2n-5〉,

G5=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-5,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)2]2,[ρ0,(ρ1ρ2)4]〉,

G6=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-5,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)2]2(ρ1ρ2)2n-6,[ρ0,(ρ1ρ2)4]〉,

G7=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-5,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)2]2,[ρ0,(ρ1ρ2)4](ρ1ρ2)2n-6〉,

G8=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-5,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)2]2(ρ1ρ2)2n-6,[ρ0,(ρ1ρ2)4](ρ1ρ2)2n-6〉.

Theorem 1.3.

For n≥10, let Γ:-(G,{ρ0,ρ1,ρ2}) be a string C-group of order 2n. Then

Γ has type {4,2n-3} if and only if G≅G1 or G2;
Γ has type {4,2n-4} if and only if G≅G3 or G4;
Γ has type {4,2n-5} if and only if G≅G5,G6,G7 or G8.

Let n<10. By [4] or [15], there is a unique string C-group of order 2n with type {4,4}, and Theorem 1.3 is true for the types {4,2n-s} with n-s≥3 and s=3,4 or 5, except for the cases when n=8 or 9 with s=5. For n=8 with s=5, there are four string C-groups with type {4,8}: two are G5 and G6, and the other two are

〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)28,(ρ0⁢ρ2)2,[(ρ0ρ1)2,(ρ1ρ2)2](ρ1ρ2)4〉,

〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)23,(ρ0⁢ρ2)2,[((ρ1ρ2)2)ρ0,ρ1ρ2](ρ1ρ2)4〉.

For n=9 with s=5, there are six string C-groups with type {4,16}: four are Gi with 5≤i≤8, and the other two are

〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)24,(ρ0⁢ρ2)2,[(ρ0ρ1)2,(ρ1ρ2)2](ρ1ρ2)4〉,

〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)24,(ρ0⁢ρ2)2,[(ρ0ρ1)2,(ρ1ρ2)2](ρ2ρ1)4,[ρ1,ρ0,ρ2,ρ1,ρ0,ρ1,ρ0]〉.

2 Background results

2.1 String C-groups

Abstract regular polytopes and string C-groups are the same mathematical objects. The link between these objects may be found for instance in [20, Chapter 2]. We take here the viewpoint of string C-groups because it is the easiest and the most efficient one to define abstract regular polytopes.

Let G be a group, and let S={ρ0,…,ρd-1} be a generating set of involutions of G. For I⊆{0,…,d-1}, let GI denote the group generated by {ρi:i∈I}. Suppose that

for any i,j∈{0,…,d-1} with |i-j|>1, ρi and ρj commute (the string property);
for any I,J⊆{0,…,d-1}, GI∩GJ=GI∩J (the intersection property).

Then the pair (G,S) is called a string C-group of rankd and the order of (G,S) is simply the order of G. If (G,S) only satisfies the string property, it is called a string group generated by involutions or sggi. By the intersection property, S is a minimal generating set of G. It is known that string C-groups are the same thing as automorphism groups of regular polytopes [20, Section 2E]. The following proposition is straightforward, and for details, one may see [6].

Proposition 2.1.

The intersection property for a string C-group (G,S) of rank 3 is equivalent to the condition that S is a minimal generating set of G and

〈ρ0,ρ1〉∩〈ρ1,ρ2〉=〈ρ1〉.

The i-faces of the regular d-polytope associated with (G,S) are the right cosets of the distinguished subgroup Gi=〈ρj∣j≠i〉 for each i=0,1,…,d-1, and two faces are incident just when they intersect as cosets. The (Schläfli) type of (G,S) is the ordered set {p1,…,pd-1}, where pi is the order of ρi-1⁢ρi. In this paper, we always assume that each pi is at least 3, for otherwise, the generated group is a direct product of two smaller groups. If that happens, the string C-group (and the corresponding abstract regular polytope) is called degenerate. The following proposition is related to degenerate string C-groups of rank 3.

Proposition 2.2.

For t≥1, let

L1=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)4,(ρ1⁢ρ2)2,(ρ0⁢ρ2)2〉,

L2=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)2,(ρ1⁢ρ2)2t,(ρ0⁢ρ2)2〉,

L3=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)2t,(ρ1⁢ρ2)2,(ρ0⁢ρ2)2〉.

Then |L1|=16, |L2|=|L3|=2t+2. In particular, the listed exponents are the true orders of the corresponding elements.

The proof of Proposition 2.2 is straightforward from the fact that

L2=〈ρ0〉×〈ρ1,ρ2〉≅ℤ2×D2t+1 and L3=〈ρ0,ρ1〉×〈ρ2〉≅D2t+1×ℤ2,

where D2t+1 denotes the dihedral group of order 2t+1.

The following proposition is called the quotient criterion for a string C-group.

Proposition 2.3 ([20, Section 2E]).

Let (G,{ρ0,ρ1,ρ2}) be an sggi, and let

Λ=(〈σ0,σ1,σ2〉,{σ0,σ1,σ2})

be a string C-group. If the mapping ρj↦σj for j=0,1,2 induces a homomorphism π:G→Λ, which is one-to-one on the subgroup 〈ρ0,ρ1〉 or on 〈ρ1,ρ2〉, then (G,{ρ0,ρ1,ρ2}) is also a string C-group.

The following proposition gives some string C-groups with type {4,4}, which is proved in [7, Section 8.3] for b≥2, but it is also true for b=1 by Magma [2].

Proposition 2.4.

For b≥1, let

M1=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)4,(ρ1⁢ρ2)4,(ρ0⁢ρ2)2,(ρ2⁢ρ1⁢ρ0)2⁢b〉,

M2=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)4,(ρ1⁢ρ2)4,(ρ0⁢ρ2)2,(ρ1⁢ρ2⁢ρ1⁢ρ0)b〉.

Then |M1|=16⁢b2 and |M2|=8⁢b2. In particular, the listed exponents are the true orders of the corresponding elements.

2.2 Permutation representation graphs and CPR graphs

In [21], Daniel Pellicer introduced CPR-graphs to give a permutation representation of string C-groups (CPR stands for C-group Permutation Representation). These graphs are also sometimes called permutation representation graphs.

Let G be a group and S:-{ρ0,…,ρd-1} a generating set of involutions of G. Let ϕ be an embedding of G into the symmetric group Sn for some n. The permutation representation graph𝒢 of G determined by ϕ is the multigraph with n vertices, and with edge labels in the set {0,…,d-1}, such that any two vertices v,w are joined by an edge of label j if and only if (v)⁢((ρj)⁢ϕ)=w.

If (G,S) is a string C-group, then the permutation representation graph defined above is called a CPR-graph by Pellicer.

2.3 Group theory

Let G be a group. For x,y∈G, we use [x,y] as an abbreviation for the commutatorx-1⁢y-1⁢x⁢y of x and y, and [H,K] for the subgroup generated by all commutators [x,y] with x∈H and y∈K, when H and K are subgroups of G. The following proposition is a basic property of commutators, and its proof is straightforward.

Proposition 2.5.

Let G be a group. Then, for any x,y,z∈G,

[x⁢y,z]=[x,z]y⁢[y,z] 𝑎𝑛𝑑 [x,y⁢z]=[x,z]⁢[x,y]z.

The commutator (or derived) subgroup G′ of a group G is the subgroup generated by all commutators [x,y] for any x,y∈G. With Proposition 2.5, it is easy to prove that if G is generated by a subset M, then G′ is generated by all conjugates in G of elements [xi,xj] with xi,xj∈M; see [16, Hilfsatz III.1.11] for example.

Proposition 2.6.

Let G be a group, M⊆G and G=〈M〉. Then

G′=〈[xi,xj]g∣xi,xj∈M,g∈G〉.

The Frattini subgroup, denoted by Φ⁢(G), of a finite group G is defined to be the intersection of all maximal subgroups of G. Let G be a finite p-group for a prime p, and set ℧1⁢(G)=〈gp∣g∈G〉. The following theorem is the well-known Burnside Basis Theorem.

Theorem 2.7 ([1, Theorem 1.12]).

Let G be a p-group and |G:Φ(G)|=pd.

G/Φ⁢(G)≅ℤpd. Moreover, if N⊲G and G/N is elementary abelian, then Φ⁢(G)≤N.
Every minimal generating set of G contains exactly d elements.
Φ⁢(G)=G′⁢℧1⁢(G). In particular, if p=2, then Φ⁢(G)=℧1⁢(G).

By Theorem 2.7 (2), we have the following important result.

Remark 2.8.

A string 2-group has C-group representations in only one rank.

The unique cardinality of every minimal generating set of a 2-group G is called the rank of G and is denoted by d⁢(G). This is quite different from almost simple groups, where, in most cases, if a group has string C-group representations of maximal rank d, then it has string C-group representations of ranks from 3 to d. The only known exception is the alternating group A11 (see [11]).

For a subgroup H of a group G, the coreCoreG⁡(H) of H in G is the largest normal subgroup of G contained in H. The following result is called Lucchini’s theorem.

Proposition 2.9 ([18, Theorem 2.20]).

Let A be a cyclic proper subgroup of a finite group G, and let K=CoreG⁡(A). Then |A:K|<|G:A|, and in particular, if |A|≥|G:A|, then K>1.

3 Proof of Theorem 1.2

Let n≥10, s,t≥2 and n-s-t≥1. Set

R(ρ0,ρ1,ρ2)={ρ02,ρ12,ρ22,(ρ0⁢ρ1)2s,(ρ1⁢ρ2)2t,(ρ0⁢ρ2)2,[(ρ0ρ1)4,ρ2],[ρ0,(ρ1ρ2)4]},

and define

H={〈ρ0,ρ1,ρ2∣R⁢(ρ0,ρ1,ρ2),[(ρ0ρ1)2,ρ2]2n-s-t-12〉,n-s-t⁢odd,〈ρ0,ρ1,ρ2∣R⁢(ρ0,ρ1,ρ2),[(ρ0ρ1)2,(ρ1ρ2)2]2n-s-t-22〉,n-s-t⁢even.

To prove Theorem 1.2, we only need to show that H is a string C-group of order 2n with Schläfli type {2s,2t}. For convenience, write o⁢(h) for the order of h in H.

Note that ρ0 commutes with (ρ1⁢ρ2)4 because [ρ0,(ρ1⁢ρ2)4]=1. Since 〈ρ1,ρ2〉 is a dihedral group, we have (ρ1⁢ρ2)ρ1=(ρ1⁢ρ2)ρ2=(ρ1⁢ρ2)-1. It follows that 〈(ρ1⁢ρ2)4〉⊴H. Similarly, 〈(ρ0⁢ρ1)4〉⊴H as [(ρ0⁢ρ1)4,ρ2]=1.

Let L2=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)2,(ρ1⁢ρ2)2t,(ρ0⁢ρ2)2〉. Clearly, ρ0 commutes with both ρ1 and ρ2 in L2, and hence [ρ0,(ρ1⁢ρ2)4]=1. It is easy to see that the generators ρ0,ρ1,ρ2 in L2 satisfy all relations in H. This implies that L2 is a homomorphic image of H. By Proposition 2.2, ρ1⁢ρ2 has order 2t in L2, and hence has order 2t in H. It follows that

|H|=o⁢((ρ1⁢ρ2)4)⋅|H/〈(ρ1⁢ρ2)4〉|=2t-2⋅|H/〈(ρ1⁢ρ2)4〉|.

Let L3=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)2s,(ρ1⁢ρ2)2,(ρ0⁢ρ2)2〉. The element ρ2 commutes with ρ0 and ρ1 in L3, and hence [(ρ0⁢ρ1)4,ρ2]=1. Since ρ0⁢ρ2=ρ2⁢ρ0, Proposition 2.5 implies

[(ρ0⁢ρ1)2,ρ2]=[ρ1⁢ρ0⁢ρ1,ρ2]=[ρ0,ρ1⁢ρ2⁢ρ1]ρ1=[ρ0,(ρ1⁢ρ2)2]ρ2⁢ρ1.

Hence [(ρ0⁢ρ1)2,ρ2]=1 in L3. Therefore, the generators ρ0,ρ1,ρ2 in L3 satisfy all relations in H. By Proposition 2.2, ρ0⁢ρ1 has order 2s in L3, and hence has order 2s in H. It follows that |H|=2s-2⋅|H/〈(ρ0⁢ρ1)4〉|.

To finish the proof of Theorem 1.2, we are left to prove that |H|=2n.

Case 1: s=2. We distinguish two cases, namely the case where n-t is odd and the case where n-t is even.

Assume that n-t is odd. Then

H=〈ρ0,ρ1,ρ2∣R⁢(ρ0,ρ1,ρ2),[(ρ0⁢ρ1)2,ρ2]2n-t-32〉.

Since ρ0⁢ρ2=ρ2⁢ρ0, we have [(ρ0⁢ρ1)2,ρ2]=(ρ1⁢ρ0⁢ρ1⁢ρ2)2. It follows that

[(ρ0⁢ρ1)2,ρ2]2n-t-32=(ρ1⁢ρ0⁢ρ1⁢ρ2)2n-t-12.

Note that (ρ0⁢ρ1)4=1 and 〈(ρ1⁢ρ2)4〉⊴H. Thus H/〈(ρ1⁢ρ2)4〉≅H1, where

H1=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)22,(ρ0⁢ρ2)2,(ρ1⁢ρ0⁢ρ1⁢ρ2)2n-t-12〉.

By Proposition 2.4, |H1|=8⋅(2n-t-12)2=2n-t+2, and hence

|H|=2t-2⋅|H/〈(ρ1⁢ρ2)4〉|=2t-2⁢|H1|=2n.

Assume that n-t is even. Then

H=〈ρ0,ρ1,ρ2∣R⁢(ρ0,ρ1,ρ2),[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]2n-t-42〉.

A similar argument as above gives rise to

H/〈(ρ1⁢ρ2)4〉≅H2=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)22,(ρ0⁢ρ2)2,[(ρ0ρ1)2,(ρ1ρ2)2]2n-t-42〉.

Noting (ρ0⁢ρ1)2=(ρ0⁢ρ1)-2 and (ρ1⁢ρ2)2=(ρ1⁢ρ2)-2 in H2, we have

[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]2n-t-42=(((ρ0⁢ρ1)2⁢(ρ1⁢ρ2)2)2)2n-t-42=(((ρ0⁢ρ1⁢ρ2)2)2)2n-t-42=(ρ0⁢ρ1⁢ρ2)2⋅2n-t-22

because ρ0⁢ρ2=ρ2⁢ρ0, and so

H2=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)22,(ρ0⁢ρ2)2,(ρ0⁢ρ1⁢ρ2)2⋅2n-t-42〉.

By Proposition 2.4,

|H2|=16⋅(2n-t-22)2=2n-t+2,

|H|=2t-2⋅|H/〈(ρ1⁢ρ2)4〉|=2t-2⋅|H2|=2n.

Case 2: s>2. Assume that n-t-s is odd. Then

H=〈ρ0,ρ1,ρ2∣R⁢(ρ0,ρ1,ρ2),[(ρ0⁢ρ1)2,ρ2]2n-t-s-12〉.

It follows that H/〈(ρ0⁢ρ1)4〉≅H3, where

H3=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2t,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)4],[(ρ0ρ1)2,ρ2]2(n-s+2)-t-32〉.

