3Nets realizing a group in a projective plane
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Abstract
In a projective plane \(\mathit{PG}(2,\mathbb{K})\) defined over an algebraically closed field \(\mathbb{K}\) of characteristic 0, we give a complete classification of 3nets realizing a finite group. An infinite family, due to Yuzvinsky (Compos. Math. 140:1614–1624, 2004), arises from plane cubics and comprises 3nets realizing cyclic and direct products of two cyclic groups. Another known infinite family, due to Pereira and Yuzvinsky (Adv. Math. 219:672–688, 2008), comprises 3nets realizing dihedral groups. We prove that there is no further infinite family. Urzúa’s 3nets (Adv. Geom. 10:287–310, 2010) realizing the quaternion group of order 8 are the unique sporadic examples.
If p is larger than the order of the group, the above classification holds in characteristic p>0 apart from three possible exceptions \(\rm{Alt}_{4}\), \(\rm{Sym}_{4}\), and \(\rm{Alt}_{5}\).
Motivation for the study of finite 3nets in the complex plane comes from the study of complex line arrangements and from resonance theory; see (Falk and Yuzvinsky in Compos. Math. 143:1069–1088, 2007; Miguel and Buzunáriz in Graphs Comb. 25:469–488, 2009; Pereira and Yuzvinsky in Adv. Math. 219:672–688, 2008; Yuzvinsky in Compos. Math. 140:1614–1624, 2004; Yuzvinsky in Proc. Am. Math. Soc. 137:1641–1648, 2009).
Keywords
3Net Dual 3net Projective plane Embedding Cubic curve1 Introduction
In a projective plane a 3net consists of three pairwise disjoint classes of lines such that every point incident with two lines from distinct classes is incident with exactly one line from each of the three classes. If one of the classes has finite size, say n, then the other two classes also have size n, called the order of the 3net.
In the present paper, combinatorial methods are used to investigate finite 3nets realizing a group. Since key examples, such as algebraic 3nets and tetrahedron type 3nets, arise naturally in the dual plane of \(\mathit{PG}(2,\mathbb{K})\), it is convenient to work with the dual concept of a 3net.
Formally, a dual 3net of order n in \(\mathit{PG}(2,\mathbb{K})\) consists of a triple (Λ _{1},Λ _{2},Λ _{3}) with Λ _{1},Λ _{2},Λ _{3} pairwise disjoint point sets of size n, called components, such that every line meeting two distinct components meets each component in precisely one point. A dual 3net (Λ _{1},Λ _{2},Λ _{3}) realizing a group is algebraic if its points lie on a plane cubic and is of tetrahedron type if its components lie on the six sides (diagonals) of a nondegenerate quadrangle in such a way that Λ _{ i }=Δ _{ i }∪Γ _{ i } with Δ _{ i } and Γ _{ i } lying on opposite sides for i=1,2,3.
The goal of this paper is to prove the following classification theorem.
Theorem 1
 (I)
G is either cyclic or the direct product of two cyclic groups, and (Λ _{1},Λ _{2},Λ _{3}) is algebraic.
 (II)
G is dihedral, and (Λ _{1},Λ _{2},Λ _{3}) is of tetrahedron type.
 (III)
G is the quaternion group of order 8.
 (IV)
G has order 12 and is isomorphic to \(\rm{Alt}_{4}\).
 (V)
G has order 24 and is isomorphic to \(\rm{Sym}_{4}\).
 (VI)
G has order 60 and is isomorphic to \(\rm{Alt}_{5}\).
A computer aided exhaustive search shows that if p=0, then (IV) (and hence (V), (VI)) does not occur; see [14].
Theorem 1 shows that every realizable finite group can act in \(\mathit{PG}(2,\mathbb{K})\) as a projectivity group. This confirms Yuzvinsky’s conjecture for p=0.
The proof of Theorem 1 uses some previous results due to Yuzvinsky [18], Urzúa [16], and Blokhuis, Korchmáros, and Mazzocca [2].
Our notation and terminology are standard; see [8]. In view of Theorem 1, \(\mathbb{K}\) denotes an algebraically closed field of characteristic p where either p=0 or p≥5, and any dual 3net in the present paper is supposed to have order n with n<p whenever p>0.
2 Some useful results on plane cubics
Proposition 1
[7, Theorem 6.104]
A nonsingular plane cubic \({\mathcal{F}}\) can be equipped with an additive group \((\mathcal{F},+)\) on the set of all its points. If an inflection point P _{0} of \({\mathcal{F}}\) is chosen to be the identity 0, then three distinct points \(P,Q,R\in\mathcal{F}\) are collinear if and only if P+Q+R=0. For a prime number d≠p, the subgroup of \((\mathcal{F},+)\) consisting of all elements g with dg=0 is isomorphic to C _{ d }×C _{ d }, while for d=p, it is either trivial or isomorphic to C _{ p } according as \(\mathcal{F}\) is supersingular or not.
Proposition 2
[17, Proposition 5.6, (1)]
Let \(\mathcal{F}\) be an irreducible singular plane cubic with its unique singular point U, and define the operation + on \(\mathcal{F}\setminus\{U\}\) in exactly the same way as on a nonsingular plane cubic. Then \((\mathcal{F},+)\) is an Abelian group isomorphic to the additive group of \(\mathbb{K}\) or to the multiplicative group of \(\mathbb{K}\), according as P is a cusp or a node.
If P is a nonsingular and noninflection point of \(\mathcal{F}\), then the tangent to \(\mathcal{F}\) at P meets \(\mathcal{F}\) in a point P′ other than P, and P′ is the tangential point of P. Every inflection point of a nonsingular cubic \(\mathcal{F}\) is the center of an involutory homology preserving \(\mathcal{F}\). A classical Lame configuration consists of two triples of distinct lines in \(\mathit{PG}(2,\mathbb{K})\), say ℓ _{1},ℓ _{2},ℓ _{3} and r _{1},r _{2},r _{3}, such that no line from one triple passes through the common point of two lines from the other triple. For 1≤j,k≤3, let R _{ jk } denote the common point of the lines ℓ _{ j } and r _{ k }. There are nine such common points, and they are called the points of the Lame configuration.
Proposition 3
(Lame’s Theorem)
If eight points from a Lame configuration lie on a plane cubic, then the ninth also does.
3 3Nets, quasigroups and loops
A latin square of order n is a table with n rows and n columns which has n ^{2} entries with n different elements none of them occurring twice within any row or column. If (L,∗) is a quasigroup of order n, then its multiplicative table, also called Cayley table, is a latin square of order n, and the converse also holds.
For two integers k,n both bigger than 1, let (G,⋅) be a group of order kn containing a normal subgroup (H,⋅) of order n. Let \(\mathcal{G}\) be a Cayley table of (G,⋅). Obviously, the rows and the columns representing the elements of (H,⋅) in \(\mathcal{G}\) form a latin square that is a Cayley table for (H,⋅). From \(\mathcal{G}\) we may extract k ^{2}−1 more latin squares using the cosets of H in G. In fact, for any two such cosets H _{1} and H _{2}, a latin square H _{1,2} is obtained by taking as rows (respectively columns) the elements of H _{1} (respectively H _{2}).
Proposition 4
The latin square H _{1,2} is a Cayley table for a quasigroup isotopic to the group H.
