A prime factor theorem for bipartite graphs

Transcription

1 A prime factor theorem for bipartite graphs Richard Hammack, Owen Puffenberger Department of Mathematics and Applied Mathematics Virginia Commonwealth University Richmond, VA 23284, USA Abstract It has long been known that the class of connected nonbipartite graphs (with loops allowed) obeys unique prime factorization over the direct product of graphs. Moreover, it is known that prime factorization is not necessarily unique in the class of connected bipartite graphs. But any prime factorization of a connected bipartite graph has exactly one bipartite factor. It has become folklore in some circles that this prime bipartite factor must be unique among all factorings, but until now this conjecture has withstood proof. This paper presents a proof. We show that if a connected bipartite graph G has two factorings G A B and G A B, where B and B are prime and bipartite, then B B. Keywords: graph direct product, bipartite graphs, graph factorization 2000 MSC: 05C60 1. Introduction We assume our reader is familiar with graph products, but we review the main definitions here to synchronize notation and terminology. See [3] for a survey. Let Γ be the set of (isomorphism classes of) graphs without loops; thus Γ Γ 0, where Γ 0 is the set of graphs with loops allowed. The direct product Corresponding author addresses: rhammack@vcu.edu (Richard Hammack), puffenbergod@vcu.edu (Owen Puffenberger) Preprint submitted to European J. Combinatorics April 14, 2014

2 of two graphs A, B Γ 0 is the graph A B with vertices V (A) V (B) and edges E(A B) = {(a, b)(a, b ) aa E(A) and bb E(B)}. Figure 1 shows an example. This product is commutative and associative in the sense that the maps (a, b) (b, a) and (a, (b, c)) ((a, b), c) are isomorphisms A B B A and A (B C) (A B) C, respectively. B A B A Figure 1: Direct product of graphs If K1 denotes a vertex on which there is a loop, then K 1 A A for any graph A, so K1 is the unit for the direct product. A nontrivial graph G Γ 0 is prime over if for any factoring G A B into graphs A, B Γ 0 it follows that one of A or B is K1 and the other is isomorphic to G. A consequence of a fundamental result by McKenzie [6] is that every connected nonbipartite graph in Γ 0 factors over uniquely into primes in Γ 0. Specifically, if G = A 1 A 2 A k and G = B 1 B 2 B l are two prime factorings of a connected nonbipartite graph G, then k = l and B i A π(i) for some permutation π of {1, 2,..., k}. McKenzie s paper involves general relational structures; for purely graph-theoretical proofs of unique prime factorization, see Imrich [5], or [3] for a more recent proof. But if G is bipartite, its prime factorization may not be unique. Figure 2 illustrates this. It shows a graph G = C 6 and prime factorings G A K 2 and G A K 2 with A / A. Indeed, we claim that if B is any prime bipartite graph, there exists a bipartite graph G with distinct prime factorings, but with common bipartite prime factor B. To see this, say B has a bipartition V (B) = X 0 X 1 and consider graphs G = A B and A B, where A and A are the (prime) graphs from Figure 2. To establish the claim we assert that there is an isomorphism 1 2

3 A A Figure 2: Two prime factorings A K 2 and A K 2 of the bipartite graph G = C 6. ϕ A B A B. Indeed, ϕ(a, b) = { (a, b) if b X 0 ( a, b) if b X 1 is such an isomorphism, as the reader is invited to check. Let us examine factorings of bipartite graphs in more detail. An oft-used theorem by Weichsel states the following: Let A and B be connected graphs. Then A B is connected if and only if at least one of A or B is not bipartite; if both A and B are bipartite, then A B has exactly two components. Moreover, A B is bipartite if and only if at least one factor is bipartite. (See Theorem 5.9 of [3] and its proof.) It follows that if a connected bipartite graph G has a prime factoring G = A 1 A 2 A k, then exactly one prime factor is bipartite. This is borne out in Figure 2; but notice that although the prime factorings of G = C 6 are different, the bipartite factor B = K 2 is the same. Examples such as this one have prompted the following conjecture. Conjecture 1. Given two prime factorings of a connected bipartite graph, the prime bipartite factors are isomorphic. The origins of this conjecture are unclear, but it has circulated in the product graph community for some time. The article [2] proves it in the 1 special case of graphs that have a bipartite factor of K 2. We here offer a completely general proof. We will show that if a connected bipartite graph factors as G A B and G A B, where B and B are prime and bipartite, then B B. Doing this involves some machinery, which the next three sections review. Section 2 recalls the notion of the Cartesian product of graphs, and theorems regarding unique factorization over this product. Section 3 discusses the notion of so-called R-thinness, an important ingredient of our proof. Section 4 lays out the elements of the Cartesian skeleton operator on 3

4 graphs, which converts certain questions about the direct product to questions about the (more easily understood) Cartesian product. All of these things are used to prove the main results in Section 5. Parts of this work were treated in the masters thesis of second author, directed by the first. Thanks to Wilfried Imrich, whose work on prime factorization of nonbipartite graphs served as inspiration for many parts of this paper. Many thanks also to the referees for their helpful comments. 2. The Cartesian Product The Cartesian product of two graphs A, B Γ is the graph A B Γ with vertices V (A) V (B) and edges E(A B) = {(a, b)(a, b ) aa E(A) and b = b, or a = a and bb E(B)}. (See Figure 3.) The Cartesian product is commutative and associative in the sense that A B B A and A (B C) (A B) C. Letting B + C denote the disjoint union of graphs B and C, we also get the distributive law A (B + C) = A B + A C. (1) Observe that this is true equality, rather than mere isomorphism. B A B Figure 3: A Cartesian product of graphs. Clearly K 1 A A for any graph A, so K 1 is the unit for the Cartesian product. A nontrivial graph G is prime over if for any factoring G A B, one of A or B is K 1. Certainly every graph can be factored into prime factors in Γ. Sabidussi and Vizing [7, 8] proved that each connected graph has a unique prime factoring. More precisely, we have the following. A 4

