Maximal perpendicularity in certain Abelian groups

Acta Univ. Sapientiae, Mathematica, 9, 1 (2017) 235–247 DOI: 10.1515/ausm-2017-0016 Maximal perpendicularity in certain Abelian groups Mika Mattila Department of Mathematics, Tampere University of Technology, Finland email: [email protected] Jorma K. Merikoski School of Information Sciences, University of Tampere, Finland email: [email protected] Pentti Haukkanen School of Information Sciences, University of Tampere, Finland email: [email protected] Timo Tossavainen Department of Art, Communication and Education, Lulea University of Technology, Sweden email: [email protected] Abstract. We define perpendicularity in an Abelian group Gas a binary relation satisfying certain five axioms. Such a relation is maximal if it is not a subrelation of any other perpendicularity in G. A motivation for the study is that the poset (P,⊆)of all perpendicularities in Gis a lattice if Ghas a unique maximal perpendicularity, and only a meet-semilattice if not. We study the cardinality of the set of maximal perpendicularities and, on the other hand, conditions on the existence of a unique maximal perpendicularity in the following cases: G∼ =Zn,Gis finite, Gis finitely generated, and G=Z⊕Z⊕ · · · . A few such conditions are found and a few conjectured. In studying Rn, we encounter perpendicularity in a vector space. 1 Introduction Over the years, the concept of “perpendicular” has been considered axiomatically from several different perspectives. Perhaps the most well-known axiomatic description of perpendicularity is presented in the classical textbook 2010 Mathematics Subject Classification: 20K99, 20K01, 20K25 Key words and phrases: Abelian group, perpendicularity 235 236 M. Mattila, J. Merikoski, P. Haukkanen, T. Tossavainen [1] by Bachmann. This approach is designed for the construction of plane geometry and it is based on studying reflections in the metric plane which is a notion to serve as a common basis of Euclidean, hyperbolic and elliptic planes. Davis [2,3] studied rings and Abelian groups with orthogonality relations. In his approach, the aim of defining an orthogonality relation on an Abelian group was to generalize the concept of a disjointness relation on a linear space introduced earlier by Veksler [7]. A more recent axiomatization of perpendicularity and parallelism is given in [4]. This axiom system was originally constructed for educational purposes and it is applicable enough for the examination of the geometry of perpendicular and parallel lines in the Euclidean plane, and certain other non-trivial planar or numeric models, too. The present approach to defining algebraic perpendicularity was originally laid down in [5]; this article is a sequel to that. Our definition is based on the idea of describing the additive properties of the elements of an inner product space for which the inner product is zero in terms of the binary operation of an Abelian group. Following the notation of [5], let G= (G, +) be an Abelian group, G6={0}, and let ⊥be a perpendicularity in G, that is, a binary relation satisfying (A1) ∀a∈G:∃b∈G:a⊥b, (A2) ∀a∈G\ {0}:a6⊥ a, (A3) ∀a, b ∈G:a⊥b⇒b⊥a, (A4) ∀a, b, c ∈G:a⊥b∧a⊥c⇒a⊥(b+c), (A5) ∀a, b ∈G:a⊥b⇒a⊥−b. The trivial perpendicularity a⊥b⇐⇒ a=0∨b=0 always exists. A perpendicularity ⊥is minimal if it is not a superrelation of any other perpendicularity in G. This clearly happens if and only if ⊥is trivial; hence, minimal perpendicularity is always unique. Similarly, a perpendicularity is maximal if it is not a subrelation of any other perpendicularity in G. A few results on minimal and maximal perpendicularities follow easily. Proposition 1 If Gis cyclic, then it has a unique maximal perpendicularity. If Gis cyclic and infinite, then it has only the trivial perpendicularity. Maximal perpendicularity in certain Abelian groups 237 Proof. See [5, Theorem 14] and [5, Example 8].  