Proof
Fix an element t _{1}∈H _{1}. In H _{1,2}, label the row representing the element h _{1}∈H _{1} with \(h_{1}'\in H\) where \(h_{1}=t_{1}\cdot h_{1}'\). Similarly, for a fixed element t _{2}∈H _{2}, label the column representing the element h _{2}∈H _{2} with \(h_{2}'\in H\) where \(h_{2}=h_{2}'\cdot t_{2}\). The entries in H _{1,2} come from the coset H _{1}⋅H _{2}. Now, label the entry h _{3} in H _{1}⋅H _{2} with the element \(h_{3}'\in H\) where \(h_{3}=t_{1} \cdot h_{3}'\cdot t_{2}\). Doing so, H _{1,2} becomes a Cayley table for the subgroup (H,⋅), whence the assertion follows. □
In terms of a dual 3net, the relationship between 3nets and quasigroups can be described as follows. Let (L,⋅) be a loop arising from an embeddable 3net, and consider its dual 3net with its components Λ _{1},Λ _{2},Λ _{3}. For i=1,2,3, the points in Λ _{ i } are bijectively labeled by the elements of L. Let (A _{1},A _{2},A _{3}) with A _{ i }∈Λ _{ i } denote the triple of the points corresponding to the element a∈L. With this notation, a⋅b=c holds in L if and only if the points A _{1},B _{2}, and C _{3} are collinear. In this way, points in Λ _{3} are naturally labeled when a⋅b is the label of C _{3}. Let (E _{1},E _{2},E _{3}) be the triple for the unit element e of L. From e⋅e=e, the points E _{1},E _{2}, and E _{3} are collinear. Since a⋅a=a only holds for a=e, the points A _{1},A _{2},A _{3} are the vertices of a (nondegenerate) triangle whenever a≠e. Furthermore, from e⋅a=a, the points E _{1}, A _{2}, and A _{3} are collinear; similarly, a⋅e=a yields that the points A _{1}, E _{2}, and A _{3} are collinear. However, the points A _{1},A _{2}, and E _{3} form a triangle in general; they are collinear if and only if a⋅a=e, i.e., a is an involution of L.
In some cases, it is useful to relabel the points of Λ _{3} replacing the above bijection A _{3}→a from Λ _{3} to L by the bijection A _{3}→a′ where a′ is the inverse of a in (L,⋅). Doing so, three points A _{1},B _{2},C _{3} with A _{1}∈Λ _{1}, B _{2}∈Λ _{2}, C _{3}∈Λ _{3} are collinear if and only if a⋅b⋅c=e with e being the unit element in (L,⋅). This new bijective labeling will be called a collinear relabeling with respect to Λ _{3}.
Let (Λ _{1},Λ _{2},Λ _{3}) be a dual 3net that realizes a group (G,⋅) of order kn containing a subgroup (H,⋅) of order n. Then the left cosets of H provide a partition of each component Λ _{ i } into k subsets. Such subsets are called left Hmembers and denoted by \(\varGamma_{i}^{(1)},\ldots ,\varGamma_{i}^{(k)}\), or simply Γ _{ i } when this does not cause confusion. The left translation map σ _{ g }:x↦x+g preserves every left Hmember. The following lemma shows that every left Hmember Γ _{1} determines a dual 3subnet of (Λ _{1},Λ _{2},Λ _{3}) that realizes H.
Lemma 1
Let (Λ _{1},Λ _{2},Λ _{3}) be a dual 3net that realizes a group (G,⋅) of order kn containing a subgroup (H,⋅) of order n. For any left coset g⋅H of H in G, let Γ _{1}=g⋅H, Γ _{2}=H, and Γ _{3}=g⋅H. Then (Γ _{1},Γ _{2},Γ _{3}) is a 3subnet of (Λ _{1},Λ _{2},Λ _{3}) that realizes H.
Proof
For any h _{1},h _{2}∈H, we have that (g⋅h _{1})⋅h _{2}=g⋅(h _{1}⋅h _{2})=g⋅h with h∈H. Hence, any line joining a point of Γ _{1} with a point of Γ _{2} meets Γ _{3}. □
Similar results hold for right cosets of H. Therefore, for any right coset H⋅g, the triple (Γ _{1},Γ _{2},Γ _{3}) with Γ _{1}=H, Γ _{2}=H⋅g, and Γ _{3}=H⋅g is a 3subnet of (Λ _{1},Λ _{2},Λ _{3}) that realizes H.
The dual 3subnets (Γ _{1},Γ _{2},Γ _{3}) introduced in Lemma 1 play a relevant role. When g ranges over G, we obtain k such dual 3nets, each being called a dual 3net realizing the subgroup H as a subgroup of G.
Obviously, left cosets and right cosets coincide if and only if H is a normal subgroup of G, and if this is the case, we may use the shorter term of coset.
Now assume that H is a normal subgroup of G. Take two Hmembers from different components, say Γ _{ i } and Γ _{ j } with 1≤i<j≤3. By Proposition 4, there exists a member Γ _{ m } from the remaining component Λ _{ m }, with 1≤m≤3 and m≠i,j, such that (Γ _{1},Γ _{2},Γ _{3}) is a dual 3net of realizing (H,⋅). Doing so, we obtain k ^{2} dual 3subnets of (Λ _{1},Λ _{2},Λ _{3}). They are all the dual 3subnets of (Λ _{1},Λ _{2},Λ _{3}) that realize the normal subgroup (H,⋅) as a subgroup of (G,⋅).
Lemma 2
Let (Λ _{1},Λ _{2},Λ _{3}) be a dual 3net that realizes a group (G,⋅) of order kn containing a normal subgroup (H,⋅) of order n. For any two cosets g _{1}⋅H and g _{2}⋅H of H in G, let Γ _{1}=g _{1}⋅H, Γ _{2}=g _{2}⋅H, and Γ _{3}=(g _{1}⋅g _{2})⋅H. Then (Γ _{1},Γ _{2},Γ _{3}) is a 3subnet of (Λ _{1},Λ _{2},Λ _{3}) that realizes H.
If g _{1} and g _{2} range independently over G, we obtain k ^{2} such dual 3nets, each being called a dual 3net realizing the normal subgroup H as a subgroup of G.
4 The infinite families of dual 3nets realizing a group
A dual 3net (Λ _{1},Λ _{2},Λ _{3}) with n≥4 is said to be algebraic if all its points lie on a (uniquely determined) plane cubic \(\mathcal{F}\), called the associated plane cubic of (Λ _{1},Λ _{2},Λ _{3}). Algebraic dual 3nets fall into three subfamilies according as the plane cubic splits into three lines, or in an irreducible conic and a line, or it is irreducible.
4.1 Proper algebraic dual 3nets
An algebraic dual 3net (Λ _{1},Λ _{2},Λ _{3}) is said to be proper if its points lie on an irreducible plane cubic \(\mathcal{F}\).
Proposition 5
Any proper algebraic dual 3net (Λ _{1},Λ _{2},Λ _{3}) realizes a group M. There is a subgroup T≅M in \((\mathcal{F},+)\) such that each component Λ _{ i } is a coset T+g _{ i } in \((\mathcal{F},+)\) where g _{1}+g _{2}+g _{3}=0.
Proof
Now we are in a position to prove that Λ _{1} is a coset of a subgroup of \((\mathcal{F},+)\). For A _{0}∈Λ _{1}, let T _{1}={A−A _{0}∣A∈Λ _{1}}. Since (A _{1}−A _{0})−(A _{2}−A _{0})=A _{1}−A _{2}, Eq. (2) ensures the existence of A ^{∗}∈Λ _{1} for which A _{1}−A _{2}=A ^{∗}−A _{0} whenever A _{1},A _{2}∈Λ _{1}. Hence, (A _{1}−A _{0})−(A _{2}−A _{0})∈T _{1}. From this we have that T _{1} is a subgroup of \((\mathcal{F},+)\), and therefore Λ _{1} is a coset T+g _{1} of T _{1} in \((\mathcal{F},+)\).
Similarly, Λ _{2}=T _{2}+g _{2} and Λ _{3}=T _{3}+g _{3} with some subgroups T _{2},T _{3} of \((\mathcal{F},+)\) and elements \(g_{2},g_{3}\in (\mathcal{F},+)\). It remains to show that T _{1}=T _{2}=T _{3}. The line through the points g _{1} and g _{2} meets Λ _{3} in a point t ^{∗}+g _{3}. Replacing g _{3} with g _{3}+t ^{∗} allows us to assume that g _{1}+g _{2}+g _{3}=0. Then three points g _{ i }+t _{ i } with t _{ i }∈T _{ i } are collinear if and only if t _{1}+t _{2}+t _{3}=0. For t _{3}=0, this yields t _{2}=−t _{1}. Hence, every element of T _{2} is in T _{1}, and the converse also holds. Hence, T _{1}=T _{2}. Now, t _{3}=−t _{1}−t _{2} yields that T _{3}=T _{1}. Therefore, T=T _{1}=T _{2}=T _{3} and Λ _{ i }=T+g _{ i } for i=1,2,3. This shows that (Λ _{1},Λ _{2},Λ _{3}) realizes a group M≅T. □
4.2 Triangular dual 3nets
An algebraic dual 3net (Λ _{1},Λ _{2},Λ _{3}) is regular if the components lie on three lines, and it is either of pencil type or triangular according as the three lines are either concurrent or they are the sides of a triangle.
