Perpendicularity in an Abelian Group

Abstract

We give a set of axioms to establish a perpendicularity relation in an Abelian group and then study the existence of perpendicularities in (ℤ_𝑛,+) and (ℚ₊,⋅) and in certain other groups. Our approach provides a justification for the use of the symbol ⟂ denoting relative primeness in number theory and extends the domain of this convention to some degree. Related to that, we also consider parallelism from an axiomatic perspective.

1. Introduction

In [1, page 115], Graham et al. made the following suggestion:

When gcd(𝑚,𝑛)=1, the integers 𝑚 and 𝑛 have no prime factors in common and we say that they are relatively prime.

This concept is so important in practice, we ought to have a special notation for it; but alas, number theorists have not agreed on a very good one yet. Therefore we cry: HEAR US, O MATHEMATICIANS OF THE WORLD! LET US NOT WAIT ANY LONGER! WE CAN MAKE MANY FORMULAS CLEARER BY ADOPTING A NEW NOTATION NOW! LET US AGREE TO WRITE “𝑚⟂𝑛”, AND TO SAY “𝑚  IS PRIME TO  𝑛,” IF  𝑚  AND  𝑛  ARE RELATIVELY PRIME. Like perpendicular lines do not have a common direction, perpendicular numbers do not have common factors.

In fact, this cry had been answered even before it was made. Namely, in studying 𝑙-groups (i.e., groups with a lattice structure), Birkhoff [2, page 295] defines that two positive elements 𝑎 and 𝑏 of an 𝑙-group are disjoint if 𝑎∧𝑏=0 and uses the notation 𝑎⟂𝑏 for disjoint elements. He also remarks that disjointness specializes to relative primeness in the 𝑙-group of positive integers.

A motivation for the present paper is to study how justified ultimately it is to use the symbol of perpendicularity to denote relative primeness. Does this practice rely only on the analogy between having no common direction and having no common factor or is there a deeper linkage to entitle this convention? This question leads us to ask which properties essentially establish the notion of perpendicularity in the algebraic context and what the most suitable algebraic context for the axiomatization of perpendicularity actually is; we have recently studied the axioms of perpendicularity from an elementary geometric point of view [3].

In an inner product space, perpendicularity obviously traces back to the inner product being zero. However, certain features of this perpendicularity can be shifted down to simpler algebraic structures. We will define perpendicularity in an Abelian group and examine it in Section 2. In Section 3, we will focus on perpendicularity in (ℤ_𝑛,+). Davis [4] defined perpendicularity in an Abelian group differently. In Section 4, we will introduce his approach and compare it with ours. Thereafter, we will consider divisibility in (ℚ₊,⋅) in Section 5 and parallelism in an Abelian group in Section 6. We will conclude our paper with a brief discussion and a supplement to the suggestion cited previously.

2. Axioms and Properties of Perpendicularity

Throughout this paper, 𝐺=(𝐺,+) is an Abelian group so that 𝐺≠{0}. Unless otherwise stated, ⟂ is a binary relation in 𝐺 satisfying(A1)∀𝑎∈𝐺∶∃𝑏∈𝐺∶𝑎⟂𝑏,(A2)∀𝑎∈𝐺⧵{0}∶𝑎⟂^╱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.

We call ⟂ a perpendicularity in 𝐺. This concept can be defined also in weaker structures by changing these axioms appropriately. For example, if 𝐺 is an Abelian monoid, then we simply omit (A5). Since the trivial perpendicularity𝑥⟂𝑦⟺𝑥=0∨𝑦=0 (1) always exists, we are mainly interested in nontrivial perpendicularities.

We call ⟂  maximal if it is not a subrelation of any other perpendicularity in 𝐺. There always exists a maximal perpendicularity. This is obvious if 𝐺 is finite and, otherwise, it follows from Zorn’s lemma.

Proposition 1 records some elementary properties of perpendicularity; we leave the proof for the reader.

Proposition 1. Perpendicularity ⟂ has the following properties:(a)∀𝑎∈𝐺∶𝑎⟂0, (b)∀𝑎∈𝐺⧵{0}∶𝑎⟂^╱−𝑎,(c)∀𝑎,𝑏₁,…,𝑏_𝑘∈𝐺,𝛾₁,…,𝛾_𝑘∈ℤ∶𝑎⟂𝑏₁,…,𝑏_𝑘⇒𝑎⟂(𝛾₁𝑏₁+⋯+𝛾_𝑘𝑏_𝑘),(d)∀𝑎,𝑏∈𝐺,𝜇,𝜈∈ℤ∶𝑎⟂𝑏⇒𝜇𝑎⟂𝜈𝑏.

The following characterization is useful in proving that a given relation is perpendicularity.

Proposition 2. A binary relation ⟂ in 𝐺 is perpendicularity if and only if it satisfies (A1) and (A2) and(A6)∀𝑎,𝑏,𝑐∈𝐺∶𝑎⟂𝑏∧𝑎⟂𝑐⇒(𝑏−𝑐)⟂𝑎.

Proof. The “only if”-part is trivial. To prove the “if”-part, we first show that our assumptions imply Proposition 1(a). Let 𝑎∈𝐺. By (A1), there is 𝑏∈𝐺 such that 𝑎⟂𝑏. Putting 𝑐∶=𝑏 in (A6) implies 0⟂𝑎; in particular 0⟂0. Further, (A6) with 𝑎∶=0, 𝑏∶=𝑎 and 𝑐∶=0 gives (𝑎−0)⟂0, that is, 𝑎⟂0. Now we can verify the remaining axioms.(A3) Assume 𝑎⟂𝑏. Apply (A6) with 𝑐∶=0; then 𝑏⟂𝑎.(A5) Assume 𝑎⟂𝑏. Apply (A6) with 𝑏∶=0 and 𝑐∶=𝑏. Then (−𝑏)⟂𝑎, and so, by (A3), 𝑎⟂−𝑏.(A4) Assume 𝑎⟂𝑏 and 𝑎⟂𝑐; then 𝑎⟂−𝑐 by (A5). Now (A6) with 𝑐∶=−𝑐 implies (𝑏−(−𝑐))⟂𝑎, that is, (𝑏+𝑐)⟂𝑎. Hence, by (A3), 𝑎⟂(𝑏+𝑐).