By Case 1, |H3|=2n-s+2, and therefore

|H|=2s-2⋅|H/〈(ρ0⁢ρ1)4〉|=2s-2⋅|H3|=2n.

Assume that n-t-s is even. Then H/〈(ρ0⁢ρ1)4〉≅H4, where

H4=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2t,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)4],[(ρ0ρ1)2,(ρ1ρ2)2]2(n-s+2)-t-42〉.

Then |H4|=2n-s+2 from Case 1, and therefore |H|=2s-2⋅|H4|=2n. ∎

Corollary 3.1.

The pairs (G1,{ρ0,ρ1,ρ2}), (G3,{ρ0,ρ1,ρ2}), (G5,{ρ0,ρ1,ρ2}), which are defined in Theorem 1.3, are string C-groups of order 2n with Schläfli type {4,2n-3}, {4,2n-4} and {4,2n-5}, respectively.

Proof.

By taking (s,t)=(2,n-3),(2,n-4),(2,n-5) in the proof of Theorem 1.2, we know that (Hi,{ρ0,ρ1,ρ2}) for i=1,3,5 are string C-groups of order 2n with Schläfli type {4,2n-3}, {4,2n-4} and {4,2n-5} respectively, where

H1=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-3,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)4],[(ρ0ρ1)2,ρ2]〉,

H3=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-4,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)4],[(ρ0ρ1)2,(ρ1ρ2)2]〉,

H5=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-5,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)4],[ρ0,(ρ1ρ2)2]2〉.

Since ρ0⁢ρ2=ρ2⁢ρ0 and ((ρ0⁢ρ1)2)ρ0=((ρ0⁢ρ1)2)ρ1=(ρ0⁢ρ1)2, by Proposition 2.5, we have the following identities in all Hi and Gi for i=1,3,5:

[(ρ0⁢ρ1)2,ρ2]=[ρ1⁢ρ0⁢ρ1,ρ2]=[ρ0,ρ1⁢ρ2⁢ρ1]ρ1=[ρ0,(ρ1⁢ρ2)2]ρ2⁢ρ1,

[ρ0,(ρ1⁢ρ2)4]=[ρ0,(ρ1⁢ρ2)2]⁢[ρ0,(ρ1⁢ρ2)2](ρ1⁢ρ2)2=[(ρ0⁢ρ1)2,ρ2]ρ1⁢ρ2⁢[(ρ0⁢ρ1)2,ρ2](ρ1⁢ρ2)3,

[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]=[(ρ0⁢ρ1)2,ρ2]⁢[(ρ0⁢ρ1)2,ρ1⁢ρ2⁢ρ1]ρ2=[(ρ0⁢ρ1)2,ρ2]⁢[(ρ0⁢ρ1)2,ρ2]ρ1⁢ρ2.

Clearly, H5=G5. In G1,

[ρ0,(ρ1⁢ρ2)4]=[(ρ0⁢ρ1)2,ρ2]ρ1⁢ρ2⁢[(ρ0⁢ρ1)2,ρ2](ρ1⁢ρ2)3=1

because [(ρ0⁢ρ1)2,ρ2]=1. Thus H1=G1. To prove H3=G3, we only need to show that [ρ0,(ρ1⁢ρ2)4]=1 in G3. Noting that [(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]=1 in G3, we have

[(ρ0⁢ρ1)2,ρ2]ρ1⁢[(ρ0⁢ρ1)2,ρ2]=[ρ0,(ρ1⁢ρ2)2]ρ2⁢[(ρ0⁢ρ1)2,ρ2]=[ρ0,(ρ2⁢ρ1)2]⁢[(ρ0⁢ρ1)2,ρ2]=ρ0⁢(ρ1⁢ρ2)2⁢ρ0⁢(ρ2⁢ρ1)2⁢{(ρ0⁢ρ1)2}⁢ρ2⁢(ρ0⁢ρ1)2⁢ρ2=ρ0⁢{(ρ0⁢ρ1)2}⁢(ρ1⁢ρ2)2⁢ρ0⁢(ρ2⁢ρ1)2⁢ρ2⁢(ρ0⁢ρ1)2⁢ρ2=ρ1⁢ρ0⁢ρ1⁢(ρ1⁢ρ2)2⁢ρ0⁢{ρ2⁢(ρ1⁢ρ2)2}⁢(ρ0⁢ρ1)2⁢ρ2=ρ1⁢ρ0⁢ρ1⁢(ρ1⁢ρ2)2⁢ρ0⁢ρ2⁢(ρ0⁢ρ1)2⁢(ρ1⁢ρ2)2⁢ρ2=ρ1⁢ρ0⁢ρ1⁢ρ1⁢ρ2⁢ρ1⁢ρ2⁢ρ0⁢ρ2⁢ρ0⁢ρ1⁢ρ0⁢ρ1⁢ρ1⁢ρ2⁢ρ1⁢ρ2⁢ρ2=1,

that is, [(ρ0⁢ρ1)2,ρ2]ρ1=[(ρ0⁢ρ1)2,ρ2]-1. On the other hand, since

1=[(ρ0⁢ρ1)2,ρ2⁢ρ2]=[(ρ0⁢ρ1)2,ρ2]⁢[(ρ0⁢ρ1)2,ρ2]ρ2,

1=[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]=[(ρ0⁢ρ1)2,ρ2]⁢[(ρ0⁢ρ1)2,ρ2]ρ1⁢ρ2,

we have [(ρ0⁢ρ1)2,ρ2]ρ1=([(ρ0⁢ρ1)2,ρ2]ρ2)-1=[(ρ0⁢ρ1)2,ρ2]. It follows that [(ρ0⁢ρ1)2,ρ2]2=1 and [(ρ0⁢ρ1)2,ρ2]ρ1⁢ρ2=[(ρ0⁢ρ1)2,ρ2]-1, and so

[ρ0,(ρ1⁢ρ2)4]=[(ρ0⁢ρ1)2,ρ2]ρ1⁢ρ2⁢[(ρ0⁢ρ1)2,ρ2](ρ1⁢ρ2)3=[(ρ0⁢ρ1)2,ρ2]-2=1,

as required. ∎

4 Proof of Theorem 1.3

To prove Theorem 1.3, we need the following lemmas.

Lemma 4.1.

Let (G,{ρ0,ρ1,ρ2}) be a string C-group of type {2s,2t}, 2≤s≤t. Let |G|=2n and 2⁢t≥n-1. Then N=〈(ρ1⁢ρ2)2t-1〉⊴G, and (G¯,{ρ0¯,ρ1¯,ρ2¯}) is a string C-group of type {2s,2t-1} and order 2n-1, where G¯=G/N and x¯=x⁢N for any x∈G.

Proof.

Let H=〈ρ0⁢ρ1,ρ1⁢ρ2〉 be the rotation subgroup of G. Then |G:H|≤2. Since {ρ0,ρ1,ρ2} is a minimal generating set of G, Theorem 2.7 (2) implies d⁢(G)=3, and since H is generated by two elements, we have |H|=2n-1.

Let M=〈ρ1⁢ρ2〉. Then |M|=2t and |M|2≥|H| as 2⁢t≥n-1. Then Proposition 2.9 implies CoreH⁡(M)>1. Since M is cyclic and |N|=2, N is characteristic in CoreH⁡(M), and so CoreH⁡(M)⊴H implies N⊴H. Noting that N lies in the center of the dihedral group 〈ρ1,ρ2〉, we have Nρ1=N, and hence N⊴G because G=〈H,ρ1〉. Clearly, G¯=2n-1.

Since t≥2, we have

N≤℧1⁢(G)=〈g2∣g∈G〉,

and by Theorem 2.7, N≤Φ⁢(G) and G/Φ⁢(G)≅ℤ23. Thus G/Φ⁢(G) has rank 3, and since G/Φ⁢(G)≅(G/N)/(Φ⁢(G)/N), G/N has rank 3, which implies that {ρ0¯,ρ1¯,ρ2¯} is a minimal generating set of G¯. It follows that ρ0¯, ρ1¯, ρ2¯ and ρ0⁢ρ2¯ are involutions. To prove that G¯ has the intersection property, by Proposition 2.1, we only need to show that 〈ρ0¯,ρ1¯〉∩〈ρ1¯,ρ2¯〉=〈ρ1¯〉.

Suppose

〈ρ0¯,ρ1¯〉∩〈ρ1¯,ρ2¯〉≠〈ρ1¯〉.

Then there exist x1∈〈ρ0,ρ1〉 and x2∈〈ρ1,ρ2〉 such that x1¯=x2¯∉〈ρ1¯〉, which implies x1∉〈ρ1〉. Since 〈ρ0,ρ1〉∩〈ρ1,ρ2〉=〈ρ1〉, we have

x1≠x2 and x1=x2⁢(ρ1⁢ρ2)2t-1

as x1¯=x2¯, which is impossible because otherwise

x1=x2⁢(ρ1⁢ρ2)2t-1∈〈ρ0,ρ1〉∩〈ρ1,ρ2〉=〈ρ1〉.

Thus 〈ρ0¯,ρ1¯〉∩〈ρ1¯,ρ2¯〉=〈ρ1¯〉, as required.

To finish the proof, we are left to show that ρ0⁢ρ1¯ and ρ1⁢ρ2¯ have order 2s and 2t-1 respectively. Since (G,{ρ0,ρ1,ρ2}) has type {2s,2t}, ρ0⁢ρ1 and ρ1⁢ρ2 have order 2s and 2t respectively. Since N≤〈ρ1⁢ρ2〉 and |N|=2, ρ1⁢ρ2¯ has order 2t-1 and ρ0⁢ρ1¯ has order 2s or 2s-1.

Suppose ρ0⁢ρ1¯ has order 2s-1. Then

(ρ0⁢ρ1)2s-1=(ρ1⁢ρ2)2t-1∈〈ρ0,ρ1〉∩〈ρ1,ρ2〉=〈ρ1〉,

and hence (ρ1⁢ρ2)2t-1=ρ1 because ρ1⁢ρ2 has order 2t. It follows that

(ρ1⁢ρ2)2t-1-1=ρ2 and (ρ1⁢ρ2)2t-2=1,

a contradiction. Thus ρ0⁢ρ1¯ has order 2s. This completes the proof. ∎

Lemma 4.2.

Let G=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ2)2〉. Then

G′=〈[ρ0,ρ1],[ρ1,ρ2],[ρ0,ρ1]ρ2〉.

Proof.

Since ρ0⁢ρ2=ρ2⁢ρ0, Proposition 2.6 implies

G′=〈[ρ0,ρ1]g,[ρ2,ρ1]h∣g,h∈G〉.

Since 〈ρ0,ρ1〉 and 〈ρ1,ρ2〉 are dihedral groups, we have

[ρ0,ρ1]ρ0=((ρ0⁢ρ1)2)ρ0=(ρ0⁢ρ1)-2=[ρ0,ρ1]-1,[ρ0,ρ1]ρ1=[ρ0,ρ1]-1,

=ρ1[ρ1,ρ2]-1 and [ρ1,ρ2]ρ2=[ρ1,ρ2]-1.

Set L=〈[ρ0,ρ1],[ρ1,ρ2],[ρ0,ρ1]ρ2〉. Since

([ρ0,ρ1]ρ2)ρ0=([ρ0,ρ1]ρ2)-1,

([ρ0,ρ1]ρ2)ρ2=[ρ0,ρ1],

([ρ0,ρ1]ρ2)ρ1=ρ1⁢ρ2⁢ρ0⁢ρ1⁢ρ0⁢ρ1⁢ρ2⁢ρ1=[ρ1,ρ2]⁢[ρ1,ρ0]ρ2⁢[ρ2,ρ1],

we have [ρ0,ρ1]g∈L for any g∈G. Since

[ρ1,ρ2]ρ0=ρ0⁢ρ1⁢ρ2⁢ρ1⁢ρ2⁢ρ0=ρ0⁢ρ1⁢ρ0⁢ρ1⁢ρ1⁢ρ2⁢ρ1⁢ρ2⁢ρ2⁢ρ1⁢ρ0⁢ρ1⁢ρ0⁢ρ2=[ρ0,ρ1]⁢[ρ1,ρ2]⁢[ρ1,ρ0]ρ2,

we have [ρ1,ρ2]h∈L for any h∈G. It follows that G′≤L, and hence G′=L. ∎

Proof of Theorem 1.3 (1).

For the sufficiency, we need to show that both G1 and G2 are string C-groups of order 2n, where n≥10 and

G1=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-3,(ρ0⁢ρ2)2,[(ρ0⁢ρ1)2,ρ2]〉,

G2=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-3,(ρ0⁢ρ2)2,[(ρ0ρ1)2,ρ2](ρ1ρ2)2n-4〉.

By Corollary 3.1, G1 is a string C-group of order 2n and we are only left with G2. However, to explain the method clearly, we prove the above fact again for G1 using a permutation representation graph that is simple and easy to understand. Let G=G1 or G2. For convenience, we write o⁢(g) for the order of g in G. We first prove the following claim.

Claim.

|G|≤2n.

Note that G/G′ is abelian and is generated by three involutions. Thus we have |G/G′|≤23. To prove the claim, it suffices to show that |G′|≤2n-3.

For G=G1, we have [(ρ0⁢ρ1)2,ρ2]=1 and ρ0⁢ρ2=ρ2⁢ρ0, which implies

[ρ0,ρ1]ρ2=[ρ0,ρ1]⁢[ρ1,ρ0]⁢[ρ0,ρ1]ρ2=[ρ0,ρ1]⁢[(ρ0⁢ρ1)2,ρ2]=[ρ0,ρ1].

Since

[ρ0,ρ1]ρ0=[ρ0,ρ1]ρ1=[ρ0,ρ1]-1,

we have 〈[ρ0,ρ1]〉⊴G, and by Lemma 4.2, we have

G′=〈[ρ0,ρ1],[ρ1,ρ2],[ρ0,ρ1]ρ2〉=〈[ρ0,ρ1],[ρ1,ρ2]〉=〈[ρ0,ρ1]〉⁢〈[ρ1,ρ2]〉.

This implies

|G′|≤|〈[ρ0,ρ1]〉|⁢|〈[ρ1,ρ2]〉|=o⁢((ρ0⁢ρ1)2)⁢o⁢((ρ1⁢ρ2)2)≤2⋅2n-4=2n-3,

as required.

For G=G2, we have [(ρ0⁢ρ1)2,ρ2]⁢(ρ1⁢ρ2)2n-4=1 and ρ0⁢ρ2=ρ2⁢ρ0, which implies

[ρ0,ρ1]ρ2=[ρ0,ρ1]⁢[(ρ0⁢ρ1)2,ρ2]=[ρ0,ρ1]⁢(ρ1⁢ρ2)-2n-4∈〈[ρ0,ρ1],[ρ1,ρ2]〉,

[ρ1,ρ2]ρ0=ρ1⁢[(ρ0⁢ρ1)2,ρ2]⁢ρ2⁢ρ1⁢ρ2=ρ1⁢(ρ1⁢ρ2)-2n-4⁢ρ1⁢(ρ1⁢ρ2)2=(ρ1⁢ρ2)2n-4⁢(ρ1⁢ρ2)2∈〈[ρ1,ρ2]〉.

It follows that 〈[ρ1,ρ2]〉⊴G because [ρ1,ρ2]ρ1=[ρ1,ρ2]ρ2=[ρ1,ρ2]-1, and by Lemma 4.2,

G′=〈[ρ0,ρ1],[ρ1,ρ2],[ρ0,ρ1]ρ2〉=〈[ρ0,ρ1],[ρ1,ρ2]〉=〈[ρ0,ρ1]〉⁢〈[ρ1,ρ2]〉.