Lemma 3
Every regular dual 3net of order n is triangular.
Proof
For a triangular dual 3net, the (uniquely determined) triangle whose sides contain the components is called the associated triangle.
Proposition 6
Every triangular dual 3net realizes a cyclic group isomorphic to a multiplicative subgroup of \(\mathbb{K}\).
Proof
Remark 2
In the proof of Proposition 6, if the unity point of the coordinate system is arbitrarily chosen, the subsets L _{1}, L _{2}, and L _{3} are not necessarily subgroups. Actually, they are cosets of (the unique) multiplicative cyclic subgroup H, say L _{1}=aH, L _{2}=bH, and L _{3}=cH, with ac=b. Furthermore, since every h∈H defines a projectivity σ _{ h }:x↦hx of the projective line, and these projectivities form a group isomorphic to H, it turns out that L _{ i } is an orbit of a cyclic projectivity group of ℓ _{ i } of order n for i=1,2,3.
Proposition 7
Let (Λ _{1},Λ _{2},Λ _{3}) be a triangular dual 3net. Then every point of (Λ _{1},Λ _{2},Λ _{3}) is the center of a unique involutory homology that preserves (Λ _{1},Λ _{2},Λ _{3}).
Proof
With the notation introduced in the proof of Proposition 6, let Φ _{1}={φ _{ ξ } φ _{ ξ′}∣ξ,ξ′∈aH} and Φ _{2}={ψ _{ η } ψ _{ η′}∣η,η′∈bH}. Then both are cyclic groups isomorphic to H. A direct computation gives the following result.
Proposition 8
Φ _{1}∩Φ _{2} is either trivial or has order 3.
Some useful consequences are stated in the following proposition.
Proposition 9
We prove another useful result.
Proposition 10
If (Γ _{1},Γ _{2},Γ _{3}) and (Σ _{1},Σ _{2},Σ _{3}) are triangular dual 3nets such that Γ _{1}=Σ _{1}, then the associated triangles share the vertices on their common side.
Proof
By Remark 2, Γ _{1} is the orbit of a cyclic projectivity group H _{1} of the line ℓ containing Γ _{1}, while the two fixed points of H _{1} on ℓ, say P _{1} and P _{2}, are vertices of the triangle containing Γ _{1},Γ _{2},Γ _{3}.
The same holds for Σ _{1} with a cyclic projectivity group H _{2} and fixed points Q _{1}, Q _{2}. From Γ _{1}=Σ _{1} we have that the projectivity group H of the line ℓ generated by H _{1} and H _{2} preserves Γ _{1}. Let M be the projectivity group generated by H _{1} and H _{2}.
Observe that M is a finite group since it has an orbit of finite size n≥3. Clearly, M≥n, and the equality holds if and only if H _{1}=H _{2}. If this is the case, then {P _{1},P _{2}}={Q _{1},Q _{2}}. Therefore, for the purpose of the proof, we may assume on the contrary that H _{1}≠H _{2} and M>n.
Now, Dickson’s classification of finite subgroups of \(\mathit{PGL}(2,{\mathbb {K}})\) applies to M. From that classification we have that M is one of the nine subgroups listed as (1),…,(9) in [12, Theorem 1], where e denotes the order of the stabilizer M _{ P } of a point P in a short Morbit, that is, an Morbit of size smaller than M. Observe that such an Morbit has size M/e. There exist finitely many short Morbits, and Σ _{1} is one of them. It may be that an Morbit is trivial as it consists of just one point.
Obviously, M is neither cyclic nor dihedral as it contains two distinct cyclic subgroups of the same order n≥3.
Also, M is not an elementary Abelian pgroup E of rank ≥2; otherwise, we would have E=M>n since the minimum size of a nontrivial Eorbit is E; see (2) in [12, Theorem 1].
From (5) in [12, Theorem 1] with p≠2,3, the possible sizes of a short \(\rm{Alt}_{4}\)orbit are 4 and 6, each larger than 3. On the other hand, \(\rm{Alt}_{4}\) has no element of order larger than 3. Therefore, \(M\not\cong\rm{Alt}_{4}\) for p≠2,3.
Similarly, from (5) in [12, Theorem 1] with p≠2,3, the possible sizes of a short \(\rm{Sym}_{4}\)orbit are 6,8,12, each larger than 4. Since \(\rm{Sym}_{4}\) has no element of order larger than 4, we have \(M\not\cong\rm{Sym}_{4}\) for p≠2,3.
Again, from (6) in [12, Theorem 1] with p≠2,5, the possible sizes of a short \(\rm{Alt}_{5}\)orbit are 10 and 12 for p=3, while 12,20,30 for p≠2,3,5. Each size exceeds 5. On the other hand, \(\rm{Alt}_{5}\) has no element of order larger than 5. Therefore, \(M\not\cong\rm{Alt}_{5}\) for p≠2,5.
The group M might be isomorphic to a subgroup L of order qk with k∣(q−1) and q=p ^{ h }, h≥1. Here L is the semidirect product of the unique (elementary Abelian) Sylow psubgroup of L by a cyclic subgroup of order k. No element in L has order larger than k when h>1 and p when h=1. From (7) in [12, Theorem 1], any nontrivial short Lorbit has size q. Therefore, M≅L implies that h=1 and n=p. But this is inconsistent with the hypothesis p>n.
Finally, M might be isomorphic to a subgroup L such that either L=PSL(2,q) or L=PGL(2,q) with q=p ^{ h }, h≥1. No element in L has order larger than q+1. From (7) and (8) in [12, Theorem 1], any short Lorbit has size either q+1 or q(q−1). For q≥3, if M≅L occurs, then n=q+1≥p+1, a contradiction with the hypothesis p>n. For q=2, we have that L=6, which is smaller than 12. Therefore, \(M\not\cong L\).
No possibility has arisen for M. Therefore, {P _{1},P _{2}}={Q _{1},Q _{2}}. □
4.3 Coniclinetype dual 3nets
An algebraic dual 3net (Λ _{1},Λ _{2},Λ _{3}) is of conicline type if two of its three components lie on an irreducible conic \(\mathcal{C}\) and the third one lies on a line ℓ. All such 3nets realize groups, and they can be described using subgroups of the projectivity group \(\mathit{PGL}(2,\mathbb{K})\) of \(\mathcal{C}\). For this purpose, some basic results on subgroups and involutions in \(\mathit{PGL}(2,\mathbb{K})\) are useful, which essentially depend on the fact that every involution in \(\mathit{PGL}(2,\mathbb{K})\) is a perspectivity whose center is a point outside \(\mathcal{C}\) and axis is the pole of the center with respect to the orthogonal polarity arising from \(\mathcal{C}\). We begin with an example.
Example 1
Take any cyclic subgroup C _{ n } of \(\mathit{PGL}(2,\mathbb{K})\) of order n≥3 with n≠p that preserves \(\mathcal{C}\). Let D _{ n } be the unique dihedral subgroup of \(\mathit{PGL}(2,\mathbb{K})\) containing C _{ n }. If j is the (only) involution in \(\mathcal {Z}(D_{n})\) and ℓ is its axis, then the centers of the other involutions in D _{ n } lie on ℓ. We have n involutions in D _{ n } other than j, and the set of the their centers is taken for Λ _{1}. Take a C _{ n }orbit \(\mathcal{O}\) on \(\mathcal{C}\) such that the tangent to \(\mathcal{C}\) at any point in \(\mathcal{O}\) is disjoint from Λ _{1}; equivalently, the D _{ n }orbit \(\mathcal{Q}\) is larger than \(\mathcal{O}\). Then \(\mathcal{Q}\) is the union of \(\mathcal{O}\) together with another C _{ n }orbit. Take these two C _{ n }orbits for Λ _{2} and Λ _{3}, respectively. Then (Λ _{1},Λ _{2},Λ _{3}) is a conicline dual 3net which realizes C _{ n }. It may be observed that ℓ is a chord of \(\mathcal {C}\) and the multiplicative group of \(\mathbb{K}\) has a subgroup of order n.