Is there a simple condition under which (A5) follows from (A1)–(A4)? The answer is positive.

Proposition 3. If all elements of 𝐺 have finite order and if ⟂ satisfies (A1)–(A4), then it satisfies (A5). If 𝐺 has at least one element of infinite order, then there exists a relation ⟂ which satisfies (A1)–(A4) but not (A5).

Proof. For the first part, assume that 𝑎,𝑏∈𝐺 satisfy 𝑎⟂𝑏, and let the order of 𝑏 be 𝑛. Then 𝑎⟂(𝑛−1)𝑏 by (A4). But (𝑛−1)𝑏=−𝑏 and (A5) follows. For the second part, let 𝑎∈𝐺 have infinite order. Then the subgroup {0,±𝑎,±2𝑎,…} is isomorphic to ℤ. The relation ⟂ defined by 𝑥⟂𝑦⟺∃𝜇,𝜈∈ℤ∶𝑥=𝜇𝑎∧𝑦=𝜈𝑎∧𝜇𝜈<0∨𝑥=0∨𝑦=0 (2) satisfies (A1)–(A4) but not (A5).

If ∅≠𝐴⊆𝐺, we define the perpendicular complement or ⟂-complement of 𝐴 as follows: 𝐴^⟂=𝑦∈𝐺∣𝑦⟂𝐴=_{𝐺⊇𝐵⟂𝐴}𝐵. (3) Here 𝑦⟂𝐴 means that 𝑦⟂𝑥 for all 𝑥∈𝐴, and 𝐵⟂𝐴 means that 𝑦⟂𝐴 for all 𝑦∈𝐵. Thus 𝐴^⟂ is the maximal set perpendicular to 𝐴. In particular, 𝐺^⟂={0} and {0}^⟂=𝐺. We also define ∅^⟂=𝐺.

Proposition 4. If 𝐴⊆𝐺, then 𝐴^⟂ is a subgroup of 𝐺. If 𝐺 is cyclic, then 𝐴^⟂ is cyclic.

Proof. The first part follows by applying the subgroup test and Proposition 2. The second part follows from the fact that any subgroup of a cyclic group is cyclic.

The next theorem tells when 𝐺 has a nontrivial perpendicularity.

Theorem 5. The following conditions are equivalent:(a)𝐺 has a nontrivial perpendicularity ⟂,(b)𝐺 has nontrivial cyclic subgroups 𝐻 and 𝐾 satisfying 𝐻∩𝐾={0},(c)𝐺 has nontrivial subgroups 𝐻 and 𝐾 satisfying 𝐻∩𝐾={0}.

Proof. (a)⇒(b). Since ⟂ is nontrivial, there exist 𝑥,𝑦∈𝐺⧵{0} such that 𝑥⟂𝑦. Then 𝐻=⟨𝑥⟩ and 𝐾=⟨𝑦⟩ apply. Here ⟨𝑎⟩ stands for the cyclic group generated by 𝑎.
(b)⇒(c). Trivial.
(c)⇒(a). Define ⟂ by 𝑥⟂𝑦⟺𝑥∈𝐻∧𝑦∈𝐾∨𝑥∈𝐾∧𝑦∈𝐻∨𝑥=0∨𝑦=0. (4)

Next we consider the maximal perpendicularity in some examples of groups. In Examples 6–9, the group operation is addition.

Example 6. Let 𝐺=ℤ₆. By Lagrange’s theorem [5, page 130, Theorem 2], the smallest 𝑛 such that ℤ_𝑛 has a nontrivial perpendicularity is 6=2⋅3 because 𝑛 must have at least two different prime factors. The nontrivial subgroups of ℤ₆ are 𝐻=⟨3⟩={0,3} and 𝐾=⟨2⟩={0,2,4}. Since 𝐺=𝐻⊕𝐾, it has exactly one nontrivial perpendicularity, defined by 0⟂0,1,2,3,4,5 and 3⟂2,4 and vice versa. Consequently, this perpendicularity is maximal.

Example 7. Let 𝐺=ℤ₂×ℤ₂, the Klein four group. Denote 0=(0,0), 𝑎=(0,1), 𝑏=(1,0), 𝑐=(1,1). The nontrivial subgroups are 𝐴={0,𝑎}, 𝐵={0,𝑏}, 𝐶={0,𝑐}. So, there are three nontrivial perpendicularities obtained as follows: Choose two elements of 𝑎, 𝑏, and 𝑐. Define that they are perpendicular to each other and to 0. Define that the remaining element is perpendicular to 0 only. All perpendicularities arising in this way are clearly maximal. We also note that 𝐺=𝐴⊕𝐵=𝐵⊕𝐶=𝐶⊕𝐴.

Example 8. Let 𝐺=ℤ. For each 𝑛≥2, the subgroup ⟨𝑛⟩=𝑛ℤ is nontrivial and there are no other nontrivial subgroups than those found in this way. Because 𝑚𝑛∈⟨𝑚⟩∩⟨𝑛⟩, there is no pair of nontrivial subgroups with intersection {0}. Hence 𝐺 has only the trivial perpendicularity.