5 Theorem 1 (Theorem 6.8 of [3]). Let G, H Γ be isomorphic connected graphs G = G 1 G k and H = H 1 H l, where each factor G i and H i is prime. Then k = l, and for any isomorphism ϕ G H, there is a permutation π of {1, 2,..., k} and isomorphisms ϕ i G π(i) H i for which ϕ(x 1, x 2,..., x k ) = (ϕ 1 (x π(1) ), ϕ 2 (x π(2) ),..., ϕ k (x π(k) )). Theorem 1 invites us to identify each H i with G π 1 (i), yielding a corollary. Corollary 1. If ϕ G 1 G k H 1 H k is an isomorphism, and each G i and H i is prime, then the vertices of each H i can be relabeled so that ϕ(x 1, x 2,..., x k ) = (x π(1), x π(2),..., x π(k) ) for a permutation π of {1,..., k}. That is, ϕ merely permutes its arguments. 3. R-Thin Graphs The notion of so-called R-thinness is an important issue in factorings over the direct product. McKenzie [6] uses this idea (in a somewhat more general form), citing an earlier use by Chang [1]. A graph G is called R-thin if no two vertices have the same neighborhood, that is, if N G (x) = N G (y) implies x = y. Said differently, any vertex is uniquely determined by its neighborhood. Maps between R-thin graphs are often conveniently rigid. For example, note that an isomorphism between R-thin bipartite graphs is completely determined by its effect on one partite set, because the neighborhoods of the other vertices are contained in that partite set. For another example, we note in passing that an analogue of Theorem 1 for the direct product holds for R-thin nonbipartite graphs. See Theorem 8.15 of [3]. More generally, one forms a relation R on the vertices of an arbitrary graph. Two vertices x and x of a graph G are in relation R, written xrx, precisely if their open neighborhoods are identical, that is, if N G (x) = N G (x ). It is a simple matter to check that R is an equivalence relation on V (G). (For example, the equivalence classes for A B in Figure 4 are {b1}, {b2}, {a1, c1} and {a2, c2}; those of A are {a, c} and {b}, and those of B are {1} and {2}.) A graph is then R-thin if and only if each R-equivalence class contains exactly one vertex. Given two R-equivalence classes X and Y (not necessarily distinct), it is easy to check that either every vertex in X is adjacent to every vertex in Y, or no vertex in X is adjacent to any in Y. 5

6 B A B B/R a2 b2 c2 2 {2} (A B)/R {b2} {a2, c2} 1 {1} a1 b1 c1 a b A c {b1} {a1, c1} {b} {a, c} A/R Figure 4: Graphs A, B and A C (left), and quotients A/R, B/R and (A B)/R (right). Given a graph G, we define a quotient graph G/R (in Γ 0 ) whose vertex set is the set of R-equivalence classes of G, and for which two classes are adjacent if they are joined by an edge of G. (And a single class carries a loop provided that an edge of G has both endpoints in that class.) Figure 4 shows quotients A/R, B/R and (A B)/R. If G is R-thin, then G/R G. It is easily verified that G/R is R-thin for any G Γ 0. Given x V (G) let [x] = {x V (G) N G (x ) = N G (x)} denote the R-equivalence class containing x. As the relation R is defined entirely in terms of adjacencies, it is clear that given an isomorphism ϕ G H we have xry in G if and only if ϕ(x)r ϕ(y) in H. Thus ϕ maps R-equivalence classes of G to R-equivalence classes of H, and in particular ϕ([x]) = [ϕ(x)]. Thus any isomorphism ϕ G H induces an isomorphism ϕ G/R H/R defined as ϕ([x]) = [ϕ(x)]. But an isomorphism ϕ G/R H/R does not necessarily imply that there is an isomorphism ϕ G H. (Consider G = P 3 and H = K 2.) However, we do have the following result in this direction. The straightforward proof can be found in Section 8.2 of [3]. Proposition 1. If there is an isomorphism ϕ G/R H/R with the property that X = ϕ(x) for each X V (G/R), then there is also an isomorphism ϕ G H. (Any such ϕ can be obtained from ϕ by declaring that ϕ restricts to a bijection X ϕ(x) for each X.) 1 Figure 4 (right) suggests an isomorphism η (A B)/R A/R B/R given by [(v, w)] ([v], [w]). Indeed, this is a general principle, as is proved in Section 8.2 of [3]. Moreover, the figure suggests [(v, w)] = [v] [w] (as sets). This too is true in general. It follows that [(v, w)] = [v] [w], and that A B is R-thin if and only if both A and B are. 6

7 Remark 1. Suppose there is an isomorphism ϕ A B A B, given as ϕ(a, b) = (ϕ A (a, b), ϕ B (a, b)). The above discussion implies the following commutative diagram of induced isomorphisms. (A B)/R ϕ (A B )/R [(a, b)] [(ϕ A (a, b), ϕ B (a, b))] η A/R B/R ϕ η A /R B /R ([a], [b]) ([ϕ A (a, b)], [ϕ B (a, b)]) Moreover, [(a, b)] = [((ϕ A (a, b), ϕ B (a, b))] = [a] [b] = [ϕ A (a, b)] [ϕ B (a, b)]. The proof of our main theorem will use these facts often. We have noted that a factoring G A B induces a corresponding factoring G/R A/R B/R. In general the converse is false: For an arbitrary G, a factoring G/R A B can not necessarily be pulled up to a factoring G = A B for which A /R A and B /R B. But there is a partial converse: Proposition 2 (Proposition 8.6 of [3]). Suppose G Γ 0 has no isolated vertices, and G/R A B. Thus there is a bijective correspondences between R-equivalence classes of G and vertices (a, b) V (A B). Let (a, b) stand for the cardinality of the R-equivalence class corresponding to (a, b). If there are maps α V (A) N and β V (B) N for which (a, b) = α(a) β(b), then there is a factoring G A B, where A /R A and B /R B. In Proposition 2, A is formed by blowing up each a V (A) to an R- class with α(a) vertices; there is an edge between between two vertices of A precisely when A has an edge joining the corresponding vertices. The graph B is formed similarly. Let K n be the complete graph on n vertices, with a loop at each vertex. Corollary 2 (Corollary 8.7 of [3]). A graph B Γ 0 factors as B = K n H if and only if n divides the cardinality of each R-equivalence class of B. In particular, the sizes of the R-equivalence classes of a prime graph are relatively prime. Our main proof uses the ideas of this section freely, sometimes without comment. We assume our reader is well versed with this material. We refer the reader who requires proofs and explanations to Section 8.2 of [3]. 7

