Research Article | Open Access
Pentti Haukkanen, Mika Mattila, Jorma K. Merikoski, Timo Tossavainen, "Perpendicularity in an Abelian Group", International Journal of Mathematics and Mathematical Sciences, vol. 2013, Article ID 983607, 8 pages, 2013. https://doi.org/10.1155/2013/983607
Perpendicularity in an Abelian Group
We give a set of axioms to establish a perpendicularity relation in an Abelian group and then study the existence of perpendicularities in and 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.
In [1, page 115], Graham et al. made the following suggestion:
When , the integers and 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 “”, AND TO SAY “ IS PRIME TO ,” IF AND 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 -groups (i.e., groups with a lattice structure), Birkhoff [2, page 295] defines that two positive elements and of an -group are disjoint if and uses the notation for disjoint elements. He also remarks that disjointness specializes to relative primeness in the -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 .
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 . Davis  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 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, is an Abelian group so that . Unless otherwise stated, is a binary relation in satisfying,,,,.
We call a perpendicularity in . This concept can be defined also in weaker structures by changing these axioms appropriately. For example, if is an Abelian monoid, then we simply omit (A5). Since the trivial perpendicularity always exists, we are mainly interested in nontrivial perpendicularities.
We call maximal if it is not a subrelation of any other perpendicularity in . There always exists a maximal perpendicularity. This is obvious if 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), (b),(c),(d).
The following characterization is useful in proving that a given relation is perpendicularity.
Proposition 2. A binary relation in is perpendicularity if and only if it satisfies (A1) and (A2) and.
Proof. The “only if”-part is trivial. To prove the “if”-part, we first show that our assumptions imply Proposition 1(a). Let . By (A1), there is such that . Putting in (A6) implies ; in particular . Further, (A6) with , and gives , that is, . Now we can verify the remaining axioms.(A3) Assume . Apply (A6) with ; then .(A5) Assume . Apply (A6) with and . Then , and so, by (A3), .(A4) Assume and ; then by (A5). Now (A6) with implies , that is, . Hence, by (A3), .
Is there a simple condition under which (A5) follows from (A1)–(A4)? The answer is positive.
Proposition 3. If all elements of have finite order and if satisfies (A1)–(A4), then it satisfies (A5). If 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 satisfy , and let the order of be . Then by (A4). But and (A5) follows. For the second part, let have infinite order. Then the subgroup is isomorphic to . The relation defined by satisfies (A1)–(A4) but not (A5).
If , we define the perpendicular complement or -complement of as follows: Here means that for all , and means that for all . Thus is the maximal set perpendicular to . In particular, and . We also define .
Proposition 4. If , then is a subgroup of . If is cyclic, then 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 has a nontrivial perpendicularity.
Theorem 5. The following conditions are equivalent:(a) has a nontrivial perpendicularity ,(b) has nontrivial cyclic subgroups and satisfying ,(c) has nontrivial subgroups and satisfying .
Proof. (a)(b). Since is nontrivial, there exist such that . Then and apply. Here stands for the cyclic group generated by .
(c)(a). Define by
Example 6. Let . By Lagrange’s theorem [5, page 130, Theorem 2], the smallest such that has a nontrivial perpendicularity is because must have at least two different prime factors. The nontrivial subgroups of are and . Since , it has exactly one nontrivial perpendicularity, defined by and and vice versa. Consequently, this perpendicularity is maximal.
Example 7. Let , the Klein four group. Denote , , , . The nontrivial subgroups are , , . So, there are three nontrivial perpendicularities obtained as follows: Choose two elements of , , and . 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 .
Example 8. Let . For each , the subgroup is nontrivial and there are no other nontrivial subgroups than those found in this way. Because , there is no pair of nontrivial subgroups with intersection . Hence has only the trivial perpendicularity.
Example 9. Let . Since has infinitely many pairs of nontrivial subgroups with intersection , it has infinitely many nontrivial perpendicularities. For example, let and and define by (4). To see that this perpendicularity is not maximal, let and and define by Then .
Example 10. Let , where denotes the set of positive rational numbers.
Every can be uniquely expressed as where for each and only a finite number of them are nonzero. The symbol stands for the set of primes. For example, if , then , , , .
Assign now In other words, if then Hence, for example, . In particular, for , applying (9) to and yields that
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
Studying perpendicularities requires that we know the structure of . Next we take a more thorough look at perpendicularity in . To that end, we begin by introducing a suitable notation to discuss the structure of 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 where are distinct and , then where The decomposition (12) is unique (up to the order of subgroups).
Although we consider mainly as an Abelian group, it is now useful to work with as a ring.
Lemma 12. For all , ,
Proof. It is enough to consider , . Regarding and as integers, we have and (14) follows.
Lemma 13. Let be a perpendicularity in . Then
Now we are ready to introduce a perpendicularity which turns out to be maximal in . Let where , . The relation , defined in by is clearly a perpendicularity.
Theorem 14. The perpendicularity is maximal and every other perpendicularity in is contained in it.
Proof. Let be another perpendicularity in . Our claim is that . By (12), we can express
where the integers and the residue class , .
Suppose against the claim of theorem that there exist such that but . Then for some . Reordering the indices so that and applying (16), we have which implies that by (14). Hence, by Proposition 1(d), that is, Consequently, by (A2). In other words, regarding also as an integer, and divides . However, since it divides neither nor , this is a contradiction. Hence, .