In particular,

|G′|≤|〈[ρ0,ρ1]〉|⁢|〈[ρ1,ρ2]〉|=o⁢((ρ0⁢ρ1)2)⁢o⁢((ρ1⁢ρ2)2)≤2⋅2n-4=2n-3,

as required.

Now we are ready to finish the sufficiency proof by considering two cases. We use another method than the quotient method, based on permutation representation graphs. We give the details for G1 as they are simpler than those of G2 and might help the reader understand the case G2.

Case 1: G=G1. The key point is to construct a permutation group A of order at least 2n on a set Ω that is an epimorphic image of G, that is, A has three generators, say a,b,c, satisfying the same relations as ρ0,ρ1,ρ2. The permutation representation graph has vertex set Ω with a-, b- and c-edges. Recall that an x-edge (x=a,b or c) connects two points in Ω if and only if x interchanges them. It is easy to have such graphs when n is small by taking Ω as the set of right cosets of the subgroup 〈ρ0,ρ2〉 in G, where ρ0, ρ1 and ρ2 produce the a-, b- and c-edges, respectively. We give in Figure 1 a permutation representation graph for G1 and explain below how it is constructed.

$Figure 1 A permutation representation graph corresponding to G1{G_{1}}$

Figure 1

A permutation representation graph corresponding to G1

Set t=2n-3 and write ij⁢tk=j⁢t+4⁢i+k where 0≤i≤t4-1,0≤j≤1 and 1≤k≤4. Then a,b,c are permutations on the set {1,2,…,2n-2}:

a=∏i=0t4-1(i02,it2)⁢(i03,it3),b=∏i=0t4-1(i01,i02)⁢(it1,it2)⁢(i03,i04)⁢(it3,it4),

c=(001)⁢(0t1)⁢((t4-1)04)⁢((t4-1)t4)⋅∏i=0t4-1(i02,i03)(it2,it3)∏i=0t4-2(i04,(i+1)01)(it4,(i+1)t1).

Here (i+1)j⁢tk=j⁢t+4⁢(i+1)+k for 0≤i≤t4-2. Note that 1-cycles are also given in the product of distinct cycles of c and this would be helpful to compute conjugations of some elements by c. It is easy to see that a is fixed under conjugacy of c, that is, ac=a. It follows that (a⁢c)2=1. We further have

a⁢b=∏i=0t4-1(i01,i02,it1,it2)⁢(i03,it4,it3,i04),

b⁢c=∏i=01(1+t⁢i,3+t⁢i,…,t-1+t⁢i,t+t⁢i,t-2+t⁢i,…,2+t⁢i),

(a⁢b)2=∏i=0t8-1(i01,it1)⁢(i02,it2)⁢(i03,it3)⁢(i04,it4).

Let A=〈a,b,c〉. Clearly, a2=b2=c2=1, (a⁢b)4=1 and (b⁢c)2n-3=1. Furthermore, (a⁢b)2 is fixed under conjugacy of c, that is, ((a⁢b)2)c=(a⁢b)2, and hence [(a⁢b)2,c]=1. Clearly, A is transitive on {1,2,…,2n-2}, and the stabilizer A1 has order at least 4 because a,c∈A1. This implies that A is a permutation group of order at least 2n and its generators a,b,c satisfy the same relations as ρ0,ρ1,ρ2 in G. Then there is an epimorphism ϕ:G↦A such that ρ0ϕ=a, ρ1ϕ=b and ρ2ϕ=c. Since |A|≥2n and |G|≤2n, ϕ is an isomorphism, implying |G|=2n.

The generators ρ0,ρ1,ρ2 in

L1:-〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)4,(ρ1⁢ρ2)2,(ρ0⁢ρ2)2〉

satisfy all relations in G. This implies that the map: ρ0↦ρ0, ρ1↦ρ1, ρ2↦ρ2 induces a homomorphism from G to L1. By Proposition 2.2, o⁢(ρ0⁢ρ1)=4 in L1, and hence o⁢(ρ0⁢ρ1)=4 in G, and by Proposition 2.3, (G,{ρ0,ρ1,ρ2}) is a string C-group.

Case 2: G=G2. As in Case 1, we give in Figure 2 a permutation representation graph for G2. Note that, as in Case 1, bc consists of two paths of length 2⁢t with alternating labels of b and c, where t=2n-4, and here all of the real complexity lies in the definition of a.

$Figure 2 A permutation representation graph corresponding to G2{G_{2}}$

Figure 2

A permutation representation graph corresponding to G2

Write c⁢i=t8-i-1 for 0≤i≤t8-1,

ij⁢tk=j⁢t+8⁢i+k and c⁢ij⁢tk=j⁢t+8⁢c⁢i+k for⁢ 0≤j≤3⁢and⁢ 1≤k≤8.

Note that 0≤i≤t8-1 if and only if 0≤c⁢i≤t8-1. Then a,b,c are permutations on the set {1,2,…,2n-2}:

a=𝑏(it2,i2⁢t2)⁢(i02,c⁢i2⁢t7)⁢(i3⁢t2,c⁢it7)⁢(i07,i3⁢t7)⋅(it3,i2⁢t3)⁢(i03,c⁢i2⁢t6)⁢(i3⁢t3,c⁢it6)⁢(i06,i3⁢t6)⋅(i04,c⁢it5)⁢(i05,c⁢it4)⁢(i2⁢t4,c⁢i3⁢t5)⁢(i2⁢t5,c⁢i3⁢t4),

b=∏j=03∏i=0t8-1(ij⁢t1,ij⁢t2)⁢(ij⁢t3,ij⁢t4)⁢(ij⁢t5,ij⁢t6)⁢(ij⁢t7,ij⁢t8),

c=𝑏(00+2⁢t⁢i1)⁢((t8-1)t+2⁢t⁢i8)⁢((t8-1)0+2⁢t⁢i8,0t+2⁢t⁢i1)⋅∏j=03(∏i=0t8-1(ij⁢t2,ij⁢t3)(ij⁢t4,ij⁢t5)(ij⁢t6,ij⁢t7)∏i=0t8-2(ij⁢t8,(i+1)j⁢t1)).

Here (i+1)j⁢t1=j⁢t+8⁢(i+1)+1 for 0≤i≤t8-2. It is easy to see that a is fixed under conjugacy of c, that is, ac=a. It follows that (a⁢c)2=1.

Let α=a,b, or c. Then α is an involution. Recall that c⁢i=t8-i-1. Since 0≤i≤t8-1 if and only if 0≤c⁢i≤t8-1, it is easy to see that if α interchanges ij1⁢tk1 and ij2⁢tk2 then α also interchanges c⁢ij1⁢tk1 and c⁢ij2⁢tk2, and if α interchanges ij1⁢tk1 and c⁢ij2⁢tk2, then α also interchanges c⁢ij1⁢tk1 and ij2⁢tk2. These facts are very helpful for the following computations:

a⁢b=𝑏(it1,it2,i2⁢t1,i2⁢t2)⁢(i01,i02,c⁢i2⁢t8,c⁢i2⁢t7)⋅(i3⁢t1,i3⁢t2,c⁢it8,c⁢it7)⁢(i07,i3⁢t8,i3⁢t7,i08)⋅(i03,c⁢i2⁢t5,i3⁢t3,c⁢it5)⁢(i05,c⁢it3,c⁢i2⁢t4,i3⁢t6)⋅(i06,i3⁢t5,c⁢i2⁢t3,c⁢it4)⁢(i04,c⁢it6,i3⁢t4,c⁢i2⁢t6),

bc=𝑏(1+2ti,3+2ti,…,2t-1+2ti,

2t+2ti,2t-2+2ti,…,2+2ti),

(a⁢b)2=𝑏(it1,i2⁢t1)⁢(i01,c⁢i2⁢t8)⁢(it8,c⁢i3⁢t1)⁢(i08,i3⁢t8)⋅(it2,i2⁢t2)⁢(i02,c⁢i2⁢t7)⁢(it7,c⁢i3⁢t2)⁢(i07,i3⁢t7)⋅(i03,i3⁢t3)⁢(it3,c⁢i3⁢t6)⁢(i06,c⁢i2⁢t3)⁢(it6,i2⁢t6)⋅(i04,i3⁢t4)⁢(it4,c⁢i3⁢t5)⁢(i05,c⁢i2⁢t4)⁢(it5,i2⁢t5),

(b⁢c)2n-4=∏i=0t-1(1+i,2⁢t-i)⁢(2⁢t+1+i,4⁢t-i).

Let A=〈a,b,c〉. Clearly, (a⁢b)4=1 and (b⁢c)2n-3=1. Since c⁢i=t8-i-1, 0≤i≤t8-2 if and only if 1≤c⁢i≤t8-1, and since c interchanges ij⁢t8 and (i+1)j⁢t1, it also interchanges c⁢ij⁢t8 and c⁢(i-1)j⁢t1, where c⁢(i-1)=t8-(i-1)-1. Thus

c⁢(a⁢b)2⁢c=t⁢b(i01,i3⁢t1)⁢(i08,c⁢i2⁢t1)⁢(it1,c⁢i3⁢t8)⁢(it8,i2⁢t8)⋅(i02,i3⁢t2)⁢(i07,c⁢i2⁢t2)⁢(it2,c⁢i3⁢t7)⁢(it7,i2⁢t7)⋅(it3,i2⁢t3)⁢(i03,c⁢i2⁢t6)⁢(it6,c⁢i3⁢t3)⁢(i06,i3⁢t6)⋅(it4,i2⁢t4)⁢(i04,c⁢i2⁢t5)⁢(it5,c⁢i3⁢t4)⁢(i05,i3⁢t5).

It is clear that (b⁢c)2n-4 interchanges i0k and c⁢it9-k as i0k+c⁢it9-k=2⁢t+1 (note that 1≤i0k≤t and t+1≤c⁢it9-k≤2⁢t), and similarly (b⁢c)2n-4 interchanges i2⁢tk and c⁢i3⁢t9-k. Then it is easy to check [(a⁢b)2,c]=(a⁢b)2⁢c⁢(a⁢b)2⁢c=(b⁢c)2n-4. It follows that the generators a,b,c of A satisfy the same relations as ρ0,ρ1,ρ2 in G, and hence A is isomorphic to G with order 2n. Clearly, A is transitive and A1 has order at least 4 because a,c∈A1. It follows that |A|≥2n, and hence |G|=2n. On the other hand, the generators ρ0,ρ1,ρ2 in

L1:-〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)4,(ρ1⁢ρ2)2,(ρ0⁢ρ2)2〉

(defined in Proposition 2.2) satisfy all relations in G. This implies o⁢(ρ0⁢ρ1)=4 in G, and by Proposition 2.3, (G,{ρ0,ρ1,ρ2}) is a string C-group.

Now we prove the necessity. Let (G,{ρ0,ρ1,ρ2}) be a string C-group of rank 3 with type {4,2n-3} and |G|=2n. Then each of ρ0,ρ1 and ρ2 has order 2, and we further have o⁢(ρ0⁢ρ1)=4, o⁢(ρ0⁢ρ2)=2 and o⁢(ρ1⁢ρ2)=2n-3. To finish the proof, we aim to show that G≅G1 or G2. Since both G1 and G2 are C-groups of order 2n of type {4,2n-3}, it suffices to show that, in G, [(ρ0⁢ρ1)2,ρ2]=1 or [(ρ0⁢ρ1)2,ρ2]⁢(ρ1⁢ρ2)2n-4=1, which will be done by induction on n. This can easily be checked to be true for n=10 by using the computational algebra package Magma [2].

Assume n≥11. Take N=〈(ρ1⁢ρ2)2n-4〉. By Lemma 4.1, we have N⊴G and (G¯=G/N,{ρ0¯,ρ1¯,ρ2¯}) (with ρi¯=N⁢ρi) is a string C-group of rank 3 of type {4,2n-4}. Since |G¯|=2n-1, by induction hypothesis, we may assume G¯=G¯1 or G¯2, where

G¯1=〈ρ0¯,ρ1¯,ρ2¯∣ρ0¯2,ρ1¯2,ρ2¯2,(ρ0¯⁢ρ1¯)22,(ρ1¯⁢ρ2¯)2n-4,(ρ0¯⁢ρ2¯)2,[(ρ0¯ρ1¯)2,ρ2¯]〉,

G¯2=〈ρ0¯,ρ1¯,ρ2¯∣ρ0¯2,ρ1¯2,ρ2¯2,(ρ0¯⁢ρ1¯)22,(ρ1¯⁢ρ2¯)2n-4,(ρ0¯⁢ρ2¯)2,[(ρ0¯ρ1¯)2,ρ2¯](ρ1¯ρ2¯)2n-5〉.

Suppose G¯=G¯2. Since N=〈(ρ1⁢ρ2)2n-4〉≅ℤ2, we have

[(ρ0⁢ρ1)2,ρ2]⁢(ρ1⁢ρ2)2n-5=1 or (ρ1⁢ρ2)2n-4,

implying [(ρ0⁢ρ1)2,ρ2]=(ρ1⁢ρ2)δ⋅2n-5, where δ=1 or -1. Since

((ρ0⁢ρ1)2)ρ0=(ρ0⁢ρ1)-2=(ρ0⁢ρ1)2 and ρ0⁢ρ2=ρ2⁢ρ0,

we have [(ρ0⁢ρ1)2,ρ2]ρ0=[(ρ0⁢ρ1)2,ρ2], and hence [ρ0,(ρ1⁢ρ2)δ⋅2n-5]=1. By Proposition 2.5,

1=[(ρ0⁢ρ1)4,ρ2]=[(ρ0⁢ρ1)2,ρ2](ρ0⁢ρ1)2⁢[(ρ0⁢ρ1)2,ρ2]=((ρ1⁢ρ2)δ⋅2n-5)(ρ0⁢ρ1)2⁢(ρ1⁢ρ2)δ⋅2n-5=(ρ1⁢ρ2)δ⋅2n-4,

which is impossible because o⁢(ρ1⁢ρ2)=2n-3.

Thus G¯=G¯1. Since N=〈(ρ1⁢ρ2)2n-4〉≅ℤ2, we have [(ρ0⁢ρ1)2,ρ2]=1 or (ρ1⁢ρ2)2n-4. For the latter, [(ρ0⁢ρ1)2,ρ2]⁢(ρ1⁢ρ2)2n-4=(ρ1⁢ρ2)2n-3=1. It follows that G≅G1 or G2. ∎

Proof of Theorem 1.3 (2).

For the sufficiency, we need to show that both G3 and G4 are string C-group of order 2n, where n≥10 and

G3=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-4,(ρ0⁢ρ2)2,[(ρ0ρ1)2,(ρ1ρ2)2]〉,

G4=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-4,(ρ0⁢ρ2)2,[(ρ0ρ1)2,(ρ1ρ2)2](ρ1ρ2)2n-5〉.

By Corollary 3.1, we only need to show that G4 is a string C-group.