Example 9. Let 𝐺=ℝ. Since ℝ has infinitely many pairs of nontrivial subgroups with intersection {0}, it has infinitely many nontrivial perpendicularities. For example, let 𝐻=ℚ and 𝐾={𝑥√2|𝑥∈ℚ} and define ⟂ by (4). To see that this perpendicularity is not maximal, let 𝐻₁={𝑥√3|𝑥∈ℚ} and 𝐾₁={𝑥√5|𝑥∈ℚ} and define ⟂^′ by 𝑥⟂^′∨∨𝑦⟺𝑥∈𝐻∧𝑦∈𝐾𝑥∈𝐾∧𝑦∈𝐻𝑥∈𝐻₁∧𝑦∈𝐾₁∨𝑥∈𝐾₁∧𝑦∈𝐻₁∨𝑥=0∨𝑦=0. (5) Then 𝑥⟂𝑦⇒𝑥⟂^′𝑦.

Example 10. Let 𝐺=(ℚ₊,⋅), where ℚ₊ denotes the set of positive rational numbers.
Every 𝑐∈ℚ₊ can be uniquely expressed as 𝑐=_𝑝∈ℙ𝑝^{𝜈𝑝(𝑐)}, (6) where 𝜈_𝑝(𝑐)∈ℤ for each 𝑝∈ℙ and only a finite number of them are nonzero. The symbol ℙ stands for the set of primes. For example, if 𝑐=8/25, then 𝜈₂(𝑐)=3, 𝜈₃(𝑐)=0, 𝜈₅(𝑐)=−2, 𝜈₇(𝑐)=𝜈₁₁(𝑐)=⋯=0.
Assign now 𝑎⟂𝑏⟺∀𝑝∈ℙ∶𝜈_𝑝(𝑎)=0∨𝜈_𝑝(𝑏)=0. (7) In other words, if 𝑚𝑎=𝑢𝑛,𝑏=𝑣,𝑚,𝑢,𝑛,𝑣∈ℤ₊,gcd(𝑚,𝑢)=gcd(𝑛,𝑣)=1, (8) then 𝑎⟂𝑏⟺gcd(𝑚𝑢,𝑛𝑣)=1. (9) Hence, for example, 8/9⟂7/5. In particular, for 𝑚,𝑛∈ℤ₊, applying (9) to 𝑚/1 and 𝑛/1 yields that 𝑚⟂𝑛⟺gcd(𝑚,𝑛)=1. (10)
So, it seems that Graham et al. were prophetically quite right with their suggestion—and not forgetting Birkhoff either! We will discuss the perpendicularity of positive rational numbers in more detail in Section 5.

3. Perpendicularity in ℤ_𝑛

Studying perpendicularities requires that we know the structure of 𝐺. Next we take a more thorough look at perpendicularity in ℤ_𝑛. To that end, we begin by introducing a suitable notation to discuss the structure of ℤ_𝑛 and record two lemmas which are useful in the search for the maximal perpendicularity. We will also use the notations introduced in Theorem 11 and the following lemmas throughout the next sections.

Theorem 11. If 𝑛=𝑝^𝛼1₁⋯𝑝^𝛼𝑟_𝑟, (11) where 𝑝₁,…,𝑝_𝑟∈ℙ are distinct and 𝛼₁,…,𝛼_𝑟>0, then ℤ_𝑛=𝐻₁⊕⋯⊕𝐻_𝑟, (12) where 𝐻_𝑖=⟨𝑒_𝑖⟩,𝑒_𝑖=𝑛𝑝^𝛼𝑖_𝑖,𝑖=1,…,𝑟. (13) The decomposition (12) is unique (up to the order of subgroups).

Proof. The claim (12) follows from [5, page 399, Corollary 1] and from the facts that ℤ_𝑝^𝛼_𝑖𝑖≅𝐻_𝑖 and 𝐻_𝑖∩𝐻_𝑗={0} for all 𝑖,𝑗=1,…,𝑟, 𝑖≠𝑗. Uniqueness follows from [5, page 399, Corollary 2].

Although we consider ℤ_𝑛 mainly as an Abelian group, it is now useful to work with ℤ_𝑛 as a ring.

Lemma 12. For all 𝑖,𝑗=1,…,𝑟, 𝑖≠𝑗, 𝑒²_𝑖≠0,𝑒_𝑖𝑒_𝑗=0. (14)

Proof. It is enough to consider 𝑖=1, 𝑗=2. Regarding 𝑒₁ and 𝑒₂ as integers, we have 𝑒²₁=𝑛²𝑝^2𝛼1₁=𝑝^2𝛼2₂⋯𝑝^2𝛼𝑟_𝑟(),𝑒≢0mod𝑛₁𝑒₂=𝑛𝑝^𝛼1₁𝑛𝑝^𝛼2₂=𝑝^𝛼2₂⋯𝑝^𝛼𝑟_𝑟𝑝^𝛼1₁𝑝^𝛼3₃⋯𝑝^𝛼𝑟_𝑟=𝑝^𝛼3₃⋯𝑝^𝛼𝑟_𝑟𝑛≡0(mod𝑛), (15) and (14) follows.

Lemma 13. Let ⟂ be a perpendicularity in ℤ_𝑛. Then ∀𝑎,𝑏,𝑐,𝑑∈ℤ_𝑛∶𝑎⟂𝑏⟹𝑐𝑎⟂𝑑𝑏. (16)

Proof. Let 𝑐=𝛾1,  𝑑=𝛿1, where 𝛾 and 𝛿 are integers with 0≤𝛾,𝛿<𝑛. Since 𝑐𝑎=(𝛾1)𝑎=𝛾(1𝑎)=𝛾𝑎 and, similarly, 𝑑𝑏=𝛿𝑏, Proposition 1(d) implies (16).