8 4. The Cartesian Skeleton We now recall the definition of the Cartesian skeleton S(G) of an arbitrary graph G in Γ 0. The Cartesian skeleton S(G) is a graph on the vertex set of G that has the property S(A B) = S(A) S(B) in the class of R-thin graphs, thereby linking the direct and Cartesian products. We construct S(G) as a certain subgraph of the Boolean square of G. The Boolean square of a graph G is the graph G s with V (G s ) = V (G) and E(G s ) = {xy N G (x) N G (y) }. Thus, xy is an edge of G s whenever G has an x, y-walk of length two. For instance, if p 3, then Kp s = Kp (i.e., K p with a loop added to each vertex). Also, K2 s = K 1 + K 1 and Ks 1 = K 1. The left side of Figure 5 shows graphs A, B and A B (bold) together with their Boolean squares A s, B s and (A B) s (dotted). If G has an x, y-walk W of even length, then G s has an x, y-walk of length W /2 on alternate vertices of W. Thus G s is connected if G is connected and has an odd cycle. (An odd cycle guarantees an even walk between any two vertices of G.) On the other hand, if G is connected and bipartite, then G s has exactly two components, and their respective vertex sets are the two partite sets of G. B B z z y y x x A A Figure 5: Left: Graphs A, B, A B and their Boolean squares A s, B s and (A B) s (dotted). Right: Graphs A, B, A B and their Cartesian skeletons S(A), S(B) and S(A B) (dotted). We now show how to form S(G) as a certain spanning subgraph of G s. Consider an arbitrary factorization G A B, by which we identify each vertex of G with an ordered pair (a, b). We say that an edge (a, b)(a, b ) of 8

9 G s is Cartesian relative to the factorization A B if either a = a and b b, or a a and b = b. For example, in Figure 5 edges xz and zy of G s are Cartesian (relative to the factorization A B), but edges xy and yy of G s are not Cartesian. We will make S(G) from G s by removing the edges of G s that are not Cartesian, but we do this in a way that does not reference the factoring A B of G. We identify two intrinsic criteria for a non-loop edge of G s that tell us if it may fail to be Cartesian relative to some factoring of G. (Note that the symbol means proper inclusion.) (i) In Figure 5 edge xy of G s is not Cartesian, and there is a z V (G) with N G (x) N G (y) N G (x) N G (z) and N G (x) N G (y) N G (y) N G (z). (ii) In Figure 5 the edge x y of G s is not Cartesian, and there is a z V (G) with N G (x ) N G (z ) N G (y ). Our aim is to remove from G s all edges that meet one of these criteria. We package the above criteria into the following definition. Definition 1. An edge xy of G s is dispensable if x = y or there exists z V (G) for which both of the following statements hold. (1) N G (x) N G (y) N G (x) N G (z) or N G (x) N G (z) N G (y), (2) N G (y) N G (x) N G (y) N G (z) or N G (y) N G (z) N G (x). Observe that the above statements (1) and (2) are symmetric in x and y. It is easy to confirm that (i) or (ii) holding for a triple x, y, z is equivalent to both of (1) and (2) holding. Now we come to the main definition of this section. Definition 2. The Cartesian skeleton of a graph G is the spanning subgraph S(G) of G s obtained by removing all dispensable edges from G s. The right side of Figure 5 is the same as its left side, except all dispensable edges of A s, B s and (A B) s are deleted. Thus the remaining dotted edges are S(A), S(B) and S(A B). Note that although S(G) was defined without regard to the factorization G = A B, we nonetheless have S(A B) = S(A) S(B). The following proposition from [4] asserts that this always holds for R-thin graphs. 9

10 Proposition 3. If A, B are R-thin graphs without isolated vertices, then S(A B) = S(A) S(B). (This is equality, not mere isomorphism; the graphs S(A B) and S(A) S(B) have identical vertex and edge sets.) As S(G) is defined entirely in terms of the adjacency structure of G, we have the following immediate consequence of Definition 2. Proposition 4. Any isomorphism ϕ G H, as a map V (G) V (H), is also an isomorphism ϕ S(G) S(H). We will also need a result concerning connectivity of Cartesian skeletons. The following result (which does not require R-thinness) is from [4]. (For another proof, see Chapter 8 of [3].) Proposition 5. Suppose G is connected. (i) If G has an odd cycle, then S(G) is connected. (ii) If G is nontrivial bipartite, then S(G) has two connected components. Their respective vertex sets are the two partite sets of G. 5. Main Result Having discussed the necessary supporting material, we can now prove our main result. Theorem 2. Suppose a connected bipartite graph G factors as G A B and G A B, where B and B are prime and bipartite. Then B B. Proof. Let G be as stated. Take an isomorphism ϕ A B A B. Then ϕ(a, b) = (ϕ A (a, b), ϕ B (a, b)) for certain component functions ϕ A and ϕ B. We now reduce to R-thin graphs in order to apply the properties of the Cartesian skeleton operator. By Remark 1 in Section 3 there is an induced isomorphism A/R B/R ϕ A /R B /R ([a], [b]) ([ϕ A (a, b)], [ϕ B (a, b)]). (2) For future reference, we note that Remark 1 of Section 3 also states [a] [b] = [ϕ A (a, b)] [ϕ B (a, b)]. (3) 10