The cyclic subgroups C _{ n } form a unique conjugacy class in \(\mathit{PGL}(2,\mathbb{K})\). For a cyclic subgroup C _{ n } of \(\mathit{PGL}(2,\mathbb{K})\) of order n, the above construction provides a unique example of a dual 3net realizing C _{ n }. Using the classification of finite subgroups of \(\mathit{PGL}(2,\mathbb{K})\) as in the proof of [2, Theorem 6.1], the following result can be proven. For details, see the preliminary version [10, 11].
Proposition 11
Up to projectivities, the conicline dual 3nets of order n are those described in Example 1.
A corollary of this is the following result.
Proposition 12
A conicline dual 3net realizes a cyclic group C _{ n }.
The result below can be proven with an argument similar to that used in the proof of Proposition 10. For details, see the preliminary version [10, 11].
Proposition 13
Let (Γ _{1},Γ _{2},Γ _{3}) and (Δ _{1},Δ _{2},Δ _{3}) be two coniclinetype dual 3nets where Γ _{3} lies on the line ℓ and Δ _{3} lies on the line s. If Γ _{1}=Δ _{1}, then ℓ=s.
4.4 Tetrahedron type dual 3nets
Proposition 14
Any tetrahedrontype dual 3net realizes a dihedral group.
Proof
Thus, \((\varGamma_{1}' \cup\varGamma_{2}', \varGamma_{3}' \cup \varDelta _{1}', \varDelta _{2}' \cup \varDelta _{3}')\) is a dual 3subnet of (Σ _{1},Σ _{2},Σ _{3}), and H is a subgroup of the dihedral group \(2.\mathbb{K}^{*}\). As H is not cyclic but it has a cyclic subgroup of index 2, we conclude that H is itself dihedral. □
5 Classification of loworder dual 3nets
An exhaustive computer aided search gives the following results. For details, see [14].
Proposition 15
Any dual 3net realizing an Abelian group of order ≤8 is algebraic. The dual of Urzúa’s 3nets are the only dual 3net that realize the quaternion group of order 8.
Proposition 16
Any dual 3net realizing an Abelian group of order 9 is algebraic.
Proposition 17
If p=0, no dual 3net realizes \(\rm{Alt}_{4}\).
6 Characterizations of the infinite families
Proposition 18
Every dual 3net realizing a cyclic group is algebraic.
Proof
For n=3, we have that 3n=9, and hence all points of the dual 3net lie on a cubic. Therefore, n≥4 is assumed.
Let (Λ _{1},Λ _{2},Λ _{3}) be a dual 3net of order n that realizes the cyclic group (L,∗). Therefore, the points of each component are labeled by I _{ n }. After a collinear relabeling with respect to Λ _{3}, consider the configuration of the following nine points: 0,1,2 from Λ _{1}, 0,1,2 from Λ _{2}, and n−1,n−2,n−3 from Λ _{3}. For the seek of a clearer notation, the point with label a in the component Λ _{ m } will be denoted by a _{ m }.
 (i)
{0_{1},1_{2},(n−1)_{3}}, {1_{1},2_{2},(n−3)_{3}}, {2_{1},0_{2},(n−2)_{3}};
 (ii)
{0_{1},2_{2},(n−2)_{3}}, {1_{1},0_{2},(n−1)_{3}}, {2_{1},1_{2},(n−3)_{3}}.
 (iii)
{1_{1},2_{2},(n−3)_{3}}, {2_{1},0_{2},(n−2)_{3}}, {3_{1},1_{2},(n−4)_{3}};
 (iv)
{1_{1},1_{2},(n−2)_{3}}, {2_{1},2_{2},(n−4)_{3}}, {3_{1},0_{2},(n−3)_{3}}.
 (v)
{0_{1},3_{2},(n−3)_{3}}, {1_{1},1_{2},(n−2)_{3}}, {2_{1},2_{2},(n−4)_{3}};
 (vi)
{0_{1},2_{2},(n−2)_{3}}, {1_{1},3_{2},(n−4)_{3}}, {2_{1},1_{2},(n−3)_{3}}.
 (vii)
{1_{1},3_{2},(n−4)_{3}}, {2_{1},1_{2},(n−3)_{3}}, {3_{1},2_{2},(n−5)_{3}};
 (viii)
{1_{1},2_{2},(n−3)_{3}}, {2_{1},3_{2},(n−5)_{3}}, {3_{1},1_{2},(n−4)_{3}}.
 (ix)
{2_{1},2_{2},(n−4)_{3}}, {3_{1},0_{2},(n−3)_{3}}, {4_{1},1_{2},(n−5)_{3}};
 (x)
{2_{1},1_{2},(n−3)_{3}}, {3_{1},2_{2},(n−5)_{3}}, {4_{1},0_{2},(n−4)_{3}}.
Proposition 19
[17, Theorem 5.4]
If an Abelian group G contains an element of order ≥10, then every dual 3net realizing G is algebraic.
Proposition 20
[17, Theorem 4.2]
No dual 3net realizes an elementary Abelian group of order 2^{ h } with h≥3.
Proposition 21
[2, Theorem 5.1]
Let (Λ _{1},Λ _{2},Λ _{3}) be a dual 3net such that at least one component lies on a line. Then (Λ _{1},Λ _{2},Λ _{3}) is either triangular or of conicline type.
Lemma 4
Let (Γ _{1},Γ _{2},Γ _{3}) be an algebraic dual 3net lying on a plane cubic \(\mathcal{F}\). If \(\mathcal{F}\) is reducible, then (Γ _{1},Γ _{2},Γ _{3}) is either triangle or of conicline type, according as \(\mathcal{F}\) splits into three lines or into a line and an irreducible conic.
Proposition 22
Every dual 3net realizing a dihedral group of order 2n with n≥3 is of tetrahedron type.
Proof
Remark 3
By Proposition 22, the dual 3nets given in [15, Sect. 6.2] are of tetrahedron type.
Proposition 23
Let G be a finite group containing a normal subgroup H of order n≥3. Assume that G can be realized by a dual 3net (Λ _{1},Λ _{2},Λ _{3}) and that every dual 3subnet of (Λ _{1},Λ _{2},Λ _{3}) realizing H as a subgroup of G is triangular. Then H is cyclic, and (Λ _{1},Λ _{2},Λ _{3}) is either triangular or of tetrahedron type.
Proof
By Proposition 6, H is cyclic. Fix an Hmember Γ _{1} from Λ _{1}, and denote by ℓ _{1} the line containing Γ _{1}. Consider all the triangles that contain some dual 3net \((\varGamma_{1},\varGamma_{2}^{j},\varGamma_{3}^{s})\) realizing H as a subgroup of G. By Proposition 10, these triangles have two common vertices, say P and Q, lying on ℓ _{1}. For the third vertex R _{ j } of the triangle containing \((\varGamma_{1},\varGamma_{2}^{j},\varGamma_{3}^{s})\), there are two possibilities, namely, either the side PR _{ j } contains \(\varGamma_{2}^{j}\) and the side QR _{ j } contains \(\varGamma_{3}^{s}\), or vice versa. Therefore, every Hmember \(\varGamma_{2}^{j}\) from Λ _{2} (as well as every Hmember \(\varGamma_{3}^{s}\) from Λ _{3}) is contained in a line passing through P or Q.
Now, replace Γ _{1} by another Horbit \(\varGamma_{1}^{i}\) lying in Λ _{1} and repeat the above argument. If ℓ _{ i } is the line containing \(\varGamma_{1}^{i}\) and P _{ i },Q _{ i } denote the vertices, then again every Hmember \(\varGamma_{2}^{j}\) from Λ _{2} (as well as every Hmember \(\varGamma_{3}^{s}\) from Λ _{3}) is contained in a line passing through P _{ i } or Q _{ i }.