Now we are ready to introduce a perpendicularity which turns out to be maximal in ℤ_𝑛. Let 𝑥=𝑥₁+⋯+𝑥_𝑟,𝑦=𝑦₁+⋯+𝑦_𝑟∈ℤ_𝑛, (17) where 𝑥_𝑖,𝑦_𝑖∈𝐻_𝑖, 𝑖=1,…,𝑟. The relation ⟂₀, defined in ℤ_𝑛 by 𝑥⟂₀𝑦⟺∀𝑖∈{1,…,𝑟}∶𝑥_𝑖=0∨𝑦_𝑖=0, (18) is clearly a perpendicularity.

Theorem 14. The perpendicularity ⟂₀ is maximal and every other perpendicularity in ℤ_𝑛 is contained in it.

Proof. Let ⟂ be another perpendicularity in ℤ_𝑛. Our claim is that 𝑥⟂𝑦⇒𝑥⟂₀𝑦. By (12), we can express 𝑥=𝜉₁𝑒₁+⋯+𝜉_𝑟𝑒_𝑟,𝑦=𝜂₁𝑒₁+⋯+𝜂_𝑟𝑒_𝑟, (19) where the integers 𝜉_𝑖,𝜂_𝑖∈{0,…,𝑝^𝛼𝑖_𝑖−1} and the residue class 𝑒_𝑖=𝑛/𝑝^𝛼𝑖_𝑖, 𝑖=1,…,𝑟.
Suppose against the claim of theorem that there exist 𝑥,𝑦∈ℤ_𝑛 such that 𝑥⟂𝑦 but 𝑥⟂^╱₀𝑦. Then 𝜉_𝑖,𝜂_𝑖≠0 for some 𝑖. Reordering the indices so that 𝑖=1 and applying (16), we have 𝑥𝑒₁⟂𝑦𝑒₁ which implies that 𝜉₁𝑒²₁⟂𝜂₁𝑒²₁ (20) by (14). Hence, by Proposition 1(d), 𝜂₁gcd𝜉₁,𝜂₁𝜉₁𝑒²₁⟂𝜉₁gcd𝜉₁,𝜂₁𝜂₁𝑒²₁, (21) that is, lcm𝜉₁,𝜂₁𝑒²₁⟂lcm𝜉₁,𝜂₁𝑒²₁. (22) Consequently, lcm(𝜉₁,𝜂₁)𝑒²₁=0 by (A2). In other words, regarding also 𝑒₁ as an integer, lcm𝜉₁,𝜂₁𝑒²₁=lcm𝜉₁,𝜂₁𝑛²𝑝^2𝛼1₁=lcm𝜉₁,𝜂₁𝑝^2𝛼2₂⋯𝑝^2𝛼𝑟_𝑟≡0(mod𝑛), (23) and 𝑝^𝛼1₁ divides lcm(𝜉₁,𝜂₁). However, since it divides neither 𝜉₁ nor 𝜂₁, this is a contradiction. Hence, 𝑥⟂₀𝑦.

Considering the direct sum (12) external, we can identify 𝑥 and 𝑦 in (17) with vectors (𝑥₁,…,𝑥_𝑟) and (𝑦₁,…,𝑦_𝑟), respectively. So, it is natural to define their “inner product” by ⟨𝑥,𝑦⟩=𝑥₁𝑦₁+⋯+𝑥_𝑟𝑦_𝑟. (24) Proposition 15 shows that this operation coincides with the ordinary multiplication in ℤ_𝑛.

Proposition 15. Given 𝑥,𝑦∈ℤ_𝑛, ⟨𝑥,𝑦⟩=𝑥𝑦. (25)

Proof. We have 𝑥𝑦=^𝑟_𝑖=1𝑥_𝑖^𝑟_𝑖=1𝑦_𝑖=^𝑟_𝑖=1𝑥_𝑖𝑦_𝑖+^𝑟_{𝑖,𝑗=1𝑖≠𝑗}𝑥_𝑖𝑦_𝑗. (26) But, recalling (19) and (14), ^𝑟_{𝑖,𝑗=1𝑖≠𝑗}𝑥_𝑖𝑦_𝑗=^𝑟_{𝑖,𝑗=1𝑖≠𝑗}𝜉_𝑖𝜂_𝑗𝑒_𝑖𝑒_𝑗=0. (27) The claim follows.

Theorem 16. Let ⟂ be a perpendicularity in ℤ_𝑛. Then ∀𝑥,𝑦∈ℤ_𝑛∶𝑥⟂𝑦⟹𝑥𝑦=0. (28)

Proof. If 𝑥⟂𝑦, then 𝑥⟂₀𝑦 by Theorem 14. So, 𝑥𝑦=0 by (18) and (25).

Does the converse of Theorem 16 hold if ⟂=⟂₀? And, related to Proposition 15, is ⟨𝑥,𝑦⟩=𝑥𝑦 a proper inner product? Namely, an inner product in a real vector space is symmetric and bilinear and it satisfies ⟨𝑥,𝑥⟩=0⇒𝑥=0. The operation ⟨𝑥,𝑦⟩=𝑥𝑦 in ℤ_𝑛 has clearly the first and second properties but what about the third one? The answers to both questions are contained in Theorem 17.

Theorem 17. The following conditions are equivalent:(a)𝛼₁=⋯=𝛼_𝑟=1,(b)∀𝑥,𝑦∈ℤ_𝑛∶𝑥𝑦=0⇒𝑥⟂₀𝑦,(c)∀𝑥∈ℤ_𝑛∶𝑥²=0⇒𝑥=0.