11 From here the proof proceeds in four parts. Part 1 applies the Cartesian skeleton operator to the above map ϕ and employs unique factorization over the Cartesian product to uncover a simple expression for ϕ. Part 2 uses primeness of B and B to further simplify this expression. Part 3 shows B/R B /R, and Part 4 lifts this to an isomorphism B B. PART 1. Applying the Cartesian skeleton operator and Proposition 4 to the above map (2), we see that ϕ is also an isomorphism ϕ S(A/R B/R) S(A /R B /R). This yields the upper-most square in the following Diagram (4). In this top square, both occurrences of ϕ are isomorphisms. Obviously, the vertical dotted arrows do not indicate isomorphisms, but they are the identity on vertex sets, and in this sense the top square of the diagram is commutative. The remainder of the diagram proceeds as follows. Proposition 3 applied to the second line yields the third line. (The vertical double lines indicate equality and the horizontal arrows are isomorphisms.) A/R B/R ϕ A /R B /R S S(A/R B/R) ϕ S S(A /R B /R) S(A/R) S(B/R) ϕ S(A /R) S(B /R) (4) S(A/R) (B 0 + B 1 ) ϕ S(A /R) (B 0 + B 1 ) S(A/R) B 0 + S(A/R) B 1 ϕ S(A /R) B 0 + S(A /R) B 1 Concerning the fourth line, note that every R-equivalence class of B (i.e., every vertex of B/R) lies entirely in a partite set of B. Any edge of B/R runs from an R-equivalence class in one partite set of B to an R-equivalence class in the other partite set of B. It follows that B/R is bipartite, and it is connected because B is. Thus the boolean square of B/R consists of two connected components whose respective vertex sets are the partite sets of 11

12 B/R. In turn, the skeleton S(B/R) consists of two connected components B 0 and B 1 whose respective vertex sets are the partite sets of B/R. (See Proposition 5.) Thus S(B/R) = B 0 + B 1. Similarly S(B /R) = B 0 + B 1 for two other graphs B 0 and B 1. This gives the fourth line of Diagram (4). The bottom line follows from the distributive property of the Cartesian product. Consider the bottom line of Diagram (4). The isomorphism ϕ restricts to isomorphisms ϕ 0 and ϕ 1 on the two components of S(A/R) B 0 +S(A/R) B 1. Arrange the indexing of their codomains as in Diagram (5), below. S(A/R) B 0 S(A/R) B 1 ϕ 0 ϕ 1 (5) S(A /R) B 0 S(A /R) B 1 Let us reflect on Diagram (5). By Diagram (4), the graphs A/R B/R and S(A/R) B 0 +S(A/R) B 1 have identical vertex sets. Moreover, the two partite sets of A/R B/R are the vertices of S(A/R) B 0 and S(A/R) B 1. Similarly, A /R B /R and S(A /R) B 0 +S(A /R) B 1 have identical vertex sets, and the two partite sets of A /R B /R are the vertices of S(A /R) B 0 and S(A /R) B 1. The map ϕ is actually two isomorphisms ϕ A/R B/R A /R B /R, ϕ S(A/R) B 0 + S(A/R) B 1 S(A /R) B 0 + S(A /R) B 1, where ϕ(x, y) is ϕ 0 (x, y) or ϕ 1 (x, y) depending on the partite set (respectively component) that (x, y) belongs to. We summarize this in the informal Diagram (6), a slight embellishment of Diagram (5). E(A/R B/R) S(A/R) B 0 S(A/R) B 1 } A/R B/R ϕ 0 ϕ 1 (6) S(A /R) B 0 S(A /R) B 1 E(A /R B /R) } A /R B /R 12

13 With this in mind, we now refine Diagram (5). Identify S(A/R) with its prime factorization S(A/R) = A 1 A 2 A i over the Cartesian product. That is, we now label (or coordinatize) each vertex of S(A/R) with the corresponding i-tuple (a 1, a 2,..., a i ) from its prime factorization. Similarly, form the prime factorizations S(A /R) = A 1 A 2 A l, B 0 = B 01 B 02 B 0j, B 1 = B 11 B 12 B 1k, B 0 = B 01 B 02 B 0m, B 1 = B 11 B 12 B 1n. With these factorizations, Diagram (5) is updated to Diagram (7), below. S(A/R) B 0 (A 1 A i ) (B 01 B 0j ) S(A/R) B 1 (A 1 A i ) (B 11 B 1k ) ϕ 0 ϕ 1 (7) (A 1 A l ) (B 01 B 0m ) S(A /R) B 0 (A 1 A l ) (B 11 B 1n ) S(A /R) B 1 We are now going to further refine Diagram (7) to a new Diagram (8). Theorem 1 says each component function of ϕ 0 is an isomorphism from a prime factor of S(A/R) B 0 to a prime factor of S(A /R) B 0. Likewise, each component function of ϕ 1 is an isomorphism from a prime factor of S(A/R) B 1 to a prime factor of S(A /R) B 1. Via commutativity of the Cartesian product, we group together all factors of S(A/R) that both ϕ 0 and ϕ 1 send to factors of S(A /R), and we denote the product of all such factors of S(A/R) as K. (If no such prime factors of S(A/R) exist, we agree that K = K 1.) Group together those factors of S(A/R) that ϕ 0 sends to factors of S(A /R) and ϕ 1 sends to B 1, and let L be the product of those factors of S(A/R). (Possibly L = K 1.) Also group together all factors of S(A/R) that ϕ 0 sends to factors of B 0 and ϕ 1 sends to factors of S(A /R), and let M be the product of all such factors. (Possibly M = K 1.) Any remaining factors of S(A/R) are those that both ϕ 0 and ϕ 1 send to factors 13