Let G=G4. Then [(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]=(ρ1⁢ρ2)2n-5. Noting

[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2⁢(ρ1⁢ρ2)2]=[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]⁢[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2](ρ1⁢ρ2)2=(ρ1⁢ρ2)2n-4=1,

we have [(ρ0⁢ρ1)2,(ρ1⁢ρ2)2n-5]=1, which implies [ρ1,((ρ1⁢ρ2)2n-5)ρ0]=1 as [ρ1,(ρ1⁢ρ2)2n-5]=1. Thus ρ1 fixes K=〈(ρ1⁢ρ2)2n-5,((ρ1⁢ρ2)2n-5)ρ0〉. Clearly, Kρ0=Kρ2=K, and so K⁢⊴⁢G. The three generators ρ0⁢K, ρ1⁢K, ρ2⁢K in G/K satisfy the same relations as ρ0, ρ1, ρ2 in G3. In fact, (ρ1⁢K⁢ρ2⁢K)2n-5=K, and hence |G/K|≤2n-1 (here we need to check that |G3|=29 for n=9, and this can be done using Magma). Furthermore,

[ρ1⁢ρ2,((ρ1⁢ρ2)2n-5)ρ0]=1

as [ρ2,((ρ1⁢ρ2)2n-5)ρ0]=1. It follows that |K|≤4 and |G|≤2n+1.

Suppose |G|=2n+1. Then

|G/K|=2n-1 and |K|=4.

It follows that o⁢(ρ1⁢ρ2⁢K)=2n-5 in G/K, and hence o⁢(ρ1⁢ρ2)=2n-4 in G. Note that 2⁢(n-4)≥n as n≥10, by Proposition 2.9, CoreG⁡(〈ρ1⁢ρ2〉)>1, so we have 〈(ρ1⁢ρ2)2n-5〉⊴G. It follows that |K|=2, a contradiction. Thus |G|≤2n.

We give in Figure 3 a permutation representation graph of G. In this case, bc consists of two paths of length 2⁢t and a circle of length 4⁢t with alternating labels b and c, where t=2n-5.

$Figure 3 A permutation representation graph corresponding to G4{G_{4}}$

Figure 3

A permutation representation graph corresponding to G4

Write

ij⁢tk=j⁢t+8⁢i+k and c⁢ij⁢tk=j⁢t+8⁢(t8-i-1)+k

for 0≤i≤t8-1, 0≤j≤7 and 1≤k≤8. The permutations a,b,c on the set {1,2,…,2n-2} are as follows:

a=𝑏(i02,i2⁢t2)⁢(it2,i3⁢t2)⁢(i4⁢t2,i6⁢t2)⁢(i5⁢t2,i7⁢t2)⋅(i03,i2⁢t3)⁢(it3,i3⁢t3)⁢(i4⁢t3,i6⁢t3)⁢(i5⁢t3,i7⁢t3)⋅(i04,c⁢i7⁢t5)⁢(i05,c⁢i7⁢t4)⁢(i2⁢t4,c⁢i3⁢t5)⁢(i2⁢t5,c⁢i3⁢t4)⋅(i6⁢t4,c⁢it5)⁢(i6⁢t5,c⁢it4)⁢(i4⁢t4,c⁢i5⁢t5)⁢(i4⁢t5,c⁢i5⁢t4)⋅(i06,i4⁢t6)⁢(it6,i5⁢t6)⁢(i2⁢t6,i6⁢t6)⁢(i3⁢t6,i7⁢t6)⋅(i07,i4⁢t7)⁢(it7,i5⁢t7)⁢(i2⁢t7,i6⁢t7)⁢(i3⁢t7,i7⁢t7),

b=∏j=07∏i=0t8-1(ij⁢t1,ij⁢t2)⁢(ij⁢t3,ij⁢t4)⁢(ij⁢t5,ij⁢t6)⁢(ij⁢t7,ij⁢t8),

c=(001)⁢(06⁢t1)⁢((t8-1)t8)⁢((t8-1)7⁢t8)⁢(02⁢t1,04⁢t1)⁢((t8-1)3⁢t8,(t8-1)5⁢t8)

⋅∏i=03((t8-1)2⁢t⁢i8,0t+2⁢t⁢i1)

⋅∏j=07(∏i=0t8-1(ij⁢t2,ij⁢t3)⁢(ij⁢t4,ij⁢t5)⁢(ij⁢t6,ij⁢t7)⋅∏i=0t8-2(ij⁢t8,(i+1)j⁢t1)).

Here (i+1)j⁢tk=j⁢t+8⁢(i+1)+k for 0≤i≤t8-2. It is easy to see that c fixes a under conjugacy, that is, ac=a. It follows that (a⁢c)2=1. Furthermore,

a⁢b=𝑏(i01,i02,i2⁢t1,i2⁢t2)⁢(it1,it2,i3⁢t1,i3⁢t2)⁢(i4⁢t1,i4⁢t2,i6⁢t1,i6⁢t2)⁢(i5⁢t1,i5⁢t2,i7⁢t1,i7⁢t2)⋅(i03,i2⁢t4,c⁢i3⁢t6,c⁢i7⁢t5)⁢(it3,i3⁢t4,c⁢i2⁢t6,c⁢i6⁢t5)⁢(i2⁢t3,i04,c⁢i7⁢t6,c⁢i3⁢t5)⋅(i3⁢t3,it4,c⁢i6⁢t6,c⁢i2⁢t5)⁢(c⁢i4⁢t3,c⁢i6⁢t4,it6,i5⁢t5)⁢(c⁢i5⁢t3,c⁢i7⁢t4,i06,i4⁢t5)⋅(c⁢i6⁢t3,c⁢i4⁢t4,i5⁢t6,it5)⁢(c⁢i7⁢t3,c⁢i5⁢t4,i4⁢t6,i05)⋅(i07,i4⁢t8,i4⁢t7,i08)⁢(it7,i5⁢t8,i5⁢t7,it8)⋅(i2⁢t7,i6⁢t8,i6⁢t7,i2⁢t8)⁢(i3⁢t7,i7⁢t8,i7⁢t7,i3⁢t8),

b⁢c=𝑏(1+6ti,3+6ti,…,2t-1+6ti,2t+6ti,2t-2+6ti,…,2+6ti)⋅(2t+1+2ti,2t+3+2ti,…,4t-1+2ti,6t-2ti,6t-2-2ti,…,4t+2-2ti),

(a⁢b)2=𝑏(i01,i2⁢t1)⁢(it1,i3⁢t1)⁢(i4⁢t1,i6⁢t1)⁢(i5⁢t1,i7⁢t1)⋅(i02,i2⁢t2)⁢(it2,i3⁢t2)⁢(i4⁢t2,i6⁢t2)⁢(i5⁢t2,i7⁢t2)⋅(i03,c⁢i3⁢t6)⁢(it3,c⁢i2⁢t6)⁢(i2⁢t3,c⁢i7⁢t6)⁢(i3⁢t3,c⁢i6⁢t6)⋅(i06,c⁢i5⁢t3)⁢(it6,c⁢i4⁢t3)⁢(i4⁢t6,c⁢i7⁢t3)⁢(i5⁢t6,c⁢i6⁢t3)⋅(i04,c⁢i3⁢t5)⁢(it4,c⁢i2⁢t5)⁢(i2⁢t4,c⁢i7⁢t5)⁢(i3⁢t4,c⁢i6⁢t5)⋅(i05,c⁢i5⁢t4)⁢(it5,c⁢i4⁢t4)⁢(i4⁢t5,c⁢i7⁢t4)⁢(i5⁢t5,c⁢i6⁢t4)⋅(i07,i4⁢t7)⁢(it7,i5⁢t7)⁢(i2⁢t7,i6⁢t7)⁢(i3⁢t7,i7⁢t7)⋅(i08,i4⁢t8)⁢(it8,i5⁢t8)⁢(i2⁢t8,i6⁢t8)⁢(i3⁢t8,i7⁢t8),

(b⁢c)2n-5=∏i=0t-1(1+i,2⁢t-i)⁢(6⁢t+1+i,8⁢t-i)⁢∏i=02⁢t-1(2⁢t+1+i,6⁢t-i),

For 0≤i≤t8-2, c interchanges ij⁢t8 and (i+1)j⁢t1, and also c⁢ij⁢t8 and c⁢(i-1)j⁢t1. Thus

c⁢(a⁢b)2⁢c=t⁢b(i01,i4⁢t1)⁢(it1,i5⁢t1)⁢(i6⁢t1,i2⁢t1)⁢(i7⁢t1,i3⁢t1)⋅(i02,c⁢i3⁢t7)⁢(i07,c⁢i5⁢t2)⁢(it2,c⁢i2⁢t7)⁢(it7,c⁢i4⁢t2)⋅(i6⁢t2,c⁢i5⁢t7)⁢(i6⁢t7,c⁢i3⁢t2)⁢(i7⁢t2,c⁢i4⁢t7)⁢(i7⁢t7,c⁢i2⁢t2)⋅(i03,i2⁢t3)⁢(it3,i3⁢t3)⁢(i6⁢t3,i4⁢t3)⁢(i7⁢t3,i5⁢t3)⋅(i04,c⁢i5⁢t5)⁢(i05,c⁢i3⁢t4)⁢(it4,c⁢i4⁢t5)⁢(it5,c⁢i2⁢t4)⋅(i6⁢t4,c⁢i3⁢t5)⁢(i6⁢t5,c⁢i5⁢t4)⁢(i7⁢t4,c⁢i2⁢t5)⁢(i7⁢t5,c⁢i4⁢t4)⋅(i06,i4⁢t6)⁢(it6,i5⁢t6)⁢(i6⁢t6,i2⁢t6)⁢(i7⁢t6,i3⁢t6)⋅(i08,i2⁢t8)⁢(it8,i3⁢t8)⁢(i6⁢t8,i4⁢t8)⁢(i7⁢t8,i5⁢t8),

c⁢(a⁢b)2⁢c⁢b=𝑏(i01,i4⁢t2,c⁢it8,c⁢i3⁢t7)⁢(c⁢it1,c⁢i5⁢t2,i08,i2⁢t7)⋅(i2⁢t1,i6⁢t2,c⁢i5⁢t8,c⁢i7⁢t7)⁢(c⁢i3⁢t1,c⁢i7⁢t2,i4⁢t8,i6⁢t7)⋅(i4⁢t1,i02,c⁢i3⁢t8,c⁢it7)⁢(c⁢i5⁢t1,c⁢it2,i2⁢t8,i07)⋅(i6⁢t1,i2⁢t2,c⁢i7⁢t8,c⁢i5⁢t7)⁢(c⁢i7⁢t1,c⁢i3⁢t2,i6⁢t8,i4⁢t7)⋅(i03,i2⁢t4,c⁢it6,c⁢i5⁢t5)⁢(c⁢it3,c⁢i3⁢t4,i06,i4⁢t5)⋅(i2⁢t3,i04,c⁢i5⁢t6,c⁢it5)⁢(c⁢i3⁢t3,c⁢it4,i4⁢t6,i05)⋅(i4⁢t3,i6⁢t4,c⁢i3⁢t6,c⁢i7⁢t5)⁢(c⁢i5⁢t3,c⁢i7⁢t4,i2⁢t6,i6⁢t5)⋅(i6⁢t3,i4⁢t4,c⁢i7⁢t6,c⁢i3⁢t5)⁢(c⁢i7⁢t3,c⁢i5⁢t4,i6⁢t6,i2⁢t5).

Let A=〈a,b,c〉. It is clear that (b⁢c)2n-5 interchanges i0j and c⁢it9-j for each 1≤j≤8 because

i0j+c⁢it9-j=2⁢t+1

(note that 1≤i0j≤t and t+1≤c⁢it9-j≤2⁢t), and similarly (b⁢c)2n-5 also interchanges i2⁢tj and c⁢i5⁢t9-j, i3⁢tj and c⁢i4⁢t9-j, and i6⁢tj and c⁢i7⁢t9-j. Thus

[c⁢(a⁢b)2⁢c,b]=(c⁢(a⁢b)2⁢c⁢b)2=(b⁢c)2n-5,

and hence

[(a⁢b)2,c⁢b⁢c]=(b⁢c)2n-5 as⁢[(b⁢c)2n-5,c]=1.

Since [(a⁢b)2,b]=1, Proposition 2.5 implies

[(a⁢b)2,(b⁢c)2]=[(a⁢b)2,c⁢b⁢c]⁢[(a⁢b)2,b]c⁢b⁢c=(b⁢c)2n-5.

It follows that the generators a,b,c of A satisfy the same relations as ρ0,ρ1,ρ2 in G, and hence A is isomorphic to G with order 2n.

Again, let L1=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)4,(ρ1⁢ρ2)2,(ρ0⁢ρ2)2〉. The generators ρ0,ρ1,ρ2 in L1 satisfy all relations in G. This means that o⁢(ρ0⁢ρ1)=4 in G, and by Proposition 2.3, (G,{ρ0,ρ1,ρ2}) is a string C-group.

To prove the necessity, let (G,{ρ0,ρ1,ρ2}) be a string C-group of rank 3 with type {4,2n-4} and |G|=2n. Then

o⁢(ρ0)=o⁢(ρ1)=o⁢(ρ2)=o⁢(ρ0⁢ρ2)=2,o⁢(ρ0⁢ρ1)=4 and o⁢(ρ1⁢ρ2)=2n-4.

To finish the proof, we aim to show that G≅G3 or G4. Since both G3 and G4 are C-groups of order 2n and of type {4,2n-4}, it suffices to show that, in G,

[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]=1 or [(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]⁢(ρ1⁢ρ2)2n-5=1,

which will be done by induction on n. This is true for n=10 by Magma.

Assume n≥11. Take N=〈(ρ1⁢ρ2)2n-5〉. By Lemma 4.1, we have N⊴G and (G¯=G/N,{ρ0¯,ρ1¯,ρ2¯}) (with ρi¯=N⁢ρi) is a string C-group of rank 3 of type {4,2n-5}. Since |G¯|=2n-1, by induction hypothesis, we may assume G¯=G¯3 or G¯4, where

G¯3=〈ρ0¯,ρ1¯,ρ2¯∣ρ0¯2,ρ1¯2,ρ2¯2,(ρ0¯⁢ρ1¯)22,(ρ1¯⁢ρ2¯)2n-5,(ρ0¯⁢ρ2¯)2,[(ρ0¯ρ1¯)2,(ρ1¯ρ2¯)2]〉,

G¯4=〈ρ0¯,ρ1¯,ρ2¯∣ρ0¯2,ρ1¯2,ρ2¯2,(ρ0¯⁢ρ1¯)22,(ρ1¯⁢ρ2¯)2n-5,(ρ0¯⁢ρ2¯)2,[(ρ0¯ρ1¯)2,(ρ1¯ρ2¯)2](ρ1¯ρ2¯)2n-6〉.

Suppose G¯=G¯4. Since N=〈(ρ1⁢ρ2)2n-5〉≅ℤ2, we have

[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]⁢(ρ1⁢ρ2)2n-6=1 or (ρ1⁢ρ2)2n-5,

which implies [(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]=(ρ1⁢ρ2)δ⋅2n-6, where δ=1 or -1. By Proposition 2.5,

[(ρ0⁢ρ1)2,(ρ1⁢ρ2)4]=[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]⁢[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2](ρ1⁢ρ2)2=(ρ1⁢ρ2)δ⋅2n-5,

=(ρ1⁢ρ2)δ⋅2n-4=1,

implying [(ρ0⁢ρ1)2,(ρ1⁢ρ2)2n-6]=1. Thus

1=[(ρ0⁢ρ1)4,(ρ1⁢ρ2)2]=[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2](ρ0⁢ρ1)2⁢[(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]=((ρ1⁢ρ2)δ⋅2n-6)(ρ0⁢ρ1)2⁢(ρ1⁢ρ2)δ⋅2n-6=(ρ1⁢ρ2)δ⋅2n-5,

which is impossible because o⁢(ρ1⁢ρ2)=2n-4.