Proof. (a)⇒(b). Assume that 𝑥⟂^╱₀𝑦. Express 𝑥 and 𝑦 as in (19). We can rearrange the indices so that, for some 𝑠∈{1,…,𝑟}, 𝜉_𝑖,𝜂_𝑖𝜉≠0,𝑖=1,…,𝑠,_𝑖=0∨𝜂_𝑖=0,𝑖=𝑠+1,…,𝑟. (29) By (25), 𝑥𝑦=𝜉₁𝜂₁𝑒²₁+⋯+𝜉_𝑠𝜂_𝑠𝑒²_𝑠. (30) If 𝜉₁𝜂₁𝑒²₁+⋯+𝜉_𝑠𝜂_𝑠𝑒²_𝑠=0, then the integer 𝜉₁𝜂₁𝑒²₁+⋯+𝜉_𝑠𝜂_𝑠𝑒²_𝑠≡0(mod𝑛), that is, 𝜉₁𝜂₁𝑛²𝑝²₁+⋯+𝜉_𝑠𝜂_𝑠𝑛²𝑝²_𝑠≡0mod𝑛=𝑝₁⋯𝑝_𝑟. (31) However, this is impossible because none of 𝑝₁,…,𝑝_𝑠 divides the left-hand side. (Namely, 𝑝_𝑖 divides every other summand except the 𝑖th one.) Therefore, 𝑥𝑦≠0 and our claim follows by contradiction.

(b)⇒(c). If 𝑥²=0, then 𝑥⟂₀𝑥 by (b) and 𝑥=0 by (A2).

(c)⇒(a). Suppose that (a) does not hold. Then, say, 𝛼₁>1. Let 𝑥=𝑛/𝑝₁. Since the integer 𝑥²=𝑛²𝑝²₁=𝑝^2𝛼1₁⋯𝑝^2𝛼𝑟_𝑟𝑝²₁=𝑝^2𝛼1₁⁻²𝑝^2𝛼2₂⋯𝑝^2𝛼𝑟_𝑟≡0(mod𝑛), (32)

𝑥²=0𝑥≠0

Corollary 18. If and only if the conditions of Theorem 17 are satisfied, then ∀𝑥,𝑦∈ℤ_𝑛∶𝑥⟂₀𝑦⟺𝑛∣𝑥𝑦, (33) where 𝑥𝑦 is the product of integers 𝑥 and 𝑦.

Example 19. Let 𝐺=ℤ₃₀. Since 30=2⋅3⋅5, the decomposition (12) is ℤ₃₀=302⊕303⊕305={0,15}⊕{0,10,20}⊕{0,6,12,18,24}. (34) For example, since 2=0⋅15+2⋅10+2⋅6 and 15=1⋅15+0⋅10+0⋅6, we have 2⟂₀15. Generally, (33) implies that 𝑥⟂₀𝑦 if and only if the corresponding integers satisfy 30∣(𝑥𝑦).

Example 20. Let 𝐺=ℤ₃₆₀. Since 360=2³⋅3²⋅5, we have ℤ₃₆₀=3602³⊕3603²⊕3605={}⊕{}⊕{}.45,90,…,31540,80,…,32072,144,…,288 (35) For example, 5⟂₀72 because 5=1⋅45+8⋅40+0⋅72 and 72=0⋅45+0⋅40+1⋅72. Now (33) is only necessary for ⟂₀ but not sufficient. For example, 10⟂^╱₀36 due to the fact that 10=2⋅45+7⋅40+0⋅72 and 36=4⋅45+0⋅40+3⋅72. However, 360|(10⋅36).

4. Another Definition of Perpendicularity

Davis [4] defined perpendicularity as a binary relation ⟂ in 𝐺 satisfying(D1)∀𝑎,𝑏∈𝐺∶𝑎⟂𝑏⇒𝑏⟂𝑎, (D2)∀𝑎∈𝐺∶0⟂𝑎, (D3)∀𝑎∈𝐺∶𝑎⟂𝑎⇒𝑎=0, (D4)∀𝑎,𝑏,𝑐∈𝐺∶𝑏⟂𝑎∧𝑐⟂𝑎⇒(𝑏+𝑐)⟂𝑎, (D5)∀𝑎,𝑏∈𝐺∶𝑎⟂𝑏⇔{𝑎}^⟂⟂∩{𝑏}^⟂⟂={0}.

He assumes that 𝐺 is an Abelian group, but the definition applies more generally to an Abelian monoid, too. It is easy to see that (D1)–(D4) are equivalent to (A1)–(A4). Axiom (D5) arises from introducing the concept of “disjointness” on a vector lattice; see [2, page 295], [6]. In fact, ⇔ can be replaced with ⇐ in (D5) due to the following observation.

Proposition 21. Assume that ⟂ satisfies (D1)–(D3) (or, equivalently, (A1)–(A3)). Then ∀𝑎,𝑏∈𝐺∶𝑎⟂𝑏⇒{𝑎}^⟂⟂∩{𝑏}^⟂⟂={0}. (36)

Proof. We show first that if ∅≠𝐴⊆𝐺, then 𝐴∩𝐴^⟂={0}. (37) If 𝑥∈𝐴∩𝐴^⟂, then 𝑥⟂𝑦 for all 𝑦∈𝐴. In particular, 𝑥⟂𝑥, and hence 𝑥=0 by (D3) and (37) follows.
Assume next that 𝑎⟂𝑏 and let 𝑥∈{𝑎}^⟂⟂∩{𝑏}^⟂⟂. Since 𝑥⟂{𝑏}^⟂ and 𝑎∈{𝑏}^⟂, we have 𝑥⟂𝑎 implying that 𝑥∈{𝑎}^⟂. Thus 𝑥∈{𝑎}^⟂∩{𝑎}^⟂⟂. But (37) applied to 𝐴={𝑎}^⟂ implies that {𝑎}^⟂∩{𝑎}^⟂⟂={0} and 𝑥=0 follows.