14 of B 0 and B 1, respectively. Let X be the product of these factors. Thus S(A/R) = K L M X, as indicated in Diagram (8). Next, group those prime factors of B 0 that ϕ 0 sends to factors of B 0, and call their product P. Similarly, group those prime factors of B 1 that ϕ 1 sends to factors of B 1, and call their product Q. (Again, see Diagram (8).) In the spirit of Corollary 1, the images of the component functions of ϕ 0 and ϕ 1 can be identified with their domains so that each solid arrow in Diagram (8) is the identity. The dashed arrow is an exception; once we have identified the image of the first component function of ϕ 0 with its domain K, we are not at liberty to say that the first component function of ϕ 1 is also the identity, as there is no reason to expect the first component functions of ϕ 0 and ϕ 1 to be the same. However, the first component function of ϕ 1 is certainly an isomorphism κ K K, as indicated in Diagram (8). S(A/R) B 0 (K L M X) ( P ) S(A/R) B 1 (K L M X) ( Q ) ϕ 0 κ ϕ 1 (8) (K L......) (M P X ) S(A /R) B 0 (K... M...) (L Q X ) S(A /R) B 1 The two sides of Diagram (8) show that S(A /R) has distinct Cartesian factors K, L and M, so S(A /R) = K L M Y for some graph Y (possibly trivial). Accordingly, the bottom line of Diagram (8) is updated as in Diagram (9). S(A/R) B 0 (K L M X) ( P ) S(A/R) B 1 (K L M X) ( Q ) ϕ 0 κ ϕ 1 (9) (K L M Y ) (M P X ) S(A /R) B 0 (K L M Y ) (L Q X ) S(A /R) B 1 14

15 Finally, Theorem 1 implies that ϕ is as indicated in Diagram (10), where we have identified the domains of the four new component functions with their images. All solid arrows are identity functions and κ is an isomorphism. S(A/R) B 0 (K L M X) (M P Y ) S(A/R) B 1 (K L M X) (L Q Y ) ϕ 0 (K L M Y ) (M P X ) S(A /R) B 0 κ ϕ 1 (K L M Y ) (L Q X ) S(A /R) B 1 (10) From Diagram (10) we see that the isomorphism ϕ has the following effect on the respective partite sets of A/R B/R. ϕ 0 ((k, l, m 1, x), (m 2, p, y)) ((k, l, m 2, y), (m 1, p, x)) ϕ 1 ((k, l 1, m, x), (l 2, q, y)) ((κ(k), l 2, m, y), (l 1, q, x)) Also from Diagram (10) we see that any edge of A/R B/R has form ((k 1, l 1, m 1, x 1 ), (m 2, p, y 1 )) ((k 2, l 2, m 3, x 2 ), (l 3, q, y 2 )) where the left endpoint is a vertex of S(A/R) B 0 and the right endpoint belongs to S(A/R) B 1. Also, as ϕ A/R B/R A /R B /R is an isomorphism, it carries this edge of A/R B/R to the edge ((k 1, l 1, m 2, y 1 ), (m 1, p, x 1 )) ((κ(k 2 ), l 3, m 3, y 2 ), (l 2, q, x 2 )), which joins S(A /R) B 0 to S(A /R) B 1. To standardize the discussion, we agree that in writing an edge of A/R B/R, the left endpoint is always assumed to be a vertex of S(A/R) B 0 and the right endpoint is a vertex of S(A/R) B 1. We adopt a similar convention for edges of A /R B /R. PART 2. Diagram (10) can be simplified considerably in the sense that primeness of B and B forces both X and Y to be trivial. We now show this. 15

16 To motivate our process, note that S(B/R) has a Cartesian factor of Y, and S(B /R) has a Cartesian factor of X, as follows: S(B/R) = B 0 + B 1 = M P Y + L Q Y = (M P + L Q) Y, S(B /R) = B 0 + B 1 = M P X + L Q X = (M P + L Q) X. We will show that these factorings of S(B/R) and S(B /R) induce corresponding factorings (over ) of the prime graphs B and B. This will imply X = K 1 = Y. We will show only that Y = K 1, for the fact that X = K 1 follows from exactly the same argument (using ϕ 1 in the place of ϕ). We carry out our plan in two steps. First we construct graphs S and T, with V (T ) = V (Y ), for which B/R S T. Then we lift to a factoring B = S T with S bipartite (nontrivial), and S /R = S and T /R = T. Primeness of B then forces T to be trivial, hence also T and consequently Y. Step 1. We now construct graphs S and T for which B/R = S T, with S nontrivial bipartite and V (T ) = V (Y ). Define T as follows: V (T ) = V (Y ) E(T ) = { y 1 y 2 ((,,, ), (,, y 1 )) ((,,, ), (,, y 2 )) E(A/R B/R) }. The asterisks indicate definite (but unspecified) vertices of factors in the top row of Diagram (10). Their exact values play no role in the definition of T. Define S to be a bipartite graph whose partite sets are the vertices of M P and L Q, respectively. Specifically, V (S) = V (M P ) V (L Q) E(S) = {(m, p)(l, q) ((,,, ), (m, p, )) ((,,, ), (l, q, )) E(A/R B/R)}. We now describe an isomorphism µ B/R S T. Recall that V (B/R) = V (B 0 ) V (B 1 ) = V (M P Y ) V (L Q Y ). Simply say µ(u, v, y) = ((u, v), y), which is clearly a bijection from V (B/R) to V (S T ). And it is a homomorphism, as follows. Take any edge (m, p, y 1 )(l, q, y 2 ) of B/R. Then A/R B/R has an edge ((,, ), (m, p, y 1 ))((,, ), (l, q, y 2 )). By definition of S and T, it follows that (m, p)(l, q) E(S) and y 1 y 2 E(T ). Then µ(m, p, y 1 )µ(l, q, y 2 ) = ((m, p), y 1 )((l, q), y 2 ) E(S T ). 16