Maximal perpendicularity is not necessarily unique even if Gis finite. For example [5, Example 7], the Klein four group has three nontrivial perpendicularities, all of them maximal. Proposition 2 A maximal perpendicularity always exists. Proof. If ⊥1⊆⊥2⊆. . . are perpendicularities in G, then ∪∞ i=1⊥iis clearly a perpendicularity in G. So, the claim follows from Zorn’s lemma.  Let (P,⊆)be the poset (partially ordered set) of all perpendicularities in G. (In fact, every nonempty family of sets is a poset under subset relation.) Proposition 3 A perpendicularity in Gis maximal if and only if it is a maximal element of P. There is a unique maximal perpendicularity in Gif and only if there is a largest element in P. The trivial perpendicularity is the unique minimal perpendicularity of G, in other words, the smallest element of P. Proof. Easy and omitted.  A motivation for the present study is that Pis a lattice if Ghas a unique maximal perpendicularity, and only a meet-semilattice if not. Below we survey the uniqueness of maximal perpendicularity in the following cases: G∼ =Zn (Section 2), Gis finite (Section 3), Gis finitely generated (Section 4), and G∼ =Z⊕Z⊕ · · · ∼ =(Q+,·)(Sections 5 and 6). In addition to solving the question about the uniqueness in certain cases, we shall conjecture a few equivalent conditions for the existence of a unique maximal perpendicularity. We complete our paper by regarding Rnboth as an additive group and as a vector space. 2G∼ =Zn,n>1 If G∼ =Z, then it has only the trivial perpendicularity by Proposition 1. The case of G∼ =Zn=Z⊕ · · · ⊕ Z(ncopies, n>1) is hence more interesting. Let us choose g1,...,gn∈Gsuch that g1= (1, 0, 0, 0, . . . , 0), g2= (γ21, 1, 0, 0, . . . , 0), g3= (γ31, γ32, 1, 0, . . . , 0), . . . gn= (γn1, γn2, . . . , γn.n−1, 1),(1) 238 M. Mattila, J. Merikoski, P. Haukkanen, T. Tossavainen where the γij’s are integers. Denote by h·i the generated subgroup. Lemma 1 If G∼ =Znand g1,...,gnare as in (1), then G=hg1i⊕···⊕hgni.(2) Proof. For any x∈G, there obviously are unique ξ1,...,ξn∈Zsatisfying x=ξ1g1+· · · +ξngn.  Let g1,...,gn,n>1, be as above. Also choose g0 1,...,g0 n∈Gas in (1) such that g0 i6=gifor at least one i∈N={1,...,n}. So, there is m∈Nwith g1=g0 1, . . . , gm−1=g0 m−1, gm6=g0 m.(3) Let a, b ∈G. Then, by Lemma 1, a=a1+· · · +an=a0 1+· · · +a0 n, b =b1+· · · +bn=b0 1+· · · +b0 n,(4) where ai, bi∈ hgiiand a0 i, b0 i∈ hg0 iifor all i∈N. Define now the relations ⊥0and ⊥0 0by a⊥0b⇐⇒ ∀i∈N:ai=0∨bi=0, a⊥0 0b⇐⇒ ∀i∈N:a0 i=0∨b0 i=0. (5) These relations are clearly perpendicularities in G. Lemma 2 Let ⊥0and ⊥0 0be as in (5). A maximal perpendicularity ⊥max in G∼ =Zn,n>1, cannot contain both of them. Proof. We proceed by contradiction. Suppose that ⊥max⊇⊥0∪ ⊥0 0.(6) We have gm⊥0g1,...,gm−1and g0 m⊥0 0g0 1,...,g0 m−1implying that g0 m⊥0 0 g1,...,gm−1by (3). Therefore gm, g0 m⊥max g1,...,gm−1 by (6). Now, applying (A3), (A4) and (A5) yields that (gm−g0 m)⊥max (ξ1g1+· · · +ξm−1gm−1) for all ξ1,...,ξm−1∈Z. But d=gm−g0 m= (δ1,...,δn)has δm=· · · =δn=0, which implies that there are ξ1,...,ξm−1∈Zsuch that d=ξ1g1+· · ·+ξm−1gm−1. So, d⊥max d violating (A2) because d6=0by (3).  