How are these two perpendicularities related? We give a partial answer. Let us denote by 𝐴 and 𝐷 the axioms (A1)–(A5) and (D1)–(D5), respectively.

Proposition 22. If all elements of 𝐺 have finite order, then 𝐷⇒𝐴. If 𝐺 has at least one element of infinite order, then there exists a relation ⟂ satisfying 𝐷 but not 𝐴.

Proof. The first claim follows from Proposition 3. Concerning the second one, ⟂ defined by (2) establishes a relation satisfying 𝐷 but not (A5).

Proposition 23. Assume that 𝐺 has elements 𝑎₁,𝑎₂,𝑎₃,𝑎₄≠0 such that ⟨𝑎_𝑖⟩∩⟨𝑎_𝑗⟩={0} whenever 𝑖≠𝑗. Then there exists a relation ⟂ satisfying 𝐴 but not 𝐷.

Proof. The relation ⟂ defined by (5) with 𝐻=⟨𝑎₁⟩, 𝐾=⟨𝑎₂⟩, 𝐻₁=⟨𝑎₃⟩, and 𝐾₁=⟨𝑎₄⟩ satisfies 𝐴. Since {𝑎₁}^⟂⟂∩{𝑎₃}^⟂⟂=⟨𝑎₁⟩∩⟨𝑎₃⟩={0} and 𝑎⟂₁^╱𝑎₃, it does not satisfy (D5).

5. Divisibility in ℚ₊

It will turn out that perpendicularity has got something to do also with divisibility in ℚ₊. To that end, we begin by noticing that every 𝑏∈ℚ₊ can be said to be a rational divisor of every 𝑎∈ℚ₊ because 𝑎=𝑐𝑏 for some 𝑐∈ℚ₊. So, this divisibility is trivial. In order to be able to discuss nontrivial divisibilities in ℚ₊, we have to consider which properties essentially establish this relation. The following three ones seem quite obvious.

Let ∣ be a relation in ℚ₊ satisfying(i)∀𝑎∈ℚ₊∶𝑎∣𝑎, (ii)∀𝑎,𝑏,𝑐∈ℚ₊∶𝑐∣𝑎∧𝑐∣𝑏⇒𝑐∣(𝑎𝑏),(iii)∀𝑎,𝑏,𝑐∈ℚ₊∶𝑐∣𝑏∧𝑏∣𝑎⇒𝑐∣𝑎.

We call ∣ a divisibility in ℚ₊. In other words, divisibility is a reflexive and transitive relation (i.e., a preorder) satisfying (ii). If 𝑏∣𝑎, then we say that 𝑏 is a divisor of 𝑎 and that 𝑎 is divisible by 𝑏. If 𝑑∣𝑎, 𝑑∣𝑏 and 𝑐∣𝑎∧𝑐∣𝑏⇒𝑐∣𝑑, then 𝑑 is a greatest common divisor of 𝑎 and 𝑏, denoted by gcd_∣(𝑎,𝑏). All these notions are meaningful also in any Abelian monoid.

Let us recall that every 𝑐∈ℚ₊ can be expressed as 𝑐=_𝑝∈ℙ𝑝^{𝜈𝑝(𝑐)}, (38) where 𝜈_𝑝(𝑐)∈ℤ for each 𝑝∈ℙ, and only a finite number of them are nonzero. If 𝜈_𝑝(𝑐)≠0, then 𝑝 is a prime factor of 𝑐.

Consider the set 𝑆 of all sequences (𝑛₂,𝑛₃,…,𝑛_𝑝,…), where the index runs through ℙ, each 𝑛_𝑝∈ℤ, and only a finite number of them are nonzero. The mapping 𝑓(𝑐)=𝜈₂(𝑐),𝜈₃(𝑐),…,𝜈_𝑝(𝑐),… (39) is an isomorphism from (ℚ₊,⋅) onto (𝑆,+) where addition is defined termwise. For example, 𝑓()845+𝑓=()+()=(),𝑓8250,2,1,0,0,…3,0,−2,0,0,…3,2,−1,0,0,…45⋅25=𝑓725=𝑓2³⋅3²⋅5⁻¹=(3,2,−1,0,0,…). (40)

Given 𝑎,𝑏∈ℚ₊, we define their “inner product” being the Euclidean inner product of the vectors 𝑓(𝑎) and 𝑓(𝑏): ⟨𝑎,𝑏⟩=⟨𝑓(𝑎),𝑓(𝑏)⟩=_𝑝∈ℙ𝜈_𝑝(𝑎)𝜈_𝑝(𝑏). (41) Since only a finite number of summands are nonzero, this sum is finite. For example, 845,()25=0⋅3+2⋅0+1⋅−2+0+0+⋯=−2. (42)

Next we define |𝑐| by setting 𝜈_𝑝(|𝑐|)=|𝜈_𝑝(𝑐)| for all 𝑝∈ℙ or, equivalently, |𝑐|=𝑓⁻¹((𝜈₂(|𝑐|),𝜈₃(|𝑐|),𝜈₅(|𝑐|),…)). For example, if 𝑐=40/63=2³⋅3⁻²⋅5¹⋅7⁻¹, then |𝑐|=2³⋅3²⋅5¹⋅7¹=2520. Letting ⟂₁ be the same relation as the one defined by (7), it can be characterized now by 𝑎⟂₁𝑏⟺⟨|𝑎|,|𝑏|⟩=0. (43)

Also the relation ⟂₂ in ℚ₊, defined by 𝑎⟂₂𝑏⟺⟨𝑎,𝑏⟩=0, (44) is a perpendicularity.