Considering the direct sum (12) external, we can identify and in (17) with vectors and , respectively. So, it is natural to define their “inner product” by Proposition 15 shows that this operation coincides with the ordinary multiplication in .
Proposition 15. Given ,
Theorem 16. Let be a perpendicularity in . Then
Does the converse of Theorem 16 hold if ? And, related to Proposition 15, is a proper inner product? Namely, an inner product in a real vector space is symmetric and bilinear and it satisfies . The operation in 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),(b),(c).
Proof. (a)(b). Assume that . Express and as in (19). We can rearrange the indices so that, for some , By (25), If , then the integer , that is, However, this is impossible because none of divides the left-hand side. (Namely, divides every other summand except the th one.) Therefore, and our claim follows by contradiction.
(b)(c). If , then by (b) and by (A2).
(c)(a). Suppose that (a) does not hold. Then, say, . Let . Since the integer
Corollary 18. If and only if the conditions of Theorem 17 are satisfied, then where is the product of integers and .
Example 20. Let . Since , we have For example, because and . Now (33) is only necessary for but not sufficient. For example, due to the fact that and . However, .
4. Another Definition of Perpendicularity
Davis  defined perpendicularity as a binary relation in satisfying, , , , .
He assumes that 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], . 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
Proof. We show first that if , then
If , then for all . In particular, , and hence by (D3) and (37) follows.
Assume next that and let . Since and , we have implying that . Thus . But (37) applied to implies that and follows.
How are these two perpendicularities related? We give a partial answer. Let us denote by and the axioms (A1)–(A5) and (D1)–(D5), respectively.
Proposition 22. If all elements of have finite order, then . If has at least one element of infinite order, then there exists a relation satisfying but not .
Proposition 23. Assume that has elements such that whenever . Then there exists a relation satisfying but not .
Proof. The relation defined by (5) with , , , and satisfies . Since and , it does not satisfy (D5).
5. Divisibility in
It will turn out that perpendicularity has got something to do also with divisibility in . To that end, we begin by noticing that every can be said to be a rational divisor of every because for some . So, this divisibility is trivial. In order to be able to discuss nontrivial divisibilities in , we have to consider which properties essentially establish this relation. The following three ones seem quite obvious.
Let be a relation in satisfying(i), (ii),(iii).
We call a divisibility in . In other words, divisibility is a reflexive and transitive relation (i.e., a preorder) satisfying (ii). If , then we say that is a divisor of and that is divisible by . If , and , then is a greatest common divisor of and , denoted by . All these notions are meaningful also in any Abelian monoid.
Let us recall that every can be expressed as where for each , and only a finite number of them are nonzero. If , then is a prime factor of .
Consider the set of all sequences , where the index runs through , each , and only a finite number of them are nonzero. The mapping is an isomorphism from onto where addition is defined termwise. For example,
Given , we define their “inner product” being the Euclidean inner product of the vectors and : Since only a finite number of summands are nonzero, this sum is finite. For example,
Next we define by setting for all or, equivalently, . For example, if , then . Letting be the same relation as the one defined by (7), it can be characterized now by
Also the relation in , defined by is a perpendicularity.
We will introduce one more nontrivial perpendicularity using divisibility. For that purpose, we first notice that the relation defined by is a divisibility, exists and is unique for all , and Assume now that so that . An alternative expression for (45) is and that for (46) is For example, if and , then . Alternatively, Since , we have by (9) 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 by setting the relation so that Then always exists and is unique, and For example, We define now the corresponding perpendicularity by writing
Summing up, we have at least three nontrivial perpendicularities in . Let us see how they relate to one another.
versus . Clearly (i.e., ). The converse does not hold. For example, but .
versus . Clearly . The converse does not hold. For example, but .
versus . These perpendicularities are independent. For example, but . On the other hand, but .
However, regarding as a submonoid of , it is obvious that in . Moreover, in , they yield the very perpendicularity proposed by Graham et al.
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 have a perpendicularity and let . We say that and are parallel and write if . The relation is clearly an equivalence. If , then , since by Proposition 1(a) but by (A2). All nonzero elements are parallel if and only if is trivial.
Now, let and let , , and be as before. Denote the corresponding parallelisms by , , and , respectively. Then if and only if and have the same prime factors. Further, if and only if and have the same prime factors and and have the same prime factors.
Let us study how these parallelisms relate to one another.
versus . We show that . Assume first that . If , then there exists such that, say, and . But now and , and so contradicting the assumption. The converse does not hold. For example, let and ; then . If , then but , and hence . In other words, .
versus . Clearly . The converse does not hold. For example, but .
versus . We show that . Given , denote by the set of such positive integers that are not divisible by any , . Let , . Factorize where , are distinct and . (If or , then the corresponding “empty product” is one.) Now (The “empty sum” is zero.) Assume that , , satisfies , that is, . Then, by (57), necessarily where . Hence , and the claim follows. The converse is not valid. For example, but .
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 and , respectively. Consider the “direction vector” of by , where if and otherwise. For example, the direction vectors of 45, 1, and are, respectively , , and .
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  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.
The authors would like to thank the referees for carefully reading the paper and kind comments.
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