Nontrivial independent sets of bipartite graphs and crossintersecting families
Abstract
Let be a connected, noncomplete bipartite graph with . An independent set of is said to be trivial if or . Otherwise, is nontrivial. By we denote the size of maximalsized nontrivial independent sets of . We prove that if the automorphism group of is transitive on and , then , where is the common degree of vertices in . We also give the structures of maximalsized nontrivial independent sets of . As applications of this result, we give the upper bound of sizes of two crossintersecting families of finite sets, finite vector spaces and permutations.
keywords:
intersecting family, crossintersecting family, symmetric system, ErdősKoRado theoremMSC: 05D05, 06A07
and
1 Introduction
Let be a finite set and, for , let denote the family of all subsets of , and let and denote the symmetric group and alternative group on , respectively. In particular, for positive integer , let denote the set , for , for , and abbreviate the symmetric group and alternative group on as and , respectively.
A family of sets is said to be intersecting if holds for all . Usually, is called intersecting if . The celebrated Erdős–Ko–Rado theorem [11], says that if is a intersecting family in , then
for . The smallest was determined by Frankl [12] for and subsequently determined by Wilson [29] for all .
The ErdősKoRado theorem has many generalizations, analogs and variations. First, the notion of intersection is generalized to tintersection, and finite sets are analogous to finite vector spaces, permutations and other mathematical objects. Second, intersecting families are generalized to crossintersecting families: are said to be crossintersecting if for all and , . Some typical but far from exhaustive results are listed as follows.
Let be a finite field of order , an dimensional vector space over , and the set of all dimensional subspaces (or subspace, for short) of . Then the cardinality of equals . For brevity, we write rather than . A subset of is said to be a intersecting family if for any . The ErdősKoRado theorem for finite vector spaces says that if is a intersecting family in, then
for . This theorem was first established by Hsieh [19] for and , then by Greene and Kleitman [15] for and , and finally by Frankl and Wilson [14] for the general case.
A subset of is said to be a intersecting family if any two permutations in agree in at least points, i.e. for any , . Deza and Frankl [10] showed that an intersecting family in has size at most and conjectured that for fixed, and sufficiently large depending on , a intersecting family in has size at most . Cameron and Ku [8] proved an intersecting family of size is a coset of the stabilizer of a point. A few alternative proofs of Cameron and Ku’s result are given in [23], [16] and [27]. Ku and Leader [22] also generalized this result to partial permutations (see also [24]). Ellis, Friedgut and Pilpel [2] proved Deza and Frankl’s conjecture on intersecting family in .
Hilton [17] investigated the crossintersecting families in : Let be crossintersecting families in with . If , then
(1) 
He also determined the structures of ’s when equality holds. Borg [5] gives a simple proof of this theorem, and generalizes it to labeled sets [4], signed sets [7] and permutations [6]. We generalized this theorem to general symmetric systems [26], which contain finite sets, finite vector spaces and permutations, etc.
Hilton and Milner [18] and Frankl and Tokushige [13] also investigated the sizes of two crossintersecting families: If and are crossintersecting families with , , then .
This theorem actually gives a upper bound of the sizes of nontrivial independent sets in a bipartite graph.
Let be a simple graph with vertex set and edge set . For , define and for . If there is no possibility of confusion, we abbreviate as . A subset of is an independent set of if . A graph is bipartite if can be partitioned into two subsets and so that every edge has one end in and one end in . In this case, we denote the bipartite graph by . An independent set of is said to be trivial if or . In any other case, is nontrivial. If every vertex in is adjacent to every vertex in , then is called a complete bipartite graph. Clearly, a complete bipartite graph has only trivial independent sets. A bipartite graph is said to be parttransitive if there is a group transitively acting on and , respectively, and preserving the adjacency relation of the graph. Clearly, if is parttransitive, then every vertex of () has the same degree, written as (). By and we denote the size and the set of maximalsized nontrivial independent sets of , respectively.
This paper contributes to and for parttransitive bipartite graphs . To do this we make a simple observation as follows.
Let be a noncomplete bipartite graph and let be a nontrivial independent set of , where and . Then and , which implies that
From this one sees that
(2) 
where
and
A subset of is called a fragment in if and . By we denote the set of all fragments contained in . is defined in a similar way and write . An element is also called a fragment if . As we shall see (Lemma 2.1) that . Therefore, in order to address our problems it suffices to determine or .
Let be a finite set, and a group transitively acting on . We say the action of on is primitive, or is primitive on , if preserves no nontrivial partition of . In any other case, the action of is imprimitive. It is easy to see that if the action of on is transitive and imprimitive, then there is a subset of such that and or for every . In this case, is called an imprimitive set in . It is well known that the action of is primitive if and only if for each , the stabilizer of , written as defined to be the set , is a maximal subgroup of (cf. [20, Theorem 1.12]). Furthermore, a subset of is said to be semiimprimitive if and or for each . Clearly, every subset of is semiimprimitve.
The following are main results of this paper.
Theorem 1.1
Let be a noncomplete bipartite graph with . If is parttransitive and every fragment in and is primitive under the action of a group . Then . Moreover,