We will introduce one more nontrivial perpendicularity using divisibility. For that purpose, we first notice that the relation 𝛿 defined by 𝑏𝛿𝑎⟺∀𝑝∈ℙ∶𝜈_𝑝(𝑏)≤𝜈_𝑝(𝑎) (45) is a divisibility, gcd_𝛿(𝑎,𝑏) exists and is unique for all 𝑎,𝑏∈ℚ₊, and gcd_𝛿(𝑎,𝑏)=_𝑝∈ℙ𝑝^{min(𝜈𝑝(𝑎),𝜈𝑝(𝑏))}. (46) Assume now that 𝑚,𝑛,𝑢,𝑣∈ℤ₊ so that gcd(𝑚,𝑢)=gcd(𝑛,𝑣)=1. An alternative expression for (45) is 𝑛𝑣𝛿𝑚𝑢⟺𝑛|𝑚∧𝑢|𝑣, (47) and that for (46) is gcd_𝛿𝑚𝑢,𝑛𝑣=gcd(𝑚,𝑛)().lcm𝑢,𝑣 (48) For example, if 𝑎=45/14=2⁻¹⋅3²⋅5¹⋅7⁻¹ and 𝑏=33/100=2⁻²⋅3¹⋅5⁻²⋅11¹, then gcd_𝛿(𝑎,𝑏)=2⁻²⋅3¹⋅5⁻²⋅7⁻¹=3/700. Alternatively, gcd_𝛿()=()𝑎,𝑏gcd45,33()=3lcm14,100.700 (49) Since gcd_𝛿(|𝑚/𝑢|,|𝑛/𝑣|)=gcd(𝑚𝑢,𝑛𝑣), we have by (9) 𝑎⟂₁𝑏⟺gcd_𝛿(|𝑎|,|𝑏|)=1. (50) This relation generalizes (10) and answers the cry of Graham et al. in a slightly wider context than what they, perhaps, had thought.

Eugeni and Rizzi [7, Section 2] defined divisibility in ℚ₊ by setting the relation 𝛾 so that 𝑛𝑣𝛾𝑚𝑢⟺𝑛|𝑚∧𝑣|𝑢. (51) Then gcd_𝛾(𝑎,𝑏) always exists and is unique, and gcd_𝛾𝑚𝑢,𝑛𝑣=gcd(𝑚,𝑛)().gcd𝑢,𝑣 (52) For example, gcd_𝛾45,1433=()100gcd45,33()=3gcd14,1002. (53) We define now the corresponding perpendicularity by writing 𝑎⟂_ER𝑏⟺gcd_𝛾()()()𝑎,𝑏=1⟺gcd𝑚,𝑛=gcd𝑢,𝑣=1. (54)

Summing up, we have at least three nontrivial perpendicularities in ℚ₊. Let us see how they relate to one another.

⟂₁  versus ⟂₂. Clearly ⟂₁⇒⟂₂ (i.e., 𝑥⟂₁𝑦⇒𝑥⟂₂𝑦). The converse does not hold. For example, 6⟂₂2/3 but 6⟂^╱₁2/3.

⟂₁  versus ⟂_ER. Clearly ⟂₁⇒⟂_ER. The converse does not hold. For example, 2/3⟂_ER3/2 but 2/3⟂^╱₁3/2.

⟂₂  versus ⟂_ER. These perpendicularities are independent. For example, 6⟂₂2/3 but 6⟂^╱_ER2/3. On the other hand, 2/3⟂_ER3/2 but 2/3⟂^╱₂3/2.

However, regarding (ℤ₊,⋅) as a submonoid of (ℚ₊,⋅), it is obvious that ⟂₁=⟂₂=⟂_ER in ℤ₊. Moreover, in ℤ₊, they yield the very perpendicularity proposed by Graham et al.

6. Parallelism

Parallelism is closely related to perpendicularity. Considering different geometric contexts we notice soon that, in general, parallelism does not have any other properties except those of equivalence. However, any equivalence relation cannot be said to stand for parallelism in any reasonable way. This leads us to ask whether it is possible or not to define parallelism in Abelian groups having a perpendicularity so that it makes sense.

Let 𝐺 have a perpendicularity ⟂ and let 𝑎,𝑏∈𝐺. We say that 𝑎 and 𝑏 are parallel and write 𝑎‖𝑏 if {𝑎}^⟂={𝑏}^⟂. The relation ‖ is clearly an equivalence. If 𝑎≠0, then 𝑎∦0, since {0}^⟂=𝐺 by Proposition 1(a) but {𝑎}^⟂≠𝐺 by (A2). All nonzero elements are parallel if and only if ⟂ is trivial.

If 𝐺=ℤ_𝑛 and ⟂=⟂₀, then, recalling (19), 𝑥‖𝑦⟺∀𝑖∈{1,…,𝑟}∶𝜉_𝑖=0⟺𝜂_𝑖⟺{𝑥}=0_⟂=𝑦^⟂=𝐻_𝑖𝑖⊕⋯⊕𝐻_𝑖𝑡, (55) where 𝜉_𝑖=𝜂_𝑖=0⇔𝑖∈{𝑖₁,…,𝑖_𝑡}. For example, consider ℤ₃₀ (see Example 19). Since 2=0⋅15+2⋅10+2⋅6 and 16=0⋅15+1⋅10+1⋅6, we have {2}^⟂={16}^⟂={0,15}, and so 2‖16.