Assume that {P,Q}≠{P _{ i },Q _{ i }}. If one of the vertices arising from Γ _{1}, say P, coincides with one of the vertices, say P _{ i }, arising from \(\varGamma_{1}^{i}\), then the line QQ _{ i } must contain either \(\varGamma_{2}^{j}\) or \(\varGamma_{3}^{s}\) from each \((\varGamma_{1},\varGamma _{2}^{j},\varGamma_{3}^{s})\). Therefore, the line QQ _{ i } must contain every Hmember from Λ _{2} or every Hmember from Λ _{3}. Hence, Λ _{2} or Λ _{3} lies on the line QQ _{ i }. By Proposition 21, (Λ _{1},Λ _{2},Λ _{3}) is either triangular or conicline type. The latter case cannot actually occur as Λ _{1} contains Γ _{1} and hence contains at least three collinear points.
Therefore, {P,Q}∩{P _{ i },Q _{ i }}=∅ may be assumed. Then the Hmembers from Λ _{2} and Λ _{3} lie on four lines, namely PP _{ i },PQ _{ i },QP _{ i },QQ _{ i }. Observe that these lines may be assumed to be pairwise distinct, otherwise Λ _{2} (or Λ _{3}) is contained in a line, and again (Λ _{1},Λ _{2},Λ _{3}) is triangular. Therefore, half of the Hmembers from Λ _{2} lie on one of these four lines, say PQ _{ i }, and half of them on QP _{ i }. Similarly, each of the lines PP _{ i } and QQ _{ i } contains half from the Hmembers from Λ _{3}.
In the above argument, any Hmember Γ _{2} from Λ _{2} may play the role of Γ _{1}. Therefore, there exist two lines such that each Hmember from Λ _{1} lies on one of them. Actually, these two lines are PQ and P _{ i } Q _{ i } since each of them contains an Hmember from Λ _{1}. In this case, (Λ _{1},Λ _{2},Λ _{3}) is of tetrahedron type. □
Since a dihedral group of order ≥8 has a unique cyclic subgroup of index 2 and such a subgroup is characteristic, Propositions 23 and 14 have the following corollary.
Proposition 24
Let G be a finite group of order n≥12 containing a normal dihedral subgroup D. If G is realized by a dual 3net, then G is itself dihedral.
7 Dual 3nets preserved by projectivities
Proposition 25
Let (Λ _{1},Λ _{2},Λ _{3}) be a dual 3net of order n≥4 realizing a group G. If every point in Λ _{1} is the center of an involutory homology that preserves Λ _{1} while interchanges Λ _{2} with Λ _{3}, then either Λ _{1} is contained in a line, or n=9. In the latter case, (Λ _{1},Λ _{2},Λ _{3}) lies on a nonsingular cubic \(\mathcal{F}\) whose inflection points are the points in Λ _{1}.
Proof
Let Φ be the projectivity group generated by all products φ _{ a′} φ _{ a } where both a, a′ range over G. Obviously, Φ leaves both Λ _{2} and Λ _{3} invariant. In particular, Φ induces a permutation group on Λ _{2}. We show that if μ∈Φ fixes Λ _{2} pointwise, then μ is trivial. Since n>3, the projectivity μ has at least four fixed points in \(\mathit{PG}(2,\mathbb{K})\). Therefore, μ is either trivial or a homology. Assume that μ is nontrivial, and let C be the center and c the axis of μ. Take a line ℓ through C that contains a point P∈Λ _{3} and assume that C is a point in Λ _{2}. Then P is the unique common point of ℓ and Λ _{3}. Since μ preserves Λ _{2}, μ must fix P. Therefore, μ fixes Λ _{3} pointwise, and hence Λ _{3} is contained in c. But then μ cannot fix any point in Λ _{2} other than C since the definition of a dual 3net implies that c is disjoint from Λ _{2}. This contradiction means that μ is trivial, that is, Φ acts faithfully on Λ _{2}.
Let Ψ be the projectivity group generated by Φ together with some φ _{ a } where a∈G. Then Ψ=2n, and Ψ comprises the elements in Φ and the involutory homologies φ _{ a } with a ranging over G. Obviously, Ψ interchanges Λ _{2} and Λ _{3}, while it leaves Λ _{1} invariant acting on Λ _{1} as a transitive permutation group.
Two cases are investigated according as Φ contains a homology or does not. Observe that Φ contains no elation since every elation has infinite order when p=0 while its order is at least p when p>0 but p>n is assumed throughout the paper.
In the former case, let ρ∈Φ be a homology with center C∈Λ _{1} and axis c. Since ρ commutes with every element in Φ, the point C is fixed by Φ, and the line is preserved by Φ. Assume that C is also the center of ϕ _{ a } with some a∈G. The group of homologies generated by ϕ _{ a } and ρ preserves every line through C, and it has order greater than 2. But then it cannot interchange Λ _{2} with Λ _{3}. Therefore, the center of every ϕ _{ a } with a∈G lies on c. This shows that Λ _{1} is contained in c.
In the case where Φ contains no homology, Φ has odd order, and δ∈Φ has three fixed points, which are the vertices of a triangle Δ. Since δ commutes with every element in Φ, the triangle Δ is left invariant by Φ.
If Φ fixes each vertex of δ, then Φ must be cyclic since, otherwise, Ψ would contain a homology. Therefore, Ψ is a dihedral group, and we show that Λ _{1} is contained in a line. For this purpose, take a generator ρ=φ _{ a } φ _{ b } of Φ, and consider the line ℓ through the centers of φ _{ a } and φ _{ b }. Obviously, ρ preserves ℓ, and this holds for every power of ρ. Hence, Ψ also preserves ℓ. Since every φ _{ c } is conjugate to φ _{ a } under Ψ, this shows that the center of φ _{ c } must lie on ℓ as well. Therefore, Λ _{1} is contained in ℓ.
We may assume that some ρ∈Φ acts on the vertices of Δ as a 3cycle. Let Δ′ be the triangle whose vertices are the fixed points of ρ. Then ρ ^{3}=1 since ρ ^{3} fixes not only the vertices of Δ′ but also those of Δ′. Therefore, Φ=〈ρ〉×Θ where Θ is the cyclic subgroup of Φ fixing each vertex of Δ. A subgroup of Θ of index ≤3 fixes each vertex of Δ′ and hence is trivial. Therefore, Θ=3 and Φ≅C _{3}×C _{3}. This shows that n=9 and if Λ _{1} is not contained in a line, then the configuration of their points, that is, the centers of the homologies in Ψ, is isomorphic to AG(2,3), the affine plane of order 3. Such a configuration can also be viewed as the set of the nine common inflection points of the nonsingular plane cubics of a pencil \(\mathcal{P}\), each cubic left invariant by Ψ. For a point P _{2}∈Λ _{2}, take that cubic \(\mathcal{F}\) in \(\mathcal{P}\) that contains P _{2}. Since the orbit of P _{2} under the action of Ψ consists of the points in Λ _{2}∪Λ _{3}, it follows that \(\mathcal {F}\) contains each point of (Λ _{1},Λ _{2},Λ _{3}). □
The following result is a corollary of Proposition 25.
Proposition 26
Let (Λ _{1},Λ _{2},Λ _{3}) be a dual 3net of order n≥4 realizing a group G. If every point of (Λ _{1},Λ _{2},Λ _{3}) is the center of an involutory homology that preserves (Λ _{1},Λ _{2},Λ _{3}), then (Λ _{1},Λ _{2},Λ _{3}) is triangular.
Proof
By Proposition 11 and Example 1, (Λ _{1},Λ _{2},Λ _{3}) is not of conicline type. For n=9, (Λ _{1},Λ _{2},Λ _{3}) does not lie on any nonsingular cubic \(\mathcal{F}\) since no nonsingular cubic has twentyseven inflection points. Therefore, the assertion follows from Proposition 25. □
A useful generalization of Proposition 26 is given in the proposition below.
Proposition 27
 (i)
Every component contains the same number of points that are centers of involutory homologies in \(\mathcal{U}\).
 (ii)
The points of (Λ _{1},Λ _{2},Λ _{3}) that are centers of involutory homologies in \(\mathcal{U}\) form a triangular dual 3subnet (Γ _{1},Γ _{2},Γ _{3}).