if , then each fragment in has size ;

if , then each fragment in has size or unless there is a semiimprimitive fragment in or .
As consequences of this theorem we give the upper bounds of sizes of two crossintersecting families of finite sets, finite vector spaces and symmetric groups.
Theorem 1.2
Let be positive integers with , , , , and . If and are crosstintersecting, then
(3) 
Moreover,

when , equality holds if and only if and for any ;

when , equality holds if and only if either and for any , or and for any , or and for some , or and for some .
Theorem 1.3
Let be an dimensional vector space over the field of order and let be positive integers with , , , , and . If and are crossintersecting, then
(4) 
Moreover, equality holds if and only if and where , or and where , subject to .
Theorem 1.4
Let and be positive integers with and . If and are crossintersecting families in , then
(5) 
where is the number of derangements in . Moreover, equality holds if and only if where .
2 Proof of Theorem 1.1
Before to start the proof of Theorem 1.1 we present two lemmas.
Lemma 2.1
Let be a noncomplete bipartite graph. Then, , and

if and only if , and ;

and are both in if , and .
Proof. Suppose and put . Clearly, . If , writing , then and . So , yielding a contradiction. Hence , and . Symmetrically, for , putting , we have and . We then obtain that and (i) holds.
Now, suppose that , and . Then and . Note that and . We have
which implies that and , hence (ii) holds. ∎
From the first statement of this lemma it follows that there is a one to one correspondence , where
Moreover, is an involution, i.e., , and . A fragment is called balanced if . Clearly, all balanced fragments have identical size .
Lemma 2.2
Let be a noncomplete and parttransitive bipartite graph under the action of a group . Suppose that such that for some . If , then and are both in .
Proof. Without loss of generality, suppose and . Since and ,
Then, by Lemma 2.1 (ii), and are both in .∎
From the above lemma we have that if every element of () is primitive and there is an () with , then () contains a singleton. In particular, when there are always two kinds of fragments in : one is for , the other is for . The former is a minimalsized fragment, and the latter is maximalsized one. We call the fragments of this kinds trivial. All others are nontrivial.
Proof of Theorem 1.1. From the above discussion we have that or contains a singleton, that is, . By counting the edges of we have , so because . Then
Equality holds if and only if hence because . This proves that and equality holds if and only if . In any cases, .
We complete the proof by two cases.
Case 1: . In this case we have seen that contains singletons while does not. Now, let be a maximalsized element of and write . Then is a minimalsized element of with and . Suppose . Since and are primitive, there are such that , , and . From this and Lemma 2.2 it follows that if , then , contradicting the maximality of ; if , then , contradicting the minimality of . This proves that for every .
Case 2: . In this case, if there is a nontrivial fragment in or in , let be a minimalsized one. Then . From Lemma 2.2 it follows that for every , is a fragment whenever . Then, the minimality of implies that , 1 or , for every , i.e., is semiimprimitve.
The proof is complete. ∎
For applications of the theorem, we make further discussions on the fragments in the rest of this section.
Note that most bipartite graphs concerning here have only trivial fragments, but there are actually bipartite graphs, which have sufficiently large nontrivial fragments. For example, let and are fixed positive integer with , and . Define to be an edge of if and only if (see Fig. 1 for and ). It is easy to verify that , where and the subscripts are computed modulo , is a fragment in .
However, as we shall see, whether or not a bipartite graph has sufficiently large fragments depends if it has a 2fragment.
Proposition 2.3
Let be a noncomplete bipartite graph with and , and let be a group parttransitively acting on . If there is a 2fragment in , then either