Now, let 𝐺=ℚ₊ and let ⟂₁, ⟂₂, and ⟂_ER be as before. Denote the corresponding parallelisms by ‖₁, ‖₂, and ‖_ER, respectively. Then 𝑎‖₁𝑏 if and only if 𝑎 and 𝑏 have the same prime factors. Further, 𝑚/𝑢‖_ER𝑛/𝑣 if and only if 𝑚 and 𝑛 have the same prime factors and 𝑢 and 𝑣 have the same prime factors.

Let us study how these parallelisms relate to one another.

‖₁  versus ‖₂. We show that ‖₂⇒‖₁. Assume first that 𝑎‖₂𝑏. If 𝑎∦₁𝑏, then there exists 𝑝₀∈ℙ such that, say, 𝜈_𝑝0(𝑎)=0 and 𝜈_𝑝0(𝑏)≠0. But now 𝑝₀⟂₂𝑎 and 𝑝⟂₀^╱₂𝑏, and so {𝑎}^⟂2≠{𝑏}^⟂2 contradicting the assumption. The converse does not hold. For example, let 𝑎=6 and 𝑏=12; then 𝑎‖₁𝑏. If 𝑥=2/3, then 𝑥⟂₂𝑎 but 𝑥⟂^╱₂𝑏, and hence {𝑎}^⟂2≠{𝑏}^⟂2. In other words, 𝑎∦₂𝑏.

‖₁  versus ‖_ER. Clearly ‖_ER⇒‖₁. The converse does not hold. For example, 2/3‖₁3/2 but 2/3∦_ER3/2.

‖₂  versus ‖_ER. We show that ‖₂⇒‖_ER. Given 𝑝₁,…,𝑝_𝑡∈ℙ, denote by 𝑁(𝑝₁,…,𝑝_𝑡) the set of such positive integers that are not divisible by any 𝑝_𝑖, 𝑖=1,…,𝑡. Let 𝑎=𝑚/𝑢∈ℚ₊, gcd(𝑚,𝑢)=1. Factorize 𝑚=𝑝^𝛼1₁⋯𝑝^𝛼ℎ_ℎ,𝑢=𝑞^𝛽1₁⋯𝑞^𝛽𝑘_𝑘, (56) where 𝑝₁,…,𝑝_ℎ,𝑞₁,…,𝑞_𝑘∈ℙ are distinct and 𝛼₁,…,𝛼_ℎ,𝛽₁,…,𝛽_𝑘>0. (If 𝑚=1 or 𝑢=1, then the corresponding “empty product” is one.) Now {𝑎}^⟂2=𝑝^𝜉1₁⋯𝑝^𝜉ℎ_ℎ𝑞^𝜂1₁⋯𝑞^𝜂𝑘_𝑘𝑥𝑦∣𝛼₁𝜉₁+⋯+𝛼_ℎ𝜉_ℎ+𝛽₁𝜂₁+⋯+𝛽_𝑘𝜂_𝑘𝑝=0,𝑥,𝑦∈𝑁₁,…,𝑝_ℎ,𝑞₁,…,𝑞_𝑘. (57) (The “empty sum” is zero.) Assume that 𝑏=𝑛/𝑣∈ℚ₊, gcd(𝑛,𝑣)=1, satisfies 𝑎‖₂𝑏, that is, {𝑎}^⟂2={𝑏}^⟂2. Then, by (57), necessarily 𝑛=𝑝^𝜌1₁⋯𝑝^𝜌ℎ_ℎ,𝑣=𝑞^𝜎1₁⋯𝑞^𝜎𝑘_𝑘, (58) where 𝜌₁,…,𝜌_ℎ,𝜎₁,…,𝜎_𝑘>0. Hence 𝑎‖_ER𝑏, and the claim follows. The converse is not valid. For example, 2/3‖_ER4/3 but 2/3∦₂4/3.

7. Discussion

This paper began with a citation by three established mathematicians and computer scientists who showed a remarkable intuition by promoting the use of the symbol of perpendicularity in number theory. Indeed, we have previously seen how this notion settles comfortably in this setting and gains new meanings at a more general level in the context of Abelian group theory. We conclude this paper with the following supplement to their proposal.

Let perpendicularity and parallelism mean here ⟂₁ and ‖₁, respectively. Consider the “direction vector” of 𝑐∈ℚ₊ by (𝑐(2),𝑐(3),…,𝑐(𝑝),…), where 𝑐(𝑝)=0 if 𝜈_𝑝(𝑐)=0 and 𝑐(𝑝)=1 otherwise. For example, the direction vectors of 45, 1, and 8/25 are, respectively (0,1,1,0,0,…), (0,0,…), and (1,0,1,0,0,…).

Now, like the directions of perpendicular lines are as different as possible, the prime factors of perpendicular (positive rational) numbers are as different as possible; that is, such numbers do not have common prime factors. In other words, the direction vectors of perpendicular numbers are as different as possible in the sense that they have no common element of value one. Like parallel lines have the same direction, parallel numbers have the same prime factors. In other words, their direction vectors are equal.

Finally, we note that perpendicularity can be axiomatized in a natural way also in many other algebraic structures. Davis [8] did that in a ring. In a vector space, perpendicularity is customarily defined based on an inner product. Another possible approach is to supplement (A1)–(A5) with suitable axioms concerning the multiplication of a vector by a scalar. It might be interesting to study under which additional conditions there exists an inner product inducing this perpendicularity.

Acknowledgment

The authors would like to thank the referees for carefully reading the paper and kind comments.