 (iii)Let M be the cyclic subgroup associated to (Γ _{1},Γ _{2},Γ _{3}). Then either (Λ _{1},Λ _{2},Λ _{3}) is also triangular, or$$G<\left \{ \begin{array}{l@{\quad}l} G:M^2,& \mathit{when}\ \mathrm{gcd}(3,G)=1;\\ 3G:M^2,& \mathit{when}\ \mathrm{gcd}(3,G)=3. \end{array} \right . $$
Proof
Let \(\mathcal{G}\) be the projectivity group preserving (Λ _{1},Λ _{2},Λ _{3}). Let (ijk) denote any permutation of (123). As we have already observed in the proof of Proposition 25, if \(\varphi\in\mathcal{G}\) is an involutory homology with center P∈Λ _{ i }, then φ preserves Λ _{ i } and interchanges Λ _{ j } with Λ _{ k }. If \(\sigma\in\mathcal {G}\) is another involutory homology with center R∈Λ _{ j }, then σφσ is also an involutory homology whose center S is the common point of Λ _{ k } and the line ℓ through P and R. In terms of dual 3subnets, this yields (i) and (ii). Let m be the order of (Γ _{1},Γ _{2},Γ _{3}). For m=2, (Γ _{1},Γ _{2},Γ _{3}) is triangular. For m=3, Γ _{1}∪Γ _{2}∪Γ _{3} is the Hesse configuration, and hence (Γ _{1},Γ _{2},Γ _{3}) is triangular. This holds for m≥4 by Proposition 26 applied to (Γ _{1},Γ _{2},Γ _{3}).
To prove (iii), assume that (Λ _{1},Λ _{2},Λ _{3}) is not triangular and take a point P from some component, say Λ _{3}, that does not lie on the sides of the triangle associated to (Γ _{1},Γ _{2},Γ _{3}). Since (Γ _{1},Γ _{2},Γ _{3}) is triangular, it can play the role of (Λ _{1},Λ _{2},Λ _{3}) in Sect. 4.2, and we use the notation introduced there. By the second assertion of Proposition 9, the point has as many as Θ distinct images, all lying in Λ _{3}. Therefore, G=Λ _{3}>Θ. Using Proposition 9, Θ can be written in function of M giving the assertion. □
Let \(\mathcal{U}_{2}\) be the set of all involutory homologies with center in Λ _{2} which interchanges Λ _{1} and Λ _{3}. There is a natural injective map Ψ from \(\mathcal{U}_{2}\) to G, where Ψ(ψ)=g holds if and only if the point g _{2}∈Λ _{2} is the center of ψ.
Proposition 28
 (i)
\(\mathcal{U}_{2}\) is closed by conjugation, that is, \(\psi\omega\psi\in\mathcal{U}_{2}\) whenever \(\psi,\omega\in \mathcal{U}_{2}\).
 (ii)
If \(g,h\in\varPsi(\mathcal{U}_{2})\), then \(gh^{1}g\in \varPsi(\mathcal{U}_{2})\).
 (iii)
If G has a cyclic subgroup H of order 6 with \(H\cap\varPsi(\mathcal{U}_{2})\geq3\) and \(1\in H\cap\varPsi(\mathcal {U}_{2})\), then either \(\varPsi(\mathcal{U}_{2})=H\), or \(\varPsi(\mathcal {U}_{2})\) is the subgroup of H of order 3.
Proof
8 Dual 3nets containing algebraic 3subnets of order n with n≥5
A key result is the following proposition.
Proposition 29
 (i)
Γ _{2} is contained in a line.
 (ii)
n=5, and there is an involutory homology with center in Γ _{2} that preserves every \(\mathcal{F}_{j}\) and interchanges Λ _{1} and Λ _{3}.
 (iii)
n=6, and there are three involutory homologies with center in Γ _{2} that preserve every \(\mathcal{F}_{j}\) and interchange Λ _{1} and Λ _{3}.
 (iv)
n=9, and Γ _{2} consists of the nine common inflection points of \(\mathcal{F}_{j}\).
We need the following technical lemma.
Lemma 5
Let A=(A,⊕), B=(B,+) be Abelian groups and consider the injective maps α,β,γ:A→B such that α(x)+β(y)+γ(z)=0 if and only if z=x⊕y. Then, α(x)=φ(x)+a, β(x)=φ(x)+b, γ(x)=−φ(x)−a−b for some injective homomorphism φ:A→B and elements a,b∈B.
Proof
Lemma 6
To prove Proposition 29, we point out that 3b∈φ(H) if and only if Γ _{2} contains three collinear points. Suppose that φ(x _{1})+b,φ(x _{2})+b,φ(x _{3})+b are three collinear points. Then φ(x _{1})+b+φ(x _{2})+b+φ(x _{3})+b=0, whence φ(x _{1}+x _{2}+x _{3})+3b=0. Therefore, 3b∈φ(H). Conversely, if φ(t)=3b, take three pairwise distinct elements x _{1},x _{2},x _{3}∈H such that x _{1}+x _{2}+x _{3}+t=0. Then φ(x _{1})+b+φ(x _{2})+b+φ(x _{3})+b=0. Therefore, the points φ(x _{1})+b,φ(x _{2})+b and φ(x _{3})+b of Γ _{2} are collinear. Notice that the element t=−x _{1}−x _{2}−x _{3}∈H is the same even if we make the computation with φ _{ j } and b _{ j }.
We separately deal with two cases.
8.1 Γ _{2} contains no three collinear points
8.2 Γ _{2} contains three collinear points
This time, 3b∈φ(H). Let φ(t)=3b with t∈H. If either \(\mathcal{F}\) or \(\mathcal{F}_{j}\) is reducible, then Γ _{2} is contained in a line. Therefore, both \(\mathcal{F}\) and \(\mathcal{F}_{j}\) are assumed to be irreducible.
It remains to investigate the case where 3b=φ(3t _{0}) for some t _{0}∈H. Replacing b by b−φ(t _{0}) shows that 3b=0 may be assumed. Therefore, the point P=φ(y)+b with y∈H is an inflection point of \(\mathcal{F}\) if and only if 3y=0. Furthermore, if 3y≠0, then Q=φ(⊖(2y))+b is the tangential point of P on \(\mathcal{F}\). Therefore, β determines the tangents of \(\mathcal{F}\) at its points in Γ _{2}. The same holds for \(\mathcal{F}_{j}\). By Lemma 6, P=β(y) is an inflection point of both \(\mathcal{F}\) and \(\mathcal{F}_{j}\) or none of them. In the latter case, \(\mathcal{F}\) and \(\mathcal{F}_{j}\) have the same tangent at P.
Let m be the number of common inflection points of \(\mathcal{F}\) and \(\mathcal{F}_{j}\) lying in Γ _{2}. Obviously, P=φ(0)+b is such a point, and hence m≥1. On the other hand, m may take only three values, namely 1, 3, and 9. If m=9, then \(\mathcal{F}\) is nonsingular, and Γ _{2} consists of all the nine inflection points of \(\mathcal{F}\). The same holds for \(\mathcal{F}_{j}\). If m=3, then \(\mathcal{F}\) and \(\mathcal{F}_{j}\) share their tangents at n−3 common points. Therefore, 2n−3≤9, whence n≤6.
If n=5, the zero of H is the only element y with 3y=0. This shows that \(\mathcal{F}\) (and \(\mathcal{F}_{j}\)) has only one inflection point P _{0} in Γ _{2} and P _{0} is not the tangential point of another point in Γ _{2}. Each of the remaining four points is the tangential point of exactly one point in Γ _{2}. These four points may be viewed as the vertices of a quadrangle P _{1} P _{2} P _{3} P _{4} such that the side P _{ i } P _{ i+1} is tangent to \(\mathcal{F}\) at P _{ i } for every i with P _{5}=P _{1}. Therefore, the intersection divisor of \(\mathcal{F}\) and \(\mathcal {F}_{j}\) is P _{0}+2P _{1}+2P _{2}+2P _{3}+2P _{4}, and \(\mathcal{F}_{j}\) is contained in a pencil \(\mathcal{P}\).
This completes the proof of Proposition 29.
In the case where H is an Abelian normal subgroup of G, we have the following result.