Maximal perpendicularity in certain Abelian groups 239 Theorem 1 There are infinitely many maximal perpendicularities in G∼ =Zn, n>1. Proof. There are infinitely many choices of the gi’s in (1). Different choices give different ⊥0’s in (5). Hence, the claim follows from Lemma 2. Is ⊥0defined by (5) maximal? The answer is negative. Namely, let a, b ∈G and write them as a=α1g1+· · · +αngn, b =β1g1+· · · +βngn, where the αi’s and βi’s are integers. Define ⊥1by a⊥1b⇐⇒ α1β1+· · · +αnβn=0. (7) Obviously ⊥1is a perpendicularity and ⊥0is its proper subset. But then, is ⊥1maximal? This question remains open, yet we conjecture as follows. Conjecture 1 A perpendicularity in G∼ =Zn,n > 1, is maximal if and only if it is of the form (7). We encounter another open question concerning the cardinality of the set S of maximal perpendicularities in G∼ =Zn,n>1. Denoting by |·|the cardinality, Theorem 1yields that |S|≥ℵ0. On the other hand, |Zn×Zn|=ℵ0, so, the cardinality of the set of all binary relations in G∼ =Znis 2ℵ0. Consequently, |S|≤2ℵ0. But which of these inequalities is equality? The following proposition tells what we already know. Proposition 4 If Conjecture 1is true, then the set of maximal perpendicularities in G∼ =Zn,n>1, has cardinality ℵ0. Proof. A maximal perpendicularity is of the form (5) by the conjecture. Because there are countably infinite choices of each gi,i>1, in (1), there are also countably infinite choices of the sequence (g1,...,gn). 3 Finite G In this section, we assume that Gis also finite (in addition to being Abelian). If Gis cyclic, then it has a unique maximal perpendicularity by Proposition 1. So, in the rest of this section, suppose that Gbe noncyclic if not mentioned otherwise. We begin by describing its structure. 240 M. Mattila, J. Merikoski, P. Haukkanen, T. Tossavainen Theorem 2 If Gis noncyclic and finite, then it has cyclic subgroups H1,...,Hr, r>1, of prime power order such that G=H1⊕ · · · ⊕ Hr.(8) These orders are unique. All decompositions (8)have the same number of summands of each order. Proof. See [6, p. 394, Theorem 1].  Let a, b ∈G, and let H1,...,Hrbe as in (8). Analogously to (4), a=a1+· · · +ar, b =b1+· · · +br, where ai, bi∈Hifor all i=1,...,r. Similarly as in (5), we now define a⊥0b⇐⇒ ∀i∈{1,...,r}:ai=0∨bi=0. (9) Further, if ∅ 6=A, B ⊆G, we write A⊥Bdenoting that x⊥yfor all x∈A, y∈B. Lemma 3 Let ⊥be a perpendicularity in G, and let a, b ∈G. If a⊥b, then hai⊥hbiand hai∩hbi={0}. Proof. Let ξ, η ∈Z. Then a⊥ηb by (A4) and (A5). Further, applying also (A3), we get ξa ⊥ηb. This proves the first claim. If z∈ hai ∩ hbi, then z=ξa =ηb for some ξ, η ∈Z. Now, the first claim implies that z⊥z; hence, z=0by (A1) verifying the second claim.  Theorem 3 Let Gbe noncyclic and finite. If |G|is square-free, then Ghas a unique maximal perpendicularity which, in fact, is ⊥0defined in (9). Proof. Let ⊥be a perpendicularity in G. We claim that ⊥⊆⊥0. We can omit the trivial perpendicularity; so, we suppose that 06=x, y ∈Gand x⊥y. By Theorem 2and square-freeness, Ghas cyclic subgroups H1,...,Hrwith prime orders p1,...,pr,r>1, respectively, such that G=H1⊕ · · · ⊕ Hr. Clearly, Hi={x∈G| |x|=pi}, i =1, . . . , r, Maximal perpendicularity in certain Abelian groups 241 where |·|denotes the order. Hence, this decomposition is unique (up to the ordering). Therefore, if His a subgroup of G, then H=Ht1⊕ · · · ⊕ Hts for certain indices t1,...,ts∈{1,...,r}. In particular, there are indices i1,...,ik and j1,...,jlsuch that hxi=Hi1⊕ · · · ⊕ Hik,hyi=Hj1⊕ · · · ⊕ Hjl. Since hxi ∩ hyi={0}by Lemma 3, we have Hiu6=Hjvfor all u, v. Therefore, x⊥0y, and the claim follows.  