17 To show that µ is actually an isomorphism, we must show that any edge ((m, p), y 1 )((l, q), y 2 ) E(S T ) is the image under µ of a corresponding edge (m, p, y 1 )(l, q, y 2 ) E(B). Thus suppose ((m, p), y 1 )((l, q), y 2 ) E(S T ). Then (m, p)(l, q) E(S) and y 1 y 2 E(T ). By definition of S and T, there are edges ((,,, ), (m, p, )) ((,,, ), (l, q, )) E(A/R B/R), ((,,, ), (,, y 1 )) ((,,, ), (,, y 2 )) E(A/R B/R). Applying ϕ to these edges (see Diagram (10)) we get ((,, m, ), (, p, )) ((, l,, ), (, q, )) E(A /R B /R), ((,,, y 1 ), (,, )) ((,,, y 2 ), (,, )) E(A /R B /R). From the definition of the direct product this yields ((, l,, ), (, p, )) ((,, m, ), (, q, )) E(A /R B /R), ((,,, y 2 ), (, p, )) ((,,, y 1 ), (, q, )) E(A /R B /R). Applying ϕ 1 to this pair of edges gives ((, l,, ), (, p, )) ((,, m, ), (, q, )) E(A/R B/R), ((,,, ), (, p, y 2 )) ((,,, ), (, q, y 1 )) E(A/R B/R). From this we see that ((, l,, ), (, p, y 2 )) ((,, m, ), (, q, y 1 )) E(A/R B/R), and, by applying ϕ, hence ((, l,, y 2 ), (, p, )) ((,, m, y 1 ), (, q, )) E(A /R B /R), ((,, m, y 1 ), (, p, )) ((, l,, y 2 ), (, q, )) E(A /R B /R). An application of ϕ 1 produces ((,,, ), (m, p, y 1 )) ((,,, ), (l, q, y 2 )) E(A/R B/R). 17

18 Therefore (m, p, y 1 )(l, q, y 2 ) E(B/R), and µ maps this edge to the edge ((m, p), y 1 )((l, q), y 2 ) of S T. This establishes that µ B/R S T is indeed an isomorphism. Step 2. Now we show that the factoring B/R = S T lifts to a factoring B = S T with S /R = S and T /R = T. Keep in mind that each vertex ((u, v), y) of S T B/R corresponds to a vertex (u, v, y) of B/R, which is an R-equivalence class of B. Thus it is meaningful to speak of the cardinality ((u, v), y) = (u, v, y). By Proposition 2 of Section 3, to show that B = S T we just need to produce functions α V (S) N and β V (T ) N for which (u, v, y) = ((u, v), y) = α(u, v) β(y). We now construct such functions. In what follows, a lower-case letter stands for a vertex of the graph denoted by its upper-case. Thus we will know that (k, l, m, x) denotes a vertex of K L M X = S(A/R), and (m, p, y) V (M P Y ) = V (B 0 ), etc. The reader may reference Diagram (10) for bookkeeping. Applying Equation (3) to the left-hand side of Diagram (10), we get (k, l, m 1, x) (m 2, p, y) = (k, l, m 2, y) (m 1, p, x). (11) Similarly, the right-hand side of the diagram reveals that (k, l 1, m, x) (l 2, q, y) = (κ(k), l 2, m, y) (l 1, q, x). (12) Replace both k s in Equation (11) with κ(k), and put m 1 = m 2 = m. In (12), put l 1 = l 2 = l. Dividing (11) by (12) yields so (κ(k), l, m, x) (m, p, y) (k, l, m, x) (l, q, y) = (κ(k), l, m, y) (m, p, x) (κ(k), l, m, y) (l, q, x), (m, p, y) (k, l, m, x) (m, p, x) = (l, q, y) (κ(k), l, m, x) (l, q, x). Notice that the expression on the left depends only on m, p, l and q, for the variable y does not appear on the right. Thus we have a function f for which (m, p, y) (l, q, y) = f(m, p, l, q). Fixing four vertices m 0, p 0, l 0, q 0, we use the above to get (m, p, y) = f(m, p, l 0, q 0 ) (l 0, q 0, y) (13) (l, q, y) = 1 f(m 0, p 0, l, q) (m 0, p 0, y) = f(m 0, p 0, l 0, q 0 ) f(m 0, p 0, l, q) (l 0, q 0, y) (14) 18

19 Equation (13) implies that for any m and p the denominator of the (fully reduced) rational number f(m, p, l 0, q 0 ) divides (l 0, q 0, y), for every y. And Equation (14) implies that for any l and q the denominator of the (fully reduced) rational number f(m 0,p 0,l 0,q 0 ) f(m 0,p 0,l,q) divides (l 0, q 0, y), for every y. Let d be the least common multiple of all such denominators. Then d divides (l 0, q 0, y) for any y and Equations (13) and (14) yield (m, p, y) = d f(m, p, l 0, q 0 ) (l, q, y) = d f(m 0, p 0, l 0, q 0 ) f(m 0, p 0, l, q) (l 0, q 0, y), d (l 0, q 0, y). d Thus any (u, v, y) V (B/R) has cardinality (u, v, y) = α(u, v) β(y) for the integer-valued functions α(u, v) = d f(u, v, l 0, q 0 ) if (u, v) V (B 0 ) d f(m 0, p 0, l 0, q 0 ) f(m 0, p 0, u, v) β(y) = (l 0, q 0, y). d if (u, v) V (B 1 ), So indeed Proposition 2 yields B = S T with S /R = S and T /R = T. As B is prime and S is nontrivial (it has two partite sets), it follows that T is a single vertex with a loop. Thus so is T = T /R. Because V (T ) = V (Y ), it follows that Y = K 1. A like argument shows X = K 1. Therefore we can eliminate X and Y from Diagram (10) to obtain Diagram (15), below, which codifies the structure of ϕ. S(A/R) B 0 (K L M) (M P ) S(A/R) B 1 (K L M) ( L Q ) ϕ 0 κ ϕ 1 (15) (K L M) (M P ) S(A /R) B 0 (K L M) ( L Q ) S(A /R) B 1 19