Proposition 30
Let G be a group containing a proper Abelian normal subgroup H of order n≥5. If a dual 3net (Λ _{1},Λ _{2},Λ _{3}) realizes G such that all its dual 3subnets realizing H as a subgroup of G are algebraic, then either (I) or (II) of Theorem 1 holds.
Proof
The essential tool in the proof is Proposition 29. Assume on the contrary that neither (I) nor (II) occurs.
If every Hmember is contained in a line, then every dual 3net realizing H as a subgroup of G is triangular. By Proposition 23, either (I) of (II) follows.
Take an Hmember not contained in a line. Since H is a normal subgroup, that Hmember can play the role of Γ _{2} in Proposition 29. Therefore, one of the three sporadic cases in Proposition 29 holds. Furthermore, from the proof of that proposition it follows that every \(\mathcal{F}_{j}\) is irreducible, and hence neither \(\varGamma_{1}^{j}\) nor \(\varGamma_{3}^{j}\) is contained in a line. Therefore, no Hmember is contained in a line. Since H is a normal subgroup, every 3subnet \((\varGamma_{1}^{i},\varGamma_{2}^{j},\varGamma_{3}^{s})\) realizing H as a subgroup of G lies in an irreducible plane cubic \(\mathcal{F}(i,j)\).
 n=9

From (iv) of Proposition 29 it follows that the cubics \(\mathcal{F}_{j}\) share their nine inflection points, which form Γ _{2}. So it is possible to avoid this case by replacing Γ _{2} with Γ _{1} so that Γ _{2} will not have any inflection point of \(\mathcal{F}\).
 n=6

Every Hmember Γ _{2} contains three collinear points, say Q _{1},Q _{2},Q _{3}, so that Q _{ r } is the center of an involutory homology ψ _{ r } interchanging Λ _{1} and Λ _{3}. Relabeling the points of the dual 3net permits us to assume that Q _{1}=1_{2}. Then for all x∈G, ψ _{1} interchanges the points x _{1} and x _{3}. The point a _{2}∈Λ _{2} is the intersection of the lines y _{1}(ya)_{3}, with y∈G. These lines are mapped to the lines (ya)_{1} y _{3}, which all contain the point (a ^{−1})_{2} of Λ _{2}. Therefore, the involutory homology ψ _{1} leaves Λ _{2} invariant. This holds for all involutory homologies with center in Λ _{1}∪Λ _{2}∪Λ _{3}. Since the Hmembers partition each component of (Λ _{1},Λ _{2},Λ _{3}) and every Hmember comprises six points, it turns out that half of the points of (Λ _{1},Λ _{2},Λ _{3}) are the centers of involutory homologies preserving (Λ _{1},Λ _{2},Λ _{3}). Therefore, Proposition 27(iii) applies. As in Proposition 27, let M denote the subgroup of G such that the dual 3subnet consisting of the centers of involutory homologies realizes M. As G:M=2, Proposition 27(iii) implies G<6, a contradiction.
 n=5

The arguments in discussing case n=6 can be adapted for case n=5. This time, Proposition 29 gives G:M=5. By Proposition 27(iii), if G contains an element of order 3, then G<75; otherwise, G<25. In the former case, the element of order 3 of G is in C _{ G }(H), and hence G contains a cyclic normal subgroup of order 15. Then, (Λ _{1},Λ _{2},Λ _{3}) is algebraic by Proposition 29. If G has no element of order 3, then G<25, and G contains a normal subgroup of order 10 that is either cyclic or dihedral. By Propositions 24 and 29 either (I) or (II) of Theorem 1 holds. □
The following result is a corollary of Proposition 30.
Theorem 4
Every dual 3net (Λ _{1},Λ _{2},Λ _{3}) realizing an Abelian group G is algebraic.
Proof
By absurd, let n be the smallest integer for which a counterexample (Λ _{1},Λ _{2},Λ _{3}) to Theorem 4 exists. Since any dual 3net of order ≤8 is algebraic by Propositions 15 and 18, we have that n≥9. Furthermore, again by Proposition 18, G has composite order. Since n is chosen to be as small as possible, by Proposition 30, G has only one prime divisor, namely either 2 or 3. Since G≥9, either G=2^{ r } with r≥4, or G=3^{ r } with r≥2. In the former case, G has a subgroup M of order 8, and every dual 3subnet realizing M is algebraic by Proposition 15. But this, together with Proposition 30, shows that (Λ _{1},Λ _{2},Λ _{3}) is not a counterexample. In the latter case, G contains no element of order 9, and hence it is an elementary Abelian group. But then (Λ _{1},Λ _{2},Λ _{3}) is algebraic by Proposition 16. □
9 Dual 3nets realizing 2groups
Proposition 31
 (i)
G is cyclic.
 (ii)
G≅C _{ m }×C _{ k } with n=mk.
 (iii)
G is a dihedral.
 (iv)
G is the quaternion group of order 8.
Proof
For n=4,8, the classification follows from Propositions 15 and 22 and from [17, Theorem 4.2]. Up to isomorphisms, there exist fourteen groups of order 16; each has a subgroup H of index 2 that is either an Abelian or a dihedral group. In the latter case, G is itself dihedral, by Proposition 24. So, Proposition 30 applies to G and H, yielding that G is Abelian. This completes the proof for n=16. By induction on h we assume that Proposition 31 holds for n=2^{ h }≥16, and we are going to show that this remains true for 2^{ h+1}. Let H be a subgroup of G of index 2. Then H=2^{ h }, and one of the cases (i), (ii), and (iii) hold for H. Therefore, the assertion follows from Propositions 30 and 24. □
10 Dual 3nets containing algebraic 3subnets of order n with 2≤n≤4
It is useful to investigate separately two cases according as n=3,4, or 2. An essential tool in the investigation is \(M=\mathcal {C}_{G}(H)\), the centralizer of H in G.
Proposition 32
Let G be a finite group containing a normal subgroup H of order n with n=3 or n=4. Then every dual 3net (Λ _{1},Λ _{2},Λ _{3}) realizing G is either algebraic or of tetrahedron type, or G is isomorphic either to the quaternion group of order 8, or to \({\rm{Alt}}_{4}\), or to \({\rm{Sym}}_{4}\).
Proof
First, we investigate the case where M>H. Take an element m∈M outside H. Then the subgroup T of G generated by m and H is Abelian and larger than H. Since H≥3, we have T≥6. If all Hmembers of (Λ _{1},Λ _{2},Λ _{3}) are contained in a line, then (Λ _{1},Λ _{2},Λ _{3}) is either triangular or of tetrahedron type by Proposition 23. Assume that Γ _{2} is an Hmember that is not contained in a line. Let \(\varGamma_{2}'\) be the Tmember containing Γ _{2}. We claim that (Λ _{1},Λ _{2},Λ _{3}) is algebraic. If not, then one of the exceptional cases (iii) or (iv) of Proposition 29 must hold. Clearly, in these cases, H=3. However, the centers of the involutory homologies mentioned in Proposition 29 correspond to the points in the Hmember Γ _{2}. As these centers must be collinear, we obtain that Γ _{2} is contained in a line, a contradiction.
Assume that M=H. Then G/H is an automorphism group H. If H is C _{3} or C _{4}, then Aut(H)=2, and G is either a dihedral group or the quaternion group of order 8. If H≅C _{2}×C _{2}, then G is a subgroup of \(\rm{Sym}_{4}\). The possibilities for G other than H and the dihedral group of order 8 are \(\rm {Alt}_{4}\) and \(\rm{Sym}_{4}\). Since all these groups are allowed in the proposition, the proof is finished. □
Proposition 33
Let G be a finite group with a central involution that contains no normal subgroup H of order 4. Then a dual 3net (Λ _{1},Λ _{2},Λ _{3}) realizing G is either algebraic or of tetrahedron type.
Proof
Let H be the normal subgroup generated by the (unique) central involution of G. Two cases are separately investigated according as a minimal normal subgroup \(\bar{N}\) of the factor group \(\bar{G}=G/H\) is solvable or not. Let σ be the natural homomorphism \(G\to \bar{G}\). Let \(N=\sigma^{1}(\bar{N})\).