We conjecture that also the converse holds. Conjecture 2 Let Gbe as in Theorem 2. The following conditions are equivalent: (a) Ghas a unique maximal perpendicularity, (b) |G|is square-free, (c) (8)is unique (up to the ordering of the Hi’s). Theorem 3states that (b)⇒(a). The following proposition states that (a)⇒(c). The part (c)⇒(b) remains open. Proposition 5 Let Gbe as above. If Ghas a unique maximal perpendicularity, then (8)is unique. Proof. Contrary to the uniqueness of (8), we suppose that there are decompositions G=H1⊕ · · · ⊕ Hr=H0 1⊕ · · · ⊕ H0 r such that {H1,...,Hr}6={H0 1,...,H0 r}. Let Hi=hgiiand H0 i=hg0 ii,i= 1,...,r. We define ⊥0as we did in (9) and ⊥0 0in an analogous manner applying G=H0 1⊕ · · · ⊕ H0 r. As in the proof of Lemma 2, we can show that no maximal perpendicularity contains both ⊥0and ⊥0 0. (In this lemma, g1=g0 1, but without any role in the proof.) The uniqueness of maximal perpendicularity is thus violated.  242 M. Mattila, J. Merikoski, P. Haukkanen, T. Tossavainen 4 Finitely generated G Next, we assume that Gis finitely generated. In Proposition 1and Theorem 2, we already studied the cases Gis cyclic and finite, respectively. Therefore, let Gnow be noncyclic and infinite. Its structure is described in Theorem 4which follows immediately from [6, p. 411, Theorem 3]. Theorem 4 If Gis noncyclic and infinite but finitely generated, then it has cyclic subgroups H1,...,Hrof prime power order and a subgroup H0∼ =Zn, n≥1, such that G=H0⊕H1⊕ · · · ⊕ Hr=H0⊕K. (10) These orders are unique. All decompositions (10)have the same number of summands of each order. Applying our previous results, it is now easy to study maximal perpendicularities in G. Theorem 5 Let Gbe as in Theorem 4. If n>1, then Ghas infinitely many maximal perpendicularities. Proof. Decompose H0as in (2) and define perpendicularities ⊥0and ⊥0 0in H0 as in (5). By Lemma 2, a maximal perpendicularity in H0cannot contain both of them. Therefore, regarding them also as relations in G, a maximal perpendicularity in Gcannot either contain both of them. Because there are infinitely many ⊥0’s, the claim follows.  The proof of the next theorem is very similar to that of Theorem 3. Actually, the proof applies also when one subgroup is infinite (but cyclic). Theorem 6 Let Gbe as above. If n=1and |K|is square-free, then Ghas a unique maximal perpendicularity. By the similarity between the above results and those in Section 3, we present an analogy to Conjecture 2. Conjecture 3 Let Gbe as above. The following conditions are equivalent: (a) Ghas a unique maximal perpendicularity, (b) n=1and |K|is square-free, Maximal perpendicularity in certain Abelian groups 243 (c) (10)is unique (up to the ordering of the Hi’s). Applying an analogous argument as in the proof of Proposition 5, we also get the following proposition. Proposition 6 Let Gbe as above. If Ghas a unique maximal perpendicularity, then (10)is unique. 5G∼ =Z⊕Z⊕ · · · Now, let G∼ =Z⊕Z⊕ · · · (i.e., the set of infinite integer sequences with only finitely many nonzero terms). We begin the examination of this case by recording a result corresponding to Theorem 1. Theorem 7 There are infinitely many maximal perpendicularities in G∼ = Z⊕Z⊕ · · · . Proof. Analogously to (1), choose g1, g2,· · · ∈ Gsuch that g1= (1, 0, 0, . . . ) and gi= (γi1,...,γi,i−1, 1, 0, 0, . . . ), i =1, 2, . . . , and g0 1, g0 2, . . . similarly. A simple modification of the proof of Theorem 1 applies.  Let a, b ∈G. Write them as a=α1g1+α2g2+. . . , b =β1g1+β2g2+. . . . Analogously to (7), we define a⊥1b⇐⇒ α1β1+α2β2+· · · =0. (11) (The sum is finite, because only finitely many αi’s and βi’s are nonzero.) Analogously to Conjecture 1, we state as follows. Conjecture 4 Let Gbe as in Theorem 7. A perpendicularity in Gis maximal if and only if it is of the form (11). The question about the cardinality of the set of all perpendicularities in G∼ = Zn,n > 1, remained open in Proposition 4since the answer depends on Conjecture 1. However, we can solve this question in the case of G=Z⊕Z⊕· · · .