Thus G¯=G¯3. In this case, [(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]=1 or (ρ1⁢ρ2)2n-5. For the latter, [(ρ0⁢ρ1)2,(ρ1⁢ρ2)2]⁢(ρ1⁢ρ2)2n-5=(ρ1⁢ρ2)2n-4=1. It follows that G≅G3 or G4. ∎

Proof of Theorem 1.3 (3).

Let n≥10, and let G=G5,G6,G7 or G8, where

G5=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-5,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)2]2,[ρ0,(ρ1ρ2)4]〉,

G6=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-5,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)2]2(ρ1ρ2)2n-6,[ρ0,(ρ1ρ2)4]〉,

G7=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-5,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)2]2,[ρ0,(ρ1ρ2)4](ρ1ρ2)2n-6〉,

G8=〈ρ0,ρ1,ρ2∣ρ02,ρ12,ρ22,(ρ0⁢ρ1)22,(ρ1⁢ρ2)2n-5,(ρ0⁢ρ2)2,[ρ0,(ρ1ρ2)2]2(ρ1ρ2)2n-6,[ρ0,(ρ1ρ2)4](ρ1ρ2)2n-6〉.

By Corollary 3.1, we only need to show that G6, G7 and G8 are string C-groups.

In all cases, [ρ0,(ρ1⁢ρ2)4]=1 or (ρ1⁢ρ2)2n-6. It follows from Proposition 2.5 that

[ρ0,(ρ1⁢ρ2)8]=[ρ0,(ρ1⁢ρ2)4]⁢([ρ0,(ρ1⁢ρ2)4])(ρ1⁢ρ2)4=1.

Noting that ((ρ1⁢ρ2)8)ρ1=(ρ1⁢ρ2)-8=((ρ1⁢ρ2)8)ρ2, we have K=〈(ρ1⁢ρ2)8〉⊴G. Clearly, |K|≤2n-8, and the three generators ρ0⁢K, ρ1⁢K, ρ2⁢K in G/K satisfy the same relations as ρ0, ρ1, ρ2 in G5 when n=8. By Magma, |G5|=28 when n=8, and hence |G|=|G:K|⋅|K|≤2n.

Case 1: G=G6. We construct a permutation representation graph of G, and in this graph, bc consists of four paths of length 2⁢t and two circles of length 4⁢t with alternating labels of b and c, where t=2n-6. We omit the drawing of the graph here because it is quite big.

Write

ij⁢tk=j⁢t+8⁢i+k and c⁢ij⁢tk=j⁢t+8⁢(t8-i-1)+k,

where 0≤i≤t8-1, 1≤k≤8, 0≤j≤15. The permutations a,b,c on the set {1,2,…,2n-2} are as follows:

a=𝑏(i02,i2⁢t2)⁢(it2,i3⁢t2)⁢(i4⁢t2,i7⁢t2)⁢(i8⁢t2,i11⁢t2)⁢(i12⁢t2,i14⁢t2)⁢(i13⁢t2,i15⁢t2)

⋅(i5⁢t2,c⁢i7⁢t7)⁢(i6⁢t2,c⁢i2⁢t7)⁢(i9⁢t2,c⁢i13⁢t7)⁢(i10⁢t2,c⁢i8⁢t7)

⋅(i07,i4⁢t7)⁢(it7,i5⁢t7)⁢(i3⁢t7,i6⁢t7)⁢(i9⁢t7,i12⁢t7)⁢(i10⁢t7,i14⁢t7)⁢(i11⁢t7,i15⁢t7)

⋅(i03,i2⁢t3)⁢(it3,i3⁢t3)⁢(i4⁢t3,i7⁢t3)⁢(i8⁢t3,i11⁢t3)⁢(i12⁢t3,i14⁢t3)⁢(i13⁢t3,i15⁢t3)

⋅(i5⁢t3,c⁢i7⁢t6)⁢(i6⁢t3,c⁢i2⁢t6)⁢(i9⁢t3,c⁢i13⁢t6)⁢(i10⁢t3,c⁢i8⁢t6)

⋅(i06,i4⁢t6)⁢(it6,i5⁢t6)⁢(i3⁢t6,i6⁢t6)⁢(i9⁢t6,i12⁢t6)⁢(i10⁢t6,i14⁢t6)⁢(i11⁢t6,i15⁢t6)

⋅(i6⁢t4,i8⁢t4)⁢(i7⁢t4,i9⁢t4)⁢(i04,c⁢i15⁢t5)⁢(it4,c⁢i14⁢t5)

⋅(i2⁢t4,c⁢i11⁢t5)⁢(i3⁢t4,c⁢i10⁢t5)⁢(i4⁢t4,c⁢i13⁢t5)⁢(i5⁢t4,c⁢i12⁢t5)

⋅(i6⁢t5,i8⁢t5)⁢(i7⁢t5,i9⁢t5)⁢(i05,c⁢i15⁢t4)⁢(it5,c⁢i14⁢t4)

⋅(i2⁢t5,c⁢i11⁢t4)⁢(i3⁢t5,c⁢i10⁢t4)⁢(i4⁢t5,c⁢i13⁢t4)⁢(i5⁢t5,c⁢i12⁢t4)

⋅(i8⁢t1,c⁢i9⁢t8)⁢(i9⁢t1,c⁢i8⁢t8)⁢(i10⁢t1,c⁢i13⁢t8)⁢(i11⁢t1,c⁢i12⁢t8)

⋅(i12⁢t1,c⁢i11⁢t8)⁢(i13⁢t1,c⁢i10⁢t8)⁢(i14⁢t1,c⁢i15⁢t8)⁢(i15⁢t1,c⁢i14⁢t8),

b=∏j=015∏i=0t8-1(ij⁢t1,ij⁢t2)⁢(ij⁢t3,ij⁢t4)⁢(ij⁢t5,ij⁢t6)⁢(ij⁢t7,ij⁢t8),

c=𝑏(08⁢i⁢t1)⁢(06⁢t+8⁢i⁢t1)⁢((t8-1)t+8⁢i⁢t8)⁢((t8-1)7⁢t+8⁢i⁢t8)⋅(02⁢t+8⁢i⁢t1,04⁢t+8⁢i⁢t1)⁢((t8-1)3⁢t+8⁢i⁢t8,(t8-1)5⁢t+8⁢i⁢t8)⋅∏i=07((t8-1)2⁢i⁢t8,0(2⁢i+1)⁢t1)⋅∏j=015(∏i=0t8-1(ij⁢t2,ij⁢t3)(ij⁢t4,ij⁢t5)(ij⁢t6,ij⁢t7)⋅∏i=0t8-2(ij⁢t8,(i+1)j⁢t1)).

Here (i+1)j⁢tk=j⁢t+8⁢(i+1)+k for 0≤i≤t8-2. For 0≤i≤t8-2, c interchanges ij⁢t8 and (i+1)j⁢t1, and also c⁢ij⁢t8 and c⁢(i-1)j⁢t1. It is easy to see that a is fixed under conjugacy by c, that is, ac=a. It follows that (a⁢c)2=1. Furthermore,

a⁢b=𝑏(i01,i02,i2⁢t1,i2⁢t2)⁢(it1,it2,i3⁢t1,i3⁢t2)

⋅(i4⁢t1,i4⁢t2,i7⁢t1,i7⁢t2)⁢(i5⁢t1,i5⁢t2,c⁢i7⁢t8,c⁢i7⁢t7)

⋅(i6⁢t1,i6⁢t2,c⁢i2⁢t8,c⁢i2⁢t7)⁢(i8⁢t1,c⁢i9⁢t7,c⁢i12⁢t8,i11⁢t2)

⋅(i9⁢t1,c⁢i8⁢t7,i10⁢t1,c⁢i13⁢t7)⁢(i11⁢t1,c⁢i12⁢t7,c⁢i9⁢t8,i8⁢t2)

⋅(i12⁢t1,c⁢i11⁢t7,c⁢i15⁢t8,i14⁢t2)⁢(i13⁢t1,c⁢i10⁢t7,c⁢i14⁢t8,i15⁢t2)

⋅(i14⁢t1,c⁢i15⁢t7,c⁢i11⁢t8,i12⁢t2)⁢(i15⁢t1,c⁢i14⁢t7,c⁢i10⁢t8,i13⁢t2)

⋅(i08,i07,i4⁢t8,i4⁢t7)⁢(it8,it7,i5⁢t8,i5⁢t7)

⋅(i3⁢t8,i3⁢t7,i6⁢t8,i6⁢t7)⁢(i8⁢t8,c⁢i9⁢t2,i13⁢t8,c⁢i10⁢t2)

⋅(i03,i2⁢t4,c⁢i11⁢t6,c⁢i15⁢t5)⁢(it3,i3⁢t4,c⁢i10⁢t6,c⁢i14⁢t5)

⋅(i2⁢t3,i04,c⁢i15⁢t6,c⁢i11⁢t5)⁢(i3⁢t3,it4,c⁢i14⁢t6,c⁢i10⁢t5)

⋅(i4⁢t3,i7⁢t4,i9⁢t3,c⁢i13⁢t5)⁢(i5⁢t3,c⁢i7⁢t5,c⁢i9⁢t6,c⁢i12⁢t5)

⋅(i6⁢t3,c⁢i2⁢t5,i11⁢t3,i8⁢t4)⁢(i7⁢t3,i4⁢t4,c⁢i13⁢t6,i9⁢t4)

⋅(i8⁢t3,i11⁢t4,c⁢i2⁢t6,i6⁢t4)⁢(i10⁢t3,c⁢i8⁢t5,c⁢i6⁢t6,c⁢i3⁢t5)

⋅(i12⁢t3,i14⁢t4,c⁢it6,c⁢i5⁢t5)⁢(i13⁢t3,i15⁢t4,c⁢i06,c⁢i4⁢t5)

⋅(i14⁢t3,i12⁢t4,c⁢i5⁢t6,c⁢it5)⁢(i15⁢t3,i13⁢t4,c⁢i4⁢t6,c⁢i05)

⋅(i5⁢t4,c⁢i12⁢t6,c⁢i9⁢t5,c⁢i7⁢t6)⁢(i10⁢t4,c⁢i3⁢t6,c⁢i6⁢t5,c⁢i8⁢t6),

b⁢c=𝑏(1+8ti,3+8ti,…,2t-1+8ti,2t+8ti,2t-2+8ti,…,2+8ti)⋅(2t+1+8ti,2t+3+8ti,…,4t-1+8ti,6t+8ti,6t-2+8ti,…,4t+2+8ti)⋅(4t+1+8ti,4t+3+8ti,…,6t-1+8ti,4t+8ti,4t-2+8ti,…,2t+2+8ti)⋅(6t+1+8ti,6t+3+8ti,…,8t-1+8ti,8t+8ti,8t-2+8ti,…,6t+2+8ti),

(b⁢c)2n-6=𝑏(1+i,2⁢t-i)⁢(6⁢t+1+i,8⁢t-i)⁢(8⁢t+1+i,10⁢t-i)⋅(14⁢t+1+i,16⁢t-i)⋅∏i=02⁢t-1(2t+1+i,6t-i)(10t+1+i,14t-i).

The above computations imply (a⁢b)4=1 and (b⁢c)2n-5=1. Furthermore,

(a⁢b)2=𝑏(i01,i2⁢t1)⁢(it1,i3⁢t1)⁢(i4⁢t1,i7⁢t1)⁢(i9⁢t1,i10⁢t1)

⋅(i5⁢t1,c⁢i7⁢t8)⁢(i6⁢t1,c⁢i2⁢t8)⁢(i8⁢t1,c⁢i12⁢t8)⁢(i11⁢t1,c⁢i9⁢t8)

⋅(i12⁢t1,c⁢i15⁢t8)⁢(i13⁢t1,c⁢i14⁢t8)⁢(i14⁢t1,c⁢i11⁢t8)⁢(i15⁢t1,c⁢i10⁢t8)

⋅(i08,i4⁢t8)⁢(it8,i5⁢t8)⁢(i3⁢t8,i6⁢t8)⁢(i8⁢t8,i13⁢t8)

⋅(i02,i2⁢t2)⁢(it2,i3⁢t2)⁢(i4⁢t2,i7⁢t2)⁢(i9⁢t2,i10⁢t2)

⋅(i5⁢t2,c⁢i7⁢t7)⁢(i6⁢t2,c⁢i2⁢t7)⁢(i8⁢t2,c⁢i12⁢t7)⁢(i11⁢t2,c⁢i9⁢t7)

⋅(i12⁢t2,c⁢i15⁢t7)⁢(i13⁢t2,c⁢i14⁢t7)⁢(i14⁢t2,c⁢i11⁢t7)⁢(i15⁢t2,c⁢i10⁢t7)

⋅(i07,i4⁢t7)⁢(it7,i5⁢t7)⁢(i3⁢t7,i6⁢t7)⁢(i8⁢t7,i13⁢t7)

⋅(i4⁢t3,i9⁢t3)⁢(i6⁢t3,i11⁢t3)⁢(i03,c⁢i11⁢t6)⁢(it3,c⁢i10⁢t6)

⋅(i2⁢t3,c⁢i15⁢t6)⁢(i3⁢t3,c⁢i14⁢t6)⁢(i5⁢t3,c⁢i9⁢t6)⁢(i7⁢t3,c⁢i13⁢t6)

⋅(i8⁢t3,c⁢i2⁢t6)⁢(i10⁢t3,c⁢i6⁢t6)⁢(i12⁢t3,c⁢it6)⁢(i13⁢t3,i06)

⋅(i14⁢t3,c⁢i5⁢t6)⁢(i15⁢t3,c⁢i4⁢t6)⁢(i3⁢t6,i8⁢t6)⁢(i7⁢t6,i12⁢t6)

⋅(i4⁢t4,i9⁢t4)⁢(i6⁢t4,i11⁢t4)⁢(i04,c⁢i11⁢t5)⁢(it4,c⁢i10⁢t5)

⋅(i2⁢t4,c⁢i15⁢t5)⁢(i3⁢t4,c⁢i14⁢t5)⁢(i5⁢t4,c⁢i9⁢t5)⁢(i7⁢t4,c⁢i13⁢t5)

⋅(i8⁢t4,c⁢i2⁢t5)⁢(i10⁢t4,c⁢i6⁢t5)⁢(i12⁢t4,c⁢it5)⁢(i13⁢t4,i05)