If \(\bar{N}\) is solvable, then \(\bar{N}\) is an elementary Abelian group of order d ^{ h } for a prime d. Furthermore, N is a normal subgroup of G and \(\bar{N}=N/H\). If N is Abelian, then N≥6, and the assertion follows from Proposition 30 and Theorem 4.
Bearing this in mind, the case where d=2 is investigated first. Then N has order 2^{ h+1} and is a normal subgroup of G. By Proposition 31, N is either Abelian, or it is the quaternion group Q _{8} of order 8. We may assume that N≅Q _{8}. By Proposition 31, N is not contained in a larger 2subgroup of G. Therefore, N is a (normal) Sylow 2subgroup of G. We may assume that G is larger than N. If M=C _{ G }(N) is also larger than N, take an element t∈M of outside N. Then t has odd order ≥3. The group T generated by N and t has order 8m, and its subgroup D generated by t together with an element of N of order 4 is a (normal) cyclic subgroup of M of order 4m. But this contradicts Proposition 30 as T is neither Abelian nor dihedral. Therefore, M=N, and hence G/N is isomorphic to a subgroup L of the automorphism group Aut(Q _{8}). Hence, G/N divides 24. On the other hand, since N is a Sylow 2subgroup of G, G/N must be odd. Therefore, G=24. Two possibilities arise according as either G≅SL(2,3) or G is the dicyclic group of order 24. The latter case cannot actually occur by Proposition 30 as the dicyclic group of order 24 has a (normal) cyclic subgroup of order 12.
To rule the case G≅SL(2,3) out, we rely on Propositions 29 and 28 since SL(2,3) has four cyclic groups of order 6. For this purpose, we show that every point in Λ _{2} is the center of an involutory homology preserving (Λ _{1},Λ _{2},Λ _{3}) whence the assertion will follow from Proposition 25 applied to Λ _{2}. With the notation in Sect. 7, (iii) Proposition 29 yields that \(\mathcal{U}_{2}\geq 3\). With the notation introduced in the proof of Proposition 18, we may assume that the point 1_{2} is the center of an involutory homology ϵ in \(\mathcal{U}_{2}\). From (iii) of Proposition 28 it follows that every (cyclic) subgroup of G of order 6 provides (at least) two involutory homologies other than ϵ. Therefore, \(\mathcal{U}_{2}\geq9\), and every point u _{2}∈Λ _{2} such that u ^{3}=1 is the center of an involutory homology in \(\mathcal{U}\). A straightforward computation shows that every element in G other than the unique involution e can be written as gh ^{−1} g with g ^{3}=h ^{3}=1. Thus, \(\mathcal{U}\geq23\). The involutory homology with center 1_{2} cannot actually be an exception. To show this, take an element g∈G of order 4. Then g _{2} is the center of an element in \(\mathcal{U}\). Since 1=g ^{2}=g⋅1⋅g, this holds for 1_{2}. Therefore, \(\mathcal{U}=24\). By (i) of Proposition 28, \(\mathcal{U}\) also preserves Λ _{2}. This completes the proof.
Now, the case of odd d is investigated. Since H=2 and d are coprime, Zassenhaus’ theorem [9, 10.1 Hauptsatz] ensures a complement \(W\cong\bar{N}\) such that N=W⋉H=W×H. Obviously, W is an Abelian normal subgroup of G of order at least 3. The assertion follows from Propositions 30 and 32.
If \(\bar{N}\) is not solvable, then it has a nonAbelian simple group \(\bar{T}\). Let \(\bar{S}_{2}\) be a Sylow 2subgroup of \(\bar{T}\). By Proposition 31, the realizable 2group S _{2} is either cyclic, or product of two cyclic groups, or dihedral, or quaternion of order 8. Thus, \(\bar{S}_{2}\) is either cyclic, or product of two cyclic groups, or dihedral. As \(\bar{T}\) is simple, \(\bar{S}_{2}\) cannot be cyclic. In the remaining cases we can use the classification of finite simple groups of 2rank 2 to deduce that either \(\bar{T}\cong \mathit{PSL}(2,q^{h})\) with an odd prime q and q ^{ h }≥5, or \(\bar{T}\cong\rm{Alt}_{7}\); cf. the Gorenstein–Walter theorem [6].
If \(H\not\leq T'\), then T=H×T′. As \(T'\cong\bar{T}\), T′ contains an elementary Abelian subgroup of order 4, and G contains an elementary Abelian group of order 8, a contradiction. Therefore, T is a central extension of either PSL(2,q ^{ h }) with q ^{ h } as before, or \(\rm{Alt}_{7}\) with a cyclic group of order 2. From a classical result of Schur [1, Chap. 33], either T≅SL(2,q), or T is the unique central extension of \(\rm{Alt}_{7}\) with a cyclic group of order 2. In the latter case, no dual 3net can actually realize T since Proposition 31 applies, a Sylow 2subgroup of T being isomorphic to a generalized quaternion group of order 16. To finish the proof, it suffices to observe that SL(2,q ^{ h }), with q ^{ h } as before, contains SL(2,3), whereas no dual 3net can realize SL(2,3) as we have already pointed out. □
11 3Nets and nonAbelian simple groups
Proposition 34
If a dual 3net realizes a nonAbelian simple group G, then \(G\cong\rm{Alt}_{5}\).
Proof
Let G be a nonAbelian simple group, and consider a Sylow 2subgroup S _{2} of G. By Proposition 31, S _{2} is dihedral since no Sylow 2subgroup of a nonAbelian simple group is either cyclic or the direct product of cyclic groups, see [5, Theorem 2.168], or the quaternion group of order 8, see [3]. From the Gorenstein–Walter theorem [6], either G≅PSL(2,q ^{ h }) with an odd prime q and q ^{ h }≥5, or \(G\cong \rm{Alt}_{7}\). In the former case, G has a subgroup T of order q ^{ h }(q ^{ h }−1)/2 containing a normal subgroup of order q ^{ h }. Here T is not Abelian and is dihedral only for q ^{ h }=5. Therefore, Theorem 4 and Proposition 30 leave only one case, namely q=5. This also shows that \({\rm{Alt}}_{7}\) cannot occur since \({\rm{Alt}}_{7}\) contains PSL(2,7). □
Notice that, by Proposition 17, computer results show that if p=0, then \(\rm{Alt}_{4}\) cannot be realized in \(\mathit{PG}(2,\mathbb{K})\). This implies that no dual 3net can realize \(\rm{Alt}_{5}\).
12 The proof of Theorem 1
Take a minimal normal subgroup H of G. If H is not solvable, then H is either a simple group or the product of isomorphic simple groups. By Proposition 20, the latter case cannot actually occur as every simple group contains an elementary Abelian subgroup of order 4. Therefore, if H is not solvable, \(H\cong{\rm{Alt}}_{5}\) may be assumed by Proposition 34. Two cases are considered separately according as the centralizer C _{ G }(H) of H in G is trivial or not. If C _{ G }(H)>1, take a nontrivial element u∈C _{ G }(H) and define U to be the subgroup of G generated by u together with a dihedral subgroup D _{5} of H of order 10. Since u centralizes D _{5}, the latter subgroup is a normal subgroup of U. Hence, D _{5} is a normal dihedral subgroup of U. By Proposition 24, M itself must be dihedral. Since the center of a dihedral group has order 2, this implies that u is an involution. Now, the subgroup generated by u, together with an elementary Abelian subgroup of H of order 4, generates an elementary Abelian subgroup of order 8. But this contradicts Proposition 20. Therefore, C _{ G }(H) is trivial, and, equivalently, G is contained in the automorphism group of H. From this it follows that either G=H or G≅PGL(2,5). In the latter case, G contains a subgroup isomorphic to the semidirect product of C _{5} by C _{4}. But this contradicts Proposition 24. Hence, if H is not solvable, then \(H\cong\rm{Alt}_{5}\).
If H is solvable, then it is an elementary Abelian group of order d≥2. If d is a power of 2, then d=2 or d=4, and Theorem 1 follows from Propositions 31 and 33. If d is a power of an odd prime, Theorem 1 is obtained by Propositions 30 and 32.
Notes
Acknowledgement
Nicola Pace is supported by FAPESP (Fundação de Amparo a Pesquisa do Estado de São Paulo), procs no. 12/035260.
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