there is an imprimitive set with , or

there is a subset , where or is an imprimitive set under the action of with , such that the quotient group is isomorphic to a subgroup of the dihedral group , where .
Proof. By definition we have that for any , is a 2fragment if and only if . We now define a simple graph , whose vertex set is , and whose edge set consists of all pairs ’s such that is a fragment in . Then, each element of induces an automorphism of . So is vertextransitive. As usual, the valency of is denoted by .
Let be a connected component of and let be the vertex set of . Then . If , then is clearly an imprimitive set in with . Suppose and let be a path in for distinct . Set and , where . Since , for some and . If , then , contradicting that is a fragment. Therefore, . So
From this it follows immediately the following.
Claim: if the induced subgraph is a path, and if is a cycle.
If is a complete graph, then from the above claim it follows that , so . Since , we have , hence is an imprimitive set in with .
If is not complete, then there are more than three elements of , say , such that the induced subgraph is a cycle, written as . By definition we see that for , and equality holds if , that is, is a fragment. Now, if , then, by the above claim, , which yields a contradiction since . Therefore, . Assume that is the least index such that , where . This means that every path of length less than on this cycle is a fragment. In this case, if , then there is an such that is an edge of . Setting , we have that for some . Then if , or if . Both the cases contradict that and are fragments. This proves that is a cycle, and hence (ii) holds. ∎
Proposition 2.4
Let be as in Proposition 2.3. If there are no 2fragments in , then every nontrivial fragment (if it exists) is balanced, and for each , there is a unique nontrivial fragment such that .
Proof. Let be a minimalsized nontrivial fragment in and . Then, is a maximalsized fragment in and . Without loss of generality, suppose that . Then and . We now prove that the equality holds, i.e., is balanced. Suppose, to the contrary, that . Set . As we have mentioned, is semiimprimitive, so or for all distinct . We now define a graph , whose vertex set is , and whose edge set consists of all pairs ’s such that for . Clearly, is vertextransitive. Since is primitive, is not an empty graph. Suppose that for some and . Then, for each , the parttransitivity of implies that there is a with , hence . From this it follows that the valency of , denoted by , is at least . Hence contains a cycle. Let be one of minimum length. Then the induced subgraph is a path from to for . By Lemma 2.1, if , then both and are fragments. Furthermore, if , then the minimality of implies , hence
i.e., is also a fragment. Now, if , then, by Lemma 2.1, is a fragment. However, it is clear that , yielding a contraction. Therefore, there is a unique index with such that , that is, is a maximalsized fragment. In this case, it is clear that . We now find a contradiction.
Set . Then for each , the induced subgraph is a path of length in , so the above argument is available here. We thus obtain at least many maximalsized fragments in containing . On the other hand, for every maximalsized fragment containing , we have that for some since , hence there are at most many maximalsized fragments in containing , yielding a contradiction because . This proves that , i.e., is balanced. Assume where . As we have seen, for each , there is a such that . Then is a maximalsized fragment containing , and the semiimprimitivity and imply if . Therefore, , , are the all maximalsized fragments containing . This proves that for every , there is only one nontrivial fragment with . ∎
3 Proof of Theorem 1.2
With the assumptions in the theorem, we put and . The bipartite graph is defined by the crossintersecting relation between and : For and , if and only if . It is easy to check that is connected since and , and is noncomplete since . Clearly, transitively acts on and , respectively, in a natural way, and preserves the crossintersecting relation. Therefore, for each , and