⋅(i14⁢t4,c⁢i5⁢t5)⁢(i15⁢t4,c⁢i4⁢t5)⁢(i3⁢t5,i8⁢t5)⁢(i7⁢t5,i12⁢t5),

c⁢(a⁢b)2⁢c=𝑏(i01,i4⁢t1)⁢(it1,i5⁢t1)⁢(i3⁢t1,i6⁢t1)⁢(i8⁢t1,i13⁢t1)⋅(i2⁢t1,c⁢i6⁢t8)⁢(i7⁢t1,c⁢i5⁢t8)⁢(i9⁢t1,c⁢i11⁢t8)⁢(i10⁢t1,c⁢i15⁢t8)⋅(i11⁢t1,c⁢i14⁢t8)⁢(i12⁢t1,i8⁢t8)⁢(i14⁢t1,c⁢i13⁢t8)⁢(i15⁢t1,c⁢i12⁢t8)⋅(i08,i2⁢t8)⁢(it8,i3⁢t8)⁢(i4⁢t8,i7⁢t8)⁢(i9⁢t8,i10⁢t8)⋅(i4⁢t2,i9⁢t2)⁢(i6⁢t2,i11⁢t2)⁢(i02,c⁢i11⁢t7)⁢(it2,c⁢i10⁢t7)⋅(i2⁢t2,c⁢i15⁢t7)⁢(i3⁢t2,c⁢i14⁢t7)⁢(i5⁢t2,c⁢i9⁢t7)⁢(i7⁢t2,c⁢i13⁢t7)⋅(i8⁢t2,c⁢i2⁢t7)⁢(i10⁢t2,c⁢i6⁢t7)⁢(i12⁢t2,c⁢it7)⁢(i13⁢t2,c⁢i07)⋅(i14⁢t2,c⁢i5⁢t7)⁢(i15⁢t2,c⁢i4⁢t7)⁢(i7⁢t7,i12⁢t7)⁢(i8⁢t7,i3⁢t7)⋅(i03,i2⁢t3)⁢(it3,i3⁢t3)⁢(i4⁢t3,i7⁢t3)⁢(i9⁢t3,i10⁢t3)⋅(i5⁢t3,c⁢i7⁢t6)⁢(i6⁢t3,c⁢i2⁢t6)⁢(i8⁢t3,c⁢i12⁢t6)⁢(i11⁢t3,c⁢i9⁢t6)⋅(i12⁢t3,c⁢i15⁢t6)⁢(i13⁢t3,c⁢i14⁢t6)⁢(i14⁢t3,c⁢i11⁢t6)⁢(i15⁢t3,c⁢i10⁢t6)⋅(i06,i4⁢t6)⁢(it6,i5⁢t6)⁢(i3⁢t6,i6⁢t6)⁢(i8⁢t6,i13⁢t6)⋅(i3⁢t4,i8⁢t4)⁢(i7⁢t4,i12⁢t4)⁢(i04,c⁢i13⁢t5)⁢(it4,c⁢i12⁢t5)⋅(i2⁢t4,c⁢i8⁢t5)⁢(i4⁢t4,c⁢i15⁢t5)⁢(i5⁢t4,c⁢i14⁢t5)⁢(i6⁢t4,c⁢i10⁢t5)⋅(i9⁢t4,c⁢i5⁢t5)⁢(i10⁢t4,c⁢it5)⁢(i11⁢t4,i05)⁢(i13⁢t4,c⁢i7⁢t5)⋅(i14⁢t4,c⁢i3⁢t5)⁢(i15⁢t4,c⁢i2⁢t5)⁢(i4⁢t5,i9⁢t5)⁢(i6⁢t5,i11⁢t5),

[(a⁢b)2,c]=𝑏(i01,c⁢i6⁢t8,c⁢it8,i7⁢t1)⁢(it1,i6⁢t1,c⁢i08,c⁢i7⁢t8)⋅(i2⁢t1,i4⁢t1,c⁢i5⁢t8,c⁢i3⁢t8)⁢(i3⁢t1,i5⁢t1,c⁢i4⁢t8,c⁢i2⁢t8)⋅(i8⁢t1,i15⁢t1,c⁢i9⁢t8,c⁢i14⁢t8)⁢(i9⁢t1,c⁢i15⁢t8,c⁢i8⁢t8,i14⁢t1)⋅(i10⁢t1,c⁢i11⁢t8,c⁢i13⁢t8,i12⁢t1)⁢(i11⁢t1,c⁢i10⁢t8,c⁢i12⁢t8,i13⁢t1)⋅(i02,c⁢i15⁢t7,c⁢it7,i14⁢t2)⁢(it2,c⁢i14⁢t7,c⁢i07,i15⁢t2)⋅(i2⁢t2,c⁢i11⁢t7,c⁢i5⁢t7,i12⁢t2)⁢(i3⁢t2,c⁢i10⁢t7,c⁢i4⁢t7,i13⁢t2)⋅(i4⁢t2,c⁢i13⁢t7,c⁢i3⁢t7,i10⁢t2)⁢(i5⁢t2,c⁢i12⁢t7,c⁢i2⁢t7,i11⁢t2)⋅(i6⁢t2,i8⁢t2,c⁢i7⁢t7,c⁢i9⁢t7)⁢(i7⁢t2,i9⁢t2,c⁢i6⁢t7,c⁢i8⁢t7)⋅(i03,i14⁢t3,c⁢it6,c⁢i15⁢t6)⁢(it3,i15⁢t3,c⁢i06,c⁢i14⁢t6)⋅(i2⁢t3,i12⁢t3,c⁢i5⁢t6,c⁢i11⁢t6)⁢(i3⁢t3,i13⁢t3,c⁢i4⁢t6,c⁢i10⁢t6)⋅(i4⁢t3,i10⁢t3,c⁢i3⁢t6,c⁢i13⁢t6)⁢(i5⁢t3,i11⁢t3,c⁢i2⁢t6,c⁢i12⁢t6)⋅(i6⁢t3,c⁢i9⁢t6,c⁢i7⁢t6,i8⁢t3)⁢(i7⁢t3,c⁢i8⁢t6,c⁢i6⁢t6,i9⁢t3)⋅(i04,c⁢i6⁢t5,c⁢it5,i7⁢t4)⁢(it4,i6⁢t4,c⁢i05,c⁢i7⁢t5)⋅(i2⁢t4,i4⁢t4,c⁢i5⁢t5,c⁢i3⁢t5)⁢(i3⁢t4,i5⁢t4,c⁢i4⁢t5,c⁢i2⁢t5)⋅(i8⁢t4,i15⁢t4,c⁢i9⁢t5,c⁢i14⁢t5)⁢(i9⁢t4,c⁢i15⁢t5,c⁢i8⁢t5,i14⁢t4)⋅(i10⁢t4,c⁢i11⁢t5,c⁢i13⁢t5,i12⁢t4)⁢(i11⁢t4,c⁢i10⁢t5,c⁢i12⁢t5,i13⁢t4),

=b𝑏(i01,ci15⁢t8,cit8,i14⁢t1)(it1,ci14⁢t8,ci08,i15⁢t1)

⋅(i2⁢t1,c⁢i11⁢t8,c⁢i5⁢t8,i12⁢t1)⁢(i3⁢t1,c⁢i10⁢t8,c⁢i4⁢t8,i13⁢t1)

⋅(i4⁢t1,c⁢i13⁢t8,c⁢i3⁢t8,i10⁢t1)⁢(i5⁢t1,c⁢i12⁢t8,c⁢i2⁢t8,i11⁢t1)

⋅(i6⁢t1,i8⁢t1,c⁢i7⁢t8,c⁢i9⁢t8)⁢(i7⁢t1,i9⁢t1,c⁢i6⁢t8,c⁢i8⁢t8)

⋅(i02,c⁢i6⁢t7,c⁢it7,i7⁢t2)⁢(it2,i6⁢t2,c⁢i07,c⁢i7⁢t7)

⋅(i2⁢t2,i4⁢t2,c⁢i5⁢t7,c⁢i3⁢t7)⁢(i3⁢t2,i5⁢t2,c⁢i4⁢t7,c⁢i2⁢t7)

⋅(i8⁢t2,i15⁢t2,c⁢i9⁢t7,c⁢i14⁢t7)⁢(i9⁢t2,c⁢i15⁢t7,c⁢i8⁢t7,i14⁢t2)

⋅(i10⁢t2,c⁢i11⁢t7,c⁢i13⁢t7,i12⁢t2)⁢(i11⁢t2,c⁢i10⁢t7,c⁢i12⁢t7,i13⁢t2)

⋅(i03,c⁢i6⁢t6,c⁢it6,i7⁢t3)⁢(it3,i6⁢t3,c⁢i06,c⁢i7⁢t6)

⋅(i2⁢t3,i4⁢t3,c⁢i5⁢t6,c⁢i3⁢t6)⁢(i3⁢t3,i5⁢t3,c⁢i4⁢t6,c⁢i2⁢t6)

⋅(i8⁢t3,i15⁢t3,c⁢i9⁢t6,c⁢i14⁢t6)⁢(i9⁢t3,c⁢i15⁢t6,c⁢i8⁢t6,i14⁢t3)

⋅(i10⁢t3,c⁢i11⁢t6,c⁢i13⁢t6,i12⁢t3)⁢(i11⁢t3,c⁢i10⁢t6,c⁢i12⁢t6,i13⁢t3)

⋅(i04,c⁢i14⁢t4,c⁢it5,c⁢i15⁢t5)⁢(it4,i15⁢t4,c⁢i05,c⁢i14⁢t5)

⋅(i2⁢t4,i12⁢t4,c⁢i5⁢t5,c⁢i11⁢t5)⁢(i3⁢t4,i13⁢t4,c⁢i4⁢t5,c⁢i10⁢t5)

⋅(i4⁢t4,i10⁢t4,c⁢i3⁢t5,c⁢i13⁢t5)⁢(i5⁢t4,i11⁢t4,c⁢i2⁢t5,c⁢i12⁢t5)

⋅(i6⁢t4,c⁢i9⁢t5,c⁢i7⁢t5,i8⁢t4)⁢(i7⁢t4,c⁢i8⁢t5,c⁢i6⁢t5,i9⁢t4).

Let A=〈a,b,c〉. Now one may see that [(a⁢b)2,c]b⁢c=[(a⁢b)2,c]b. By Proposition 2.5, [a,(b⁢c)2]=[(a⁢b)2,c]b⁢c and [a,(c⁢b)2]=[(a⁢b)2,c]b. It follows that [a,(b⁢c)2]=[a,(c⁢b)2], and hence [a,(b⁢c)4]=1. For 1≤j≤8, it is clear that (b⁢c)2n-6 interchanges i0j and c⁢it9-j as i0j+c⁢it9-j=2⁢t+1 (note that 1≤i0j≤t and t+1≤c⁢it9-j≤2⁢t), and similarly (b⁢c)2n-6 also interchanges i2⁢tj and c⁢i5⁢t9-j, i4⁢tj and c⁢i3⁢t9-j, i6⁢tj and c⁢i7⁢t9-j, i8⁢tj and c⁢i9⁢t9-j, i10⁢tj and c⁢i13⁢t9-j, i12⁢tj and c⁢i11⁢t9-j, and i14⁢tj and c⁢i15⁢t9-j. This implies

(b⁢c)2n-6=([(a⁢b)2,c]b)2=([(a⁢b)2,c]b⁢c)2=[a,(b⁢c)2]2.

It follows that the generators a,b,c of A satisfy the same relations as ρ0,ρ1,ρ2 in G, and hence A is isomorphic to G with order 2n.

Case 2: G=G7. We construct a permutation representation graph of G, and in this graph, bc consists of one path of length 32⁢t with alternating labels of c and b, where t=2n-5. Again, the graph is too big to be drawn in this paper.

Write ij⁢tk=j⁢t+16⁢i+k, where 0≤i≤t16-1, 1≤k≤16 and 0≤j≤1. The permutations a,b,c on the set {1,2,…,2n-5}, are as follows:

a=∏i=0t16-1(i05,it5)⁢(i06,it6)⁢(i07,it7)⁢(i08,it8)⁢(i09,it9)⁢(i010,it10)⁢(i011,it11)⁢(i012,it12)

b=(001)⁢(0t1)⁢((t16-1)016,(t16-1)t16)⋅𝑏(𝑏(ij⁢t2,ij⁢t3)(ij⁢t4,ij⁢t5)(ij⁢t6,ij⁢t7)(ij⁢t8,ij⁢t9)(ij⁢t10,ij⁢t11)⋅(ij⁢t12,ij⁢t13)(ij⁢t14,ij⁢t15)𝑏(ij⁢t16,(i+1)j⁢t1)),

c=𝑏𝑏(ij⁢t1,ij⁢t2)⁢(ij⁢t3,ij⁢t4)⁢(ij⁢t5,ij⁢t6)⁢(ij⁢t7,ij⁢t8)⋅(ij⁢t9,ij⁢t10)⁢(ij⁢t11,ij⁢t12)⁢(ij⁢t13,ij⁢t14)⁢(ij⁢t15,ij⁢t16).

Here (i+1)j⁢tk=j⁢t+16⁢(i+1)+k for 0≤i≤t16-2. It is easy to see that a is fixed under conjugacy of c, that is, ac=a. It follows that (a⁢c)2=1. Furthermore,

a⁢b=((t16-1)016,(t16-1)t16)⋅𝑏(i02,i03)(it2,it3)(i04,i05,it4,it5)(i06,it7)(i07,it6)(i08,it9)(i09,it8)⋅(i010,it11)⁢(i011,it10)⁢(i012,it13,it12,i013)⁢(i014,i015)⁢(it15,it14)⋅∏j=01∏i=0t16-2(ij⁢t16,(i+1)j⁢t1),

c⁢b=(1,3,5,…,t-1,2t,2t-2,…,t+2,t+1,t+3,…,2t-1,t,t-2,…,4,2)

The above computations imply (a⁢b)4=1 and (b⁢c)2n-5=1. Moreover, we have

(a⁢b)2=∏i=0t16-1(i04,it4)⁢(i05,it5)⁢(i012,it12)⁢(i013,it13),

c⁢(a⁢b)2⁢c=∏i=0t16-1(i03,it3)⁢(i06,it6)⁢(i011,it11)⁢(i014,it14),

[(a⁢b)2,c]=𝑏(i03,it3)⁢(i04,it4)⁢(i05,it5)⁢(i06,it6)⋅(i011,it11)⁢(i012,it12)⁢(i013,it13)⁢(i014,it14),

[(a⁢b)2,c]b=𝑏(i02,it2)⁢(i04,it4)⁢(i05,it5)⁢(i07,it7)⋅(i010,it10)⁢(i012,it12)⁢(i013,it13)⁢(i015,it15),

[(a⁢b)2,c]b⁢c=𝑏(i01,it1)⁢(i03,it3)⁢(i06,it6)⁢(i08,it8)⋅(i09,it9)⁢(i011,it11)⁢(i014,it14)⁢(i016,it16).

Let A=〈a,b,c〉. Since [a,c]=1, by Proposition 2.5, we have

[a,(b⁢c)2]=[(a⁢b)2,c]b⁢c,

and hence [a,(b⁢c)2]2=1. The element (b⁢c)2n-6 interchanges i0k and itk as

itk-i0k=t

(note that 1≤i0k≤t and t+1≤itk≤2⁢t), which implies

[(a⁢b)2,c]b⁢[(a⁢b)2,c]b⁢c=(b⁢c)2n-6.

Clearly, [(a⁢b)2,c]c=[(a⁢b)2,c]. Again, by Proposition 2.5,

[a,(b⁢c)4]=[a,(b⁢c)2]⁢[a,(b⁢c)2](b⁢c)2=[(a⁢b)2,c]b⁢c⁢[(a⁢b)2,c](b⁢c)3=([(a⁢b)2,c]c⁢b⁢[(a⁢b)2,c]b⁢c)(b⁢c)2=([(a⁢b)2,c]b⁢[(a⁢b)2,c]b⁢c)(b⁢c)2=(b⁢c)2n-6.