20 PART 3. Now we describe an isomorphism θ B/R B /R. To better describe it we first forge some notation. In Diagram (15) both S(A/R) and S(A /R) are now identified with the product K L M, so a tuple (k, l, m) could denote a vertex of either one of these graphs. To avoid ambiguity, we distinguish the tuple with a prime if it denotes a vertex in S(A /R). Thus we have (k, l, m) V (S(A/R)), but (k, l, m) V (S(A /R)). Similarly, both B 0 and B 0 are now identified with the product M P, so we say (m, p) V (B 0) but (m, p) V (B 0 ). And as both B 1 and B 1 are identified with L Q we write (l, q) V (B 1 ), but (l, q) V (B 1 ). We can define a map θ V (B/R) V (B /R) simply as θ(u, v) = (u, v), that is, θ sends any vertex (m, p) in the partite set V (B 0 ) of B/R to the vertex (m, p) in the partite set V (B 0 ) of B /R; and it sends any (l, q) in the partite set V (B 1 ) of B/R to (l, q) in the partite set V (B 1 ) of B /R. Clearly θ is a bijection. We now show it is a homomorphism. Let (m, p)(l, q) E(B/R). In what follows we show θ(m, p) θ(l, q) E(B /R). Begin with an edge Apply ϕ to this to get ((,, ), (m, p)) ((,, ), (l, q)) E(A/R B/R). ((,, m), (, p) ) ((, l, ), (, q) ) E(A /R B /R). Interchanging the endpoints in the first factor gives ((, l, ), (, p) ) ((,, m), (, q) ) E(A /R B /R). Applying ϕ 1 to this edge produces ((, l, ), (, p)) ((,, m), (, q)) E(A/R B/R). Again, interchange the endpoints in the first factor: ((,, m), (, p)) ((, l, ), (, q)) E(A/R B/R). Applying ϕ to this edge yields ((,, ), (m, p) ) ((,, ), (l, q) ) E(A /R B /R). 20

21 From this, (m, p) (l, q) E(B /R), which is to say θ(m, p) θ(l, q) E(B /R). Thus θ is a homomorphism. Given (m, p) (l, q) E(B /R), we can do the same steps as above, but in reverse order, to get θ 1 (m, p) θ 1 (l, q) E(B/R). Therefore θ B/R B /R is an isomorphism. In summary, the map θ B/R B /R defined as θ(u, v) = (u, v) is an isomorphism. PART 4. Now we finish the proof by lifting the isomorphism θ B/R B /R just obtained to an isomorphism θ B B. By Proposition 1 (in Section 3), it suffices to show (u, v) = θ(u, v), that is, (u, v) = (u, v) for each (u, v) V (B/R). In other words, we need to show that both (m, p) = (m, p) and (l, q) = (l, q) always hold. As soon as this is done we will have an isomorphism θ B B and the proof will be finished. Let s adapt Equations (11) and (12) to the new Diagram 15. With X and Y gone, and with our prime notation for distinguishing vertices of A/R B/R from those of A /R B /R, these equations are (k, l, m 1 ) (m 2, p) = (k, l, m 2 ) (m 1, p), (16) (k, l 1, m) (l 2, q) = (κ(k), l 2, m) (l 1, q). (17) Using Equation 16, we see that (k, l, m 1 ) (k, l, m 2 ) = (k, l, m 1) (m 1, p) (k, l, m 2 ) (m 1, p) = (k, l, m 1) (m 1, p) (k, l, m 1 ) (m 2, p) = (m 1, p) (m 2, p). Comparing the left end to the right end, we see that the left-most expression does not depend on k. Thus we will be justified in the third line of the next computation, where we switch one k for another in an occurrence of this left most expression. Fix vertices k 0, l 0, m 0 and q 0 of K, L, M and Q, respectively. Now take an arbitrary (m, p) V (B 0 ) V (B/R) and corresponding (m, p) V (B 0 ) V (B /R) and use Equations (16) and (17) as follows. (m, p) (m, p) = (k 0, l 0, m 0 ) (l 0, q 0 ) (k 0, l 0, m) (l 0, q 0 ) = (κ(k 0), l 0, m 0 ) (l 0, q 0 ) (κ(k 0 ), l 0, m) (l 0, q 0 ) = (k 0, l 0, m 0 ) (l 0, q 0 ) (k 0, l 0, m) (l 0, q 0 ) = (k 0, l 0, m 0 ) (k 0, l 0, m 0 ). 21 (k 0, l 0, m) (m, p) (k 0, l 0, m 0 ) (m, p) (k 0, l 0, m) (m, p) (k 0, l 0, m 0 ) (m, p) (k 0, l 0, m) (m, p) (k 0, l 0, m 0 ) (m, p)