It follows that the generators a,b,c of A satisfy the same relations as ρ0,ρ1,ρ2 in G, and hence A is a quotient group of G. In particular, o⁢(b⁢c)=2n-5 in A, and hence o⁢(ρ1⁢ρ2)=2n-5 in G. It follows that

|G|=o⁢(ρ1⁢ρ2)8⋅|G/〈(ρ1⁢ρ2)8〉|=2n-8⋅256=2n.

Case 3: G=G8. We construct a permutation representation graph of G. In this graph, bc consists of two paths of length 2⁢t and a circle of length 4⁢t alternating labels of c and b, where t=2n-6. Again, the graph is too big to be drawn in this paper.

Write

ij⁢tk=j⁢t+16⁢i+k and c⁢ij⁢tk=j⁢t+16⁢(t16-i-1)+k,

where 0≤i≤t16-1, 1≤k≤16 and 0≤j≤7. The permutations a,b,c on the set {1,2,…,2n-3} are as follows:

a=𝑏(i01,i4⁢t1)⁢(it1,i5⁢t1)⁢(i2⁢t1,i6⁢t1)⁢(i3⁢t1,i7⁢t1)

⋅(i02,i4⁢t2)⁢(it2,i5⁢t2)⁢(i2⁢t2,i6⁢t2)⁢(i3⁢t2,i7⁢t2)

⋅(it3,i2⁢t3)⁢(i5⁢t3,i6⁢t3)⁢(i03,c⁢i4⁢t14)⁢(i3⁢t3,c⁢it14)

⋅(i4⁢t3,c⁢i6⁢t14)⁢(i7⁢t3,c⁢i3⁢t14)⁢(i014,i5⁢t14)⁢(i2⁢t14,i7⁢t14)

⋅(it4,i2⁢t4)⁢(i5⁢t4,i6⁢t4)⁢(i04,c⁢i4⁢t13)⁢(i3⁢t4,c⁢it13)

⋅(i4⁢t4,c⁢i6⁢t13)⁢(i7⁢t4,c⁢i3⁢t13)⁢(i013,i5⁢t13)⁢(i2⁢t13,i7⁢t13)

⋅(it5,i4⁢t5)⁢(i3⁢t5,i6⁢t5)⁢(i05,c⁢i2⁢t12)⁢(i2⁢t5,c⁢i6⁢t12)

⋅(i5⁢t5,c⁢it12)⁢(i7⁢t5,c⁢i5⁢t12)⁢(i012,i3⁢t12)⁢(i4⁢t12,i7⁢t12)

⋅(it6,i4⁢t6)⁢(i3⁢t6,i6⁢t6)⁢(i06,c⁢i2⁢t11)⁢(i2⁢t6,c⁢i6⁢t11)

⋅(i5⁢t6,c⁢it11)⁢(i7⁢t6,c⁢i5⁢t11)⁢(i011,i3⁢t11)⁢(i4⁢t11,i7⁢t11)

⋅(i07,c⁢i5⁢t10)⁢(it7,c⁢i4⁢t10)⁢(i2⁢t7,c⁢it10)⁢(i3⁢t7,c⁢i010)

⋅(i4⁢t7,c⁢i7⁢t10)⁢(i5⁢t7,c⁢i6⁢t10)⁢(i6⁢t7,c⁢i3⁢t10)⁢(i7⁢t7,c⁢i2⁢t10)

⋅(i08,c⁢i5⁢t9)⁢(it8,c⁢i4⁢t9)⁢(i2⁢t8,c⁢it9)⁢(i3⁢t8,c⁢i09)

⋅(i4⁢t8,c⁢i7⁢t9)⁢(i5⁢t8,c⁢i6⁢t9)⁢(i6⁢t8,c⁢i3⁢t9)⁢(i7⁢t8,c⁢i2⁢t9)

⋅(i015,i2⁢t15)⁢(it15,i3⁢t15)⁢(i4⁢t15,i6⁢t15)⁢(i5⁢t15,i7⁢t15)

⋅(i016,i2⁢t16)⁢(it16,i3⁢t16)⁢(i4⁢t16,i6⁢t16)⁢(i5⁢t16,i7⁢t16),

b=(001)⁢((t16-1)t16)⁢(06⁢t1)⁢((t16-1)7⁢t16)⁢(02⁢t1,04⁢t1)⁢((t16-1)3⁢t16,(t16-1)5⁢t16)⋅∏i=03((t16-1)2⁢t⁢i16,0(2⁢i+1)⁢t1)⋅𝑏(𝑏(ij⁢t2,ij⁢t3)(ij⁢t4,ij⁢t5)(ij⁢t6,ij⁢t7)(ij⁢t8,ij⁢t9)(ij⁢t10,ij⁢t11)⋅(ij⁢t12,ij⁢t13)(ij⁢t14,ij⁢t15)𝑏(ij⁢t16,(i+1)j⁢t1)),

c=𝑏𝑏(ij⁢t1,ij⁢t2)⁢(ij⁢t3,ij⁢t4)⁢(ij⁢t5,ij⁢t6)⁢(ij⁢t7,ij⁢t8)⋅(ij⁢t9,ij⁢t10)⁢(ij⁢t11,ij⁢t12)⁢(ij⁢t13,ij⁢t14)⁢(ij⁢t15,ij⁢t16).

Here (i+1)j⁢tk=j⁢t+16⁢(i+1)+k for 0≤i≤t16-2. For 0≤i≤t16-2, b interchanges ij⁢t16 and (i+1)j⁢t1, and also c⁢ij⁢t16 and c⁢(i-1)j⁢t1. It is easy to see that a is fixed under conjugacy of c, that is, ac=a. It follows that (a⁢c)2=1. Furthermore,

a⁢b=(001,02⁢t1,06⁢t1,04⁢t1)⁢(0t1,(t16-1)4⁢t16,07⁢t1,(t16-1)2⁢t16)

⋅(03⁢t1,(t16-1)6⁢t16,05⁢t1,(t16-1)016)

⋅((t16-1)t16,(t16-1)5⁢t16,(t16-1)7⁢t16,(t16-1)3⁢t16)

⋅𝑏((i+1)01,i4⁢t16,(i+1)6⁢t1,i2⁢t16)

⋅((i+1)t1,i5⁢t16,(i+1)7⁢t1,i3⁢t16)

⋅((i+1)2⁢t1,i6⁢t16,(i+1)4⁢t1,i016)

⋅((i+1)3⁢t1,i7⁢t16,(i+1)5⁢t1,it16)

⋅𝑏(it2,i5⁢t3,i6⁢t2,i2⁢t3)⁢(i2⁢t2,i6⁢t3,i5⁢t2,it3)

⋅(i02,i4⁢t3,ci6⁢t15,ci4⁢t14)(i3⁢t2,i7⁢t3,ci3⁢t15,cit14)

⋅(i4⁢t2,i03,ci4⁢t15,ci6⁢t14)(i7⁢t2,i3⁢t3,cit15,ci3⁢t14)

⋅(i014,i5⁢t15,i7⁢t14,i2⁢t15)(i2⁢t14,i7⁢t15,i5⁢t14,i015)

⋅(i04,ci4⁢t12,ci7⁢t13,ci2⁢t12)(it4,i2⁢t5,ci6⁢t13,i4⁢t5)

⋅(i2⁢t4,it5,i4⁢t4,ci6⁢t12)(i3⁢t4,cit12,i5⁢t4,i6⁢t5)

⋅(i6⁢t4,i5⁢t5,cit13,i3⁢t5)(i7⁢t4,ci3⁢t12,ci013,ci5⁢t12)

⋅(i05,ci2⁢t13,ci7⁢t12,ci4⁢t13)(i7⁢t5,ci5⁢t13,ci012,ci3⁢t13)

⋅(i06,ci2⁢t10,i7⁢t6,ci5⁢t10)(it6,i4⁢t7,ci7⁢t11,ci4⁢t10)

⋅(i2⁢t6,ci6⁢t10,i5⁢t6,cit10)(i3⁢t6,i6⁢t7,ci3⁢t11,ci010)

⋅(i4⁢t6,it7,ci4⁢t11,ci7⁢t10)(i6⁢t6,i3⁢t7,ci011,ci3⁢t10)

⋅(i07,ci5⁢t11,i7⁢t7,ci2⁢t11)(i2⁢t7,cit11,i5⁢t7,ci6⁢t11)

⋅(i08,ci5⁢t8,i6⁢t8,ci3⁢t8)(it8,ci4⁢t8,i7⁢t8,ci2⁢t8)

⋅(i09,ci3⁢t9,i6⁢t9,ci5⁢t9)(it9,ci2⁢t9,i7⁢t9,ci4⁢t9),

c⁢b=𝑏(1+6ti,3+6ti,…,2t-1+6ti,2t+6ti,2t-2+6ti,…,2+6ti)⋅(2t+1+2ti,2t+3+2ti,…,4t-1+2ti,6t-2ti,6t-2-2ti,…,4t+2-2ti),

(c⁢b)2n-6=∏i=0t-1(1+i,2⁢t-i)⁢(6⁢t+1+i,8⁢t-i)⁢∏i=02⁢t-1(2⁢t+1+i,6⁢t-i).

The above computations imply

(a⁢b)4=1,(b⁢c)2n-5=1 and (b⁢c)2n-6=(c⁢b)2n-6.

Furthermore,

(a⁢b)2=𝑏(i01,i6⁢t1)⁢(it1,i7⁢t1)⁢(i2⁢t1,i4⁢t1)⁢(i3⁢t1,i5⁢t1)

⋅(i016,i6⁢t16)⁢(it16,i7⁢t16)⁢(i2⁢t16,i4⁢t16)⁢(i3⁢t16,i5⁢t16)

⋅(it2,i6⁢t2)⁢(i2⁢t2,i5⁢t2)⁢(i02,c⁢i6⁢t15)⁢(i3⁢t2,c⁢i3⁢t15)

⋅(i4⁢t2,c⁢i4⁢t15)⁢(i7⁢t2,c⁢it15)⁢(i015,i7⁢t15)⁢(i2⁢t15,i5⁢t15)

⋅(it3,i6⁢t3)⁢(i2⁢t3,i5⁢t3)⁢(i03,c⁢i6⁢t14)⁢(i3⁢t3,c⁢i3⁢t14)

⋅(i4⁢t3,c⁢i4⁢t14)⁢(i7⁢t3,c⁢it14)⁢(i014,i7⁢t14)⁢(i2⁢t14,i5⁢t14)

⋅(i2⁢t4,i4⁢t4)⁢(i3⁢t4,i5⁢t4)⁢(i04,c⁢i7⁢t13)⁢(it4,c⁢i6⁢t13)

⋅(i6⁢t4,c⁢it13)⁢(i7⁢t4,c⁢i013)⁢(i2⁢t13,i4⁢t13)⁢(i3⁢t13,i5⁢t13)

⋅(i2⁢t5,i4⁢t5)⁢(i3⁢t5,i5⁢t5)⁢(i05,c⁢i7⁢t12)⁢(it5,c⁢i6⁢t12)

⋅(i6⁢t5,c⁢it12)⁢(i7⁢t5,c⁢i012)⁢(i2⁢t12,i4⁢t12)⁢(i3⁢t12,i5⁢t12)

⋅(i06,i7⁢t6)⁢(i2⁢t6,i5⁢t6)⁢(it6,c⁢i7⁢t11)⁢(i3⁢t6,c⁢i3⁢t11)

⋅(i4⁢t6,c⁢i4⁢t11)⁢(i6⁢t6,c⁢i011)⁢(it11,i6⁢t11)⁢(i2⁢t11,i5⁢t11)

⋅(i07,i7⁢t7)⁢(i2⁢t7,i5⁢t7)⁢(it7,c⁢i7⁢t10)⁢(i3⁢t7,c⁢i3⁢t10)

⋅(i4⁢t7,c⁢i4⁢t10)⁢(i6⁢t7,c⁢i010)⁢(it10,i6⁢t10)⁢(i2⁢t10,i5⁢t10)

⋅(i08,i6⁢t8)⁢(it8,i7⁢t8)⁢(i2⁢t8,i4⁢t8)⁢(i3⁢t8,i5⁢t8)

⋅(i09,i6⁢t9)⁢(it9,i7⁢t9)⁢(i2⁢t9,i4⁢t9)⁢(i3⁢t9,i5⁢t9)

c⁢(a⁢b)2⁢c=𝑏(it1,i6⁢t1)⁢(i2⁢t1,i5⁢t1)⁢(i01,c⁢i6⁢t16)⁢(i3⁢t1,c⁢i3⁢t16)⋅(i4⁢t1,c⁢i4⁢t16)⁢(i7⁢t1,c⁢it16)⁢(i016,i7⁢t16)⁢(i2⁢t16,i5⁢t16)⋅(i02,i6⁢t2)⁢(it2,i7⁢t2)⁢(i2⁢t2,i4⁢t2)⁢(i3⁢t2,i5⁢t2)⋅(i015,i6⁢t15)⁢(it15,i7⁢t15)⁢(i2⁢t15,i4⁢t15)⁢(i3⁢t15,i5⁢t15)⋅(i2⁢t3,i4⁢t3)⁢(i3⁢t3,i5⁢t3)⁢(i03,c⁢i7⁢t14)⁢(it3,c⁢i6⁢t14)⋅(i6⁢t3,c⁢it14)⁢(i7⁢t3,c⁢i014)⁢(i2⁢t14,i4⁢t14)⁢(i3⁢t14,i5⁢t14)⋅(it4,i6⁢t4)⁢(i2⁢t4,i5⁢t4)⁢(i04,c⁢i6⁢t13)⁢(i3⁢t4,c⁢i3⁢t13)⋅(i4⁢t4,c⁢i4⁢t13)⁢(i7⁢t4,c⁢it13)⁢(i013,i7⁢t13)⁢(i2⁢t13,i5⁢t13)⋅(i05,i7⁢t5)⁢(i2⁢t5,i5⁢t5)⁢(it5,c⁢i7⁢t12)⁢(i3⁢t5,c⁢i3⁢t12)⋅(i4⁢t5,c⁢i4⁢t12)⁢(i6⁢t5,c⁢i012)⁢(it12,i6⁢t12)⁢(i2⁢t12,i5⁢t12)⋅(i2⁢t6,i4⁢t6)⁢(i3⁢t6,i5⁢t6)⁢(i06,c⁢i7⁢t11)⁢(it6,c⁢i6⁢t11)⋅(i6⁢t6,c⁢it11)⁢(i7⁢t6,i011)⁢(i2⁢t11,i4⁢t11)⁢(i3⁢t11,i5⁢t11)⋅(i07,i6⁢t7)⁢(it7,i7⁢t7)⁢(i2⁢t7,i4⁢t7)⁢(i3⁢t7,i5⁢t7)⋅(i010,i6⁢t10)⁢(it10,i7⁢t10)⁢(i2⁢t10,i4⁢t10)⁢(i3⁢t10,i5⁢t10)⋅(i08,i7⁢t8)⁢(i2⁢t8,i5⁢t8)⁢(it8,c⁢i7⁢t9)⁢(i3⁢t8,c⁢i3⁢t9)⋅(i4⁢t8,c⁢i4⁢t9)⁢(i6⁢t8,c⁢i09)⁢(it9,i6⁢t9)⁢(i2⁢t9,i5⁢t9)

[(a⁢b)2,c]=𝑏(i01,it1,c⁢it16,c⁢i016)⁢(i2⁢t1,c⁢i4⁢t16,c⁢i5⁢t16,i3⁢t1)⋅(i4⁢t1,i5⁢t1,c⁢i3⁢t16,c⁢i2⁢t16)⁢(i6⁢t1,c⁢i6⁢t16,c⁢i7⁢t16,i7⁢t1)⋅(i02,c⁢i015,c⁢it15,it2)⁢(i2⁢t2,i3⁢t2,c⁢i5⁢t15,c⁢i4⁢t15)⋅(i4⁢t2,c⁢i2⁢t15,c⁢i3⁢t15,i5⁢t2)⁢(i6⁢t2,i7⁢t2,c⁢i7⁢t15,c⁢i6⁢t15)⋅(i03,it3,c⁢it14,c⁢i014)⁢(i2⁢t3,i3⁢t3,c⁢i5⁢t14,c⁢i4⁢t14)⋅(i4⁢t3,c⁢i2⁢t14,c⁢i3⁢t14,i5⁢t3)⁢(i6⁢t3,c⁢i6⁢t14,c⁢i7⁢t14,i7⁢t3)⋅(i04,c⁢i013,c⁢it13,it4)⁢(i2⁢t4,c⁢i4⁢t13,c⁢i5⁢t13,i3⁢t4)⋅(i4⁢t4,i5⁢t4,c⁢i3⁢t13,c⁢i2⁢t13)⁢(i6⁢t4,i7⁢t4,c⁢i7⁢t13,c⁢i6⁢t13)⋅(i05,it5,c⁢it12,c⁢i012)⁢(i2⁢t5,c⁢i4⁢t12,c⁢i5⁢t12,i3⁢t5)⋅(i4⁢t5,i5⁢t5,c⁢i3⁢t12,c⁢i2⁢t12)⁢(i6⁢t5,c⁢i6⁢t12,c⁢i7⁢t12,i7⁢t5)⋅(i06,c⁢i011,c⁢it11,it6)⁢(i2⁢t6,i3⁢t6,c⁢i5⁢t11,c⁢i4⁢t11)⋅(i4⁢t6,c⁢i2⁢t11,c⁢i3⁢t11,i5⁢t6)⁢(i6⁢t6,i7⁢t6,c⁢i7⁢t11,c⁢i6⁢t11)⋅(i07,it7,c⁢it10,c⁢i010)⁢(i2⁢t7,i3⁢t7,c⁢i5⁢t10,c⁢i4⁢t10)⋅(i4⁢t7,c⁢i2⁢t10,c⁢i3⁢t10,i5⁢t7)⁢(i6⁢t7,c⁢i6⁢t10,c⁢i7⁢t10,i7⁢t7)⋅(i08,c⁢i09,c⁢it9,it8)⁢(i2⁢t8,c⁢i4⁢t9,c⁢i5⁢t9,i3⁢t8)⋅(i4⁢t8,i5⁢t8,c⁢i3⁢t9,c⁢i2⁢t9)⁢(i6⁢t8,i7⁢t8,c⁢i7⁢t9,c⁢i6⁢t9),

=b𝑏(i01,ci016,cit16,it1)(i2⁢t1,ci4⁢t16,ci5⁢t16,i3⁢t1)

⋅(i4⁢t1,i5⁢t1,c⁢i3⁢t16,c⁢i2⁢t16)⁢(i6⁢t1,i7⁢t1,c⁢i7⁢t16,c⁢i6⁢t16)

⋅(i02,it2,c⁢it15,c⁢i015)⁢(i2⁢t2,i3⁢t2,c⁢i5⁢t15,c⁢i4⁢t15)

⋅(i4⁢t2,c⁢i2⁢t15,c⁢i3⁢t15,i5⁢t2)⁢(i6⁢t2,c⁢i6⁢t15,c⁢i7⁢t15,i7⁢t2)

⋅(i03,c⁢i014,c⁢it14,it3)⁢(i2⁢t3,i3⁢t3,c⁢i5⁢t14,c⁢i4⁢t14)

⋅(i4⁢t3,c⁢i2⁢t14,c⁢i3⁢t14,i5⁢t3)⁢(i6⁢t3,i7⁢t3,c⁢i7⁢t14,c⁢i6⁢t14)

⋅(i04,it4,c⁢it13,c⁢i013)⁢(i2⁢t4,c⁢i4⁢t13,c⁢i5⁢t13,i3⁢t4)

⋅(i4⁢t4,i5⁢t4,c⁢i3⁢t13,c⁢i2⁢t13)⁢(i6⁢t4,c⁢i6⁢t13,c⁢i7⁢t13,i7⁢t4)

⋅(i05,c⁢i012,c⁢it12,it5)⁢(i2⁢t5,c⁢i4⁢t12,c⁢i5⁢t12,i3⁢t5)

⋅(i4⁢t5,i5⁢t5,c⁢i3⁢t12,c⁢i2⁢t12)⁢(i6⁢t5,i7⁢t5,c⁢i7⁢t12,c⁢i6⁢t12)

⋅(i06,it6,c⁢it11,c⁢i011)⁢(i2⁢t6,i3⁢t6,c⁢i5⁢t11,c⁢i4⁢t11)

⋅(i4⁢t6,c⁢i2⁢t11,c⁢i3⁢t11,i5⁢t6)⁢(i6⁢t6,c⁢i6⁢t11,c⁢i7⁢t11,i7⁢t6)

⋅(i07,c⁢i010,c⁢it10,it7)⁢(i2⁢t7,i3⁢t7,c⁢i5⁢t10,c⁢i4⁢t10)

⋅(i4⁢t7,c⁢i2⁢t10,c⁢i3⁢t10,i5⁢t7)⁢(i6⁢t7,i7⁢t7,c⁢i7⁢t10,c⁢i6⁢t10)

⋅(i08,it8,c⁢it9,c⁢i09)⁢(i2⁢t8,c⁢i4⁢t9,c⁢i5⁢t9,i3⁢t8)

⋅(i4⁢t8,i5⁢t8,c⁢i3⁢t9,c⁢i2⁢t9)⁢(i6⁢t8,c⁢i6⁢t9,c⁢i7⁢t9,i7⁢t8),

Let A=〈a,b,c〉. Now it is easy to see that

[(a⁢b)2,c]c=[(a⁢b)2,c]-1 and [(a⁢b)2,c]b⁢c=([(a⁢b)2,c]b)-1.

Every 4-cycle in the product of distinct 4-cycles of [(a⁢b)2,c] is either a 4-cycle or the inverse of a 4-cycle in [(a⁢b)2,c]b, and [(a⁢b)2,c]⁢[(a⁢b)2,c]b is an involution, which fixes 2n-4 points including the point 1. Then

[(a⁢b)2,c]⁢[(a⁢b)2,c]b=[(a⁢b)2,c]b⁢[(a⁢b)2,c].

It is clear that (c⁢b)2n-6 interchanges i0j and c⁢it16-j for each 1≤j≤16 because i0j+c⁢it16-j=2⁢t+1 (note that 1≤i0j≤t and t+1≤c⁢it16-j≤2⁢t), and similarly (c⁢b)2n-6 also interchanges i2⁢tj and c⁢i5⁢t16-j, i3⁢tj and c⁢i4⁢t16-j, and i6⁢tj and c⁢i7⁢t16-j. It follows that (c⁢b)2n-6=[(a⁢b)2,c]2=([(a⁢b)2,c]b)2.

Since [a,c]=1, by Proposition 2.5, we have

[a,(b⁢c)2]2=([(a⁢b)2,c]b⁢c)2=([(a⁢b)2,c]b)-2=(c⁢b)-2n-6=(b⁢c)2n-6,

[a,(b⁢c)4]=[a,(b⁢c)2]⁢[a,(b⁢c)2](b⁢c)2=([(a⁢b)2,c]⁢[(a⁢b)2,c]b⁢c⁢b⁢c)b⁢c=([(a⁢b)2,c]⁢([(a⁢b)2,c]-b)b⁢c)b⁢c=([(a⁢b)2,c]2)b⁢c=((c⁢b)2n-6)b⁢c=(b⁢c)2n-6.

This implies that the generators a,b,c of A satisfy the same relations as ρ0,ρ1,ρ2 in G, and hence A is a quotient group of G. In particular, o⁢(c⁢b)=2n-5 in A, and hence o⁢(ρ1⁢ρ2)=2n-5 in G. It follows that

|G|=o⁢(ρ1⁢ρ2)8⋅|G/〈(ρ1⁢ρ2)8〉|=2n-8⋅256=2n.

Now we prove the necessity. Let (G,{ρ0,ρ1,ρ2}) be a string C-group of rank 3 with type {4,2n-5} and |G|=2n. Then each of ρ0,ρ1 and ρ2 has order 2, and we further have o⁢(ρ0⁢ρ1)=4, o⁢(ρ0⁢ρ2)=2 and o⁢(ρ1⁢ρ2)=2n-5. To finish the proof, we only need to prove G≅G5,G6,G7 or G8. Since G5,G6,G7 and G8 are C-groups of order 2n of type {4,2n-5}, it suffices to show that, in G,

=21or[ρ0,(ρ2ρ1)2]2(ρ1ρ2)2n-6 =1,

[ρ0,(ρ2⁢ρ1)4]=1or[ρ0,(ρ2ρ1)4](ρ1ρ2)2n-6 =1,

which will be done by induction on n. This is true for n=10 by Magma.

Assume n≥11. Take N=〈(ρ1⁢ρ2)2n-6〉. By Lemma 4.1, we have N⊴G and (G¯=G/N,{ρ0¯,ρ1¯,ρ2¯}) (with ρi¯=N⁢ρi) is a string C-group of rank 3 of type {4,2n-6}. Since |G¯|=2n-1, by induction hypothesis, we may assume G¯=G¯5, G¯6, G¯7 or G¯8, where

G¯5=〈ρ0¯,ρ1¯,ρ2¯∣ρ0¯2,ρ1¯2,ρ2¯2,(ρ0¯⁢ρ1¯)4,(ρ1¯⁢ρ2¯)2n-6,[ρ0¯,(ρ1¯ρ2¯)2]2,[ρ0¯,(ρ1¯ρ2¯)4]〉,

G¯6=〈ρ0¯,ρ1¯,ρ2¯∣ρ0¯2,ρ1¯2,ρ2¯2,(ρ0¯⁢ρ1¯)4,(ρ1¯⁢ρ2¯)2n-6,[ρ0¯,(ρ1¯ρ2¯)2]2(ρ1¯ρ2¯)2n-7,[ρ0¯,(ρ1¯ρ2¯)4]〉,

G¯7=〈ρ0¯,ρ1¯,ρ2¯∣ρ0¯2,ρ1¯2,ρ2¯2,(ρ0¯⁢ρ1¯)4,(ρ1¯⁢ρ2¯)2n-6,[ρ0¯,(ρ1¯ρ1¯)2]2,[ρ0¯,(ρ1¯ρ2¯)4](ρ1¯ρ2¯)2n-7〉,

G¯8=〈ρ0¯,ρ1¯,ρ2¯∣ρ0¯2,ρ1¯2,ρ2¯2,(ρ0¯⁢ρ1¯)4,(ρ1¯⁢ρ2¯)2n-6,[ρ0¯,(ρ1¯ρ2¯)2]2(ρ1¯ρ2¯)2n-7,[ρ0¯,(ρ1¯ρ2¯)4](ρ1¯ρ2¯)2n-7〉.

Then

[ρ0¯,(ρ1¯⁢ρ2¯)4]=1 or [ρ0¯,(ρ1¯⁢ρ2¯)4]⁢(ρ1¯⁢ρ2¯)2n-7=1,

and since N=〈(ρ1⁢ρ2)2n-6〉≅ℤ2, we have [ρ0,(ρ1⁢ρ2)4]=(ρ1⁢ρ2)δ⋅2n-7, where δ=0,±1,2. It follows that

[ρ0,(ρ1⁢ρ2)8]=[ρ0,(ρ1⁢ρ2)4]⁢[ρ0,(ρ1⁢ρ2)4](ρ1⁢ρ2)4=(ρ1⁢ρ2)δ⋅2n-6,

and similarly [ρ0,(ρ1⁢ρ2)16]=(ρ1⁢ρ2)δ⋅2n-5=1. Since n≥11, we have

[ρ0,(ρ1⁢ρ2)2n-7]=1,

that is, ((ρ1⁢ρ2)2n-7)ρ0=(ρ1⁢ρ2)2n-7.

Suppose G¯=G¯7 or G¯8. Then [ρ0¯,(ρ1¯⁢ρ2¯)4]⁢(ρ1¯⁢ρ2¯)2n-7=1, that is,

[ρ0,(ρ1⁢ρ2)4]=(ρ1⁢ρ2)γ⋅2n-7 for⁢γ=±1.

It follows that

1=[ρ02,(ρ1⁢ρ2)4]=[ρ0,(ρ1⁢ρ2)4]ρ0⁢[ρ0,(ρ1⁢ρ2)4]=((ρ1⁢ρ2)γ⋅2n-7)ρ0⁢(ρ1⁢ρ2)γ⋅2n-7=(ρ1⁢ρ2)γ⋅2n-6,

which contradicts o⁢(ρ1⁢ρ2)=2n-5.

Suppose G¯=G¯6. Then [ρ0¯,(ρ1¯⁢ρ2¯)2]2⁢(ρ1¯⁢ρ2¯)2n-7=1, that is,

[ρ0,(ρ1⁢ρ2)2]2=(ρ1⁢ρ2)γ⁢2n-7 for⁢γ=±1.

Recall that ((ρ1⁢ρ2)γ⁢2n-7)ρ0=(ρ1⁢ρ2)γ⁢2n-7. On the other hand,

1=[ρ02,(ρ1⁢ρ2)2]=[ρ0,(ρ1⁢ρ2)2]⁢[ρ0,(ρ1⁢ρ2)2]ρ0,

and hence

([ρ0,(ρ1⁢ρ2)2]2)ρ0=([ρ0,(ρ1⁢ρ2)2]ρ0)2=([ρ0,(ρ1⁢ρ2)2]2)-1,

that is, ((ρ1⁢ρ2)γ⁢2n-7)ρ0=(ρ1⁢ρ2)-γ⁢2n-7. It follows that

(ρ1⁢ρ2)γ⁢2n-7=(ρ1⁢ρ2)-γ⁢2n-7 and (ρ1⁢ρ2)γ⁢2n-6=1,

a contradiction.

Thus G¯=G¯5. Since N=〈(ρ1⁢ρ2)2n-6〉≅ℤ2, we have [ρ0,(ρ1⁢ρ2)2]2=1 or (ρ1⁢ρ2)2n-6 and [ρ0,(ρ1⁢ρ2)4]=1 or (ρ1⁢ρ2)2n-6. It follows that G≅G5,G6,G7 or G8. ∎

Communicated by Andrea Lucchini

Funding statement: This work was supported by the National Natural Science Foundation of China (11571035, 11731002) and the 111 Project of China (B16002).

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Received: 2018-07-30

Revised: 2018-11-07

Published Online: 2019-01-30

Published in Print: 2019-07-01

Articles in the same Issue

https://doi.org/10.1515/jgth-2018-0155