22 Establish a constant C = (k 0, l 0, m 0 ) / (k 0, l 0, m 0 ). The above computation shows (m, p) = C (m, p). Next we follow a parallel computation to get (l, q) = D (l, q) for some constant D. First note that (k, l 1, m) (k, l 2, m) = (k, l 1, m) (l 1, q) (k, l 2, m) (l 1, q) = (κ 1 (k), l 1, m) (l 1, q) (κ 1 (k), l 1, m) (l 2, q) = (l 1, q) (l 2, q), so the expression on the left does not depend on k. Now we have (l, q) (l, q) = (k 0, l 0, m 0 ) (m 0, p 0 ) (k 0, l, m 0 ) (m 0, p 0 ) = (k 0, l 0, m 0 ) (m 0, p 0 ) (k 0, l, m 0 ) (m 0, p 0 ) = (κ(k 0), l 0, m 0 ) (m 0, p 0 ) (κ(k 0 ), l, m 0 ) (m 0, p 0 ) = (κ(k 0), l 0, m 0 ) = D. (k 0, l 0, m 0 ) (k 0, l, m 0 ) (l, q) (k 0, l 0, m 0 ) (l, q) (κ(k 0), l, m 0 ) (l, q) (k 0, l 0, m 0 ) (l, q) (κ(k 0), l, m 0 ) (l, q) (k 0, l 0, m 0 ) (l, q) Thus (m, p) = C (m, p) and (l, q) = D (l, q) for all m, p and l, q. Denote the partite sets of B as X 0 and X 1, and those of B as X 0 and Y 0. Arrange the indexing so that (m, p) X 0, (l, q) X 1, (m, p) X 0 and (l, q) X 1. Then X 0 = C X 0 and X 1 = D X 1. Because ϕ A B A B carries the partite set V (A) X 0 bijectively to the partite set V (A ) X 0, and because it carries the partite set V (A) X 1 bijectively to the partite set V (A ) X 1, we have V (A) X 0 = V (A ) X 0 = V (A ) C X 0, V (A) X 1 = V (A ) X 1 = V (A ) D X 1. From this, C = D. Thus we have (m, p) = C (m, p) and (l, q) = C (l, q). We must show C = 1. Corollary 2 says the greatest common divisor of the multiset of numbers (m, p), (l, q) is 1. This implies C is an integer. But also the gcd of the multiset of numbers (m, p), (l, q) is 1, so C = 1. Thus (m, p) = (m, p) and (l, q) = (l, q). Proposition 1 now implies that the isomorphism θ B/R B /R where (u, v) (u, v) lifts to an isomorphism θ B B. The theorem is proved. 22

23 It is easy to quickly get a slightly stronger (if somewhat more technical) result that could be potentially useful. The proof uses the notation set up in the proof of our main theorem, just completed, and can be viewed as a continuation of that theorem and proof. Theorem 3. Suppose a connected bipartite graph G factors as G = A B and G = A B, where B and B are prime and bipartite. Suppose B has partite sets X 0 and X 1. Then there is an isomorphism µ A B A B with the following form: µ(a, b) = { (g 0(a), θ(b)) if b X 0 (g 1 (a), θ(b)) if b X 1, where g i V (A) V (A ) are bijections and θ B B is an isomorphism. Proof. Note that the isomorphism ϕ A B A B that induced ϕ in Diagram (15) has the form of µ only if M = K 1 = L. But let us define a map µ V (A/R B/R) V (A /R B /R) as in Diagram 18 below, where the second component of µ is the isomorphism θ B/R B /R from Part 3 of the proof of our main theorem. (Compare with Diagram (15).) Note that µ has the salient structure of µ in the statement of the present theorem. S(A/R) B 0 (K L M) (M P ) S(A/R) B 1 (K L M) ( L Q ) µ 0 κ µ 1 (18) (K L M) (M P ) S(A /R) B 0 (K L M) ( L Q ) S(A /R) B 1 We first claim that µ A/R B/R A /R B /R is an isomorphism, an easy task because µ is clearly a bijection on vertices and its second component is already a known isomorphism. Note that µ is a homomorphism: Take an edge ((k 1, l 1, m 1 ), (m 2, p)) ((k 2, l 2, m 3 ), (l 3, q)) E(A/R B/R). (19) 23

24 Then (m 2, p)(l 3, q) E(B/R), but as θ is an isomorphism we immediately get (m 2, p) (l 3, q) E(B /R). Applying ϕ to the edge (19), ((k 1, l 1, m 2 ), (m 1, p) ) ((κ(k 2 ), l 3, m 3 ), (l 2, q) ) E(A /R B /R). Then (m 1, p) (l 2, q) E(B /R), but θ 1 is an isomorphism, so (m 1, p)(l 2, q) E(B/R). Combining with the edge (19), we get ((k 1, l 1, m 1 ), (m 1, p)) ((k 2, l 2, m 3 ), (l 2, q)) E(A/R B/R). Applying ϕ to this, ((k 1, l 1, m 1 ), (m 1, p) ) ((κ(k 2 ), l 2, m 3 ), (l 2, q) ) E(A /R B /R). As we earlier deduced (m 2, p) (l 3, q) E(B /R), we now have ((k 1, l 1, m 1 ), (m 2, p) ) ((κ(k 2 ), l 2, m 3 ), (l 3, q) ) E(A /R B /R), which is to say µ((k 1, l 1, m 1 ), (m 2, p)) µ((k 2, l 2, m 3 ), (l 3, q)) E(A /R B /R). Thus µ is a homomorphism. By the symmetry of Diagram 18 the reverse of this process makes µ 1 a homomorphism. Hence µ is an isomorphism. Recall that in Part 4 of the proof of Theorem 2 we showed that the equations (m, p) = (m, p) and (l, q) = (l, q) always hold. In Part 4 we also showed (m, p) (m, p) = (k 0, l 0, m 0 ) (k 0, l 0, m 0 ) and (l, q) (l, q) = (κ(k 0), l 0, m 0 ), (k 0, l 0, m 0 ) so (k 0, l 0, m 0 ) = (k 0, l 0, m 0 ) = (κ(k 0 ), l 0, m 0 ) for any k 0, l 0, m 0. Define a bijection g 0 V (A) V (A ) so that it carries each R-class (k 0, l 0, m 0 ) bijectively to (k 0, l 0, m 0 ). Also say g 1 V (A) V (A ) carries each (k 0, l 0, m 0 ) bijectively to (κ(k 0 ), l 0, m 0 ). As in the proof of Theorem 2, let θ B B be an isomorphism that sends each R-class (u, v) bijectively to (u, v). With these, let µ A B A B be as in the statement of the present theorem. By the comments in Section 3, the map µ has been defined so that it respects R-classes and induces an isomorphism µ A/R B/R A /R B /R. Therefore µ is an isomorphism. The proof is done. 24

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