#### Abstract

We provide a sufficient condition for the existence of a cycle in a connected graph which is edge-disjoint from two connected subgraphs and of such that is connected.

#### 1. Introduction

Prompted by Hobbs' conjecture, Jackson [1] proved that if is a 2-connected simple graph of minimum degree at least and , then contains a cycle such that is 2-connected. Lemos and Oxley [2] generalized this result as follows.

Theorem 1. *Let be a 2-connected simple graph and let be a subgraph of such that is either 2-connected or . Suppose that for all in and has a cycle that is edge-disjoint from . Then there exists a cycle in that is edge-disjoint from such that is 2-connected. *

Borse and Waphare [3] obtained the following result for the class of connected graphs which is analogous to the above theorem and is an improvement of a result due to Sinclair [4].

Theorem 2. * Let be a connected simple graph and let be a connected subgraph of such that there is a cycle in that is edge-disjoint from . Suppose that for all . Then there exists a cycle in that is edge-disjoint from such that is connected.*

The problem of the existence of cycles in graphs deletion of whose edges preserve connectedness is well studied in the literature (see [1–10]). In this paper, we improve Theorem 2 as follows.

Theorem 3. * Let be a connected simple graph and let and be connected subgraphs of . Suppose that there are at least two cycles in and further, no two edges of form an edge-cut of . Then there exists a cycle in that is edge-disjoint from both and such that is connected.*

Let be a connected graph of minimum degree at least three and let and be connected subgraphs of . If has only one cycle, say , then is not connected if a subset of forms an edge-cut of . Therefore, in Theorem 3, we must assume that contains at least two cycles. The following example shows that the condition in Theorem 3 regarding an edge-cut of is necessary. Let and be two disjoint copies of with . Suppose that is an edge of for . Let for . Let be the graph obtained from and by adding two new vertices , and seven new edges , , , , , , and ; see Figure 1. Then is simple, connected and the minimum degree of is 3. Let , , and . Then , and are the only cycles of . For , it is easy to see that is not connected as the set is a 2-edge-cut of for each .

As a consequence of Theorem 3, we get the following result.

Corollary 4. * Let be an Eulerian 3-edge-connected simple graph containing at least four mutually edge-disjoint cycles. Then there exist three mutually edge-disjoint cycles , and such that is connected for .*

We refer to [11] for the terms that are not defined here.

#### 2. Proof

In this section, we prove Theorem 3 and its corollary. First we prove the following lemma.

Lemma 5. * Let , and be as in the hypothesis of Theorem 3. If there is an edge of joining and , then there exists a cycle in that is edge-disjoint from both and such that is connected. *

*Proof. *Let be an edge of , having one end vertex in and the other end vertex in . Let . Then is a connected subgraph of . Let and be cycles in . If or does not contain , then it is a cycle in . If belongs to both and , then contains a cycle. Thus, in any case, there is a cycle in . Since is connected for every subset of edges of the graph with , for all . Hence, by Theorem 2, there exists a cycle in such that is connected. Obviously, is edge-disjoint from both and .

The number of edges of a cycle or a path is denoted by . If and are distinct vertices of a cycle of a graph , then and denote the edge-disjoint paths along which join to .

*Proof of Theorem 3. *Suppose that the result is false and let be a counterexample. Then there exist connected subgraphs and of satisfying the hypothesis of the theorem such that for every cycle in that is edge-disjoint from , the graph is not connected. Choose and such that is maximum. By a hypothesis, for all . Therefore, by Theorem 2, is not connected. Thus is vertex disjoint from .*Claim 1*. If is a cycle in , then every component of contains or .

Let be a cycle in . Since is connected, it is contained in a component of for . If ; then, by Theorem 2, there exists a cycle in such that is connected, which is a contradiction. Therefore . We prove that and are the only components of . Let . It is sufficient to prove that is an empty set. Assume that is nonempty. Since is connected, each component of contains a vertex of . If a component of contains exactly one vertex of , then the edges incident with that vertex form a -edge-cut of , which is a contradiction. Hence each component of contains at least two vertices of . Hence . Therefore there is a component of different from and such that . Let . Let be a path in between two vertices of . Then contains at least two cycles which are edge-disjoint with . By the maximality of , we have and . By Lemma 5, there is no edge between and . Suppose that . Then . It follows that has exactly three components and hence we get two edges of , with each having one end in and the other in such that is an edge-cut of , a contradiction. Thus .

Let . Then . By traversing from along on both sides, let be the first vertex of on one side and let be the first vertex of on the other side. By Lemma 5, . We may assume that . Then . Let . Then is connected and . By the maximality of , contains at most one cycle. Let be the component of containing . Then and . If , then contains two cycles, which is a contradiction. Hence . This implies that there is a component of such that . Traverse the path beginning at , let be the second vertex of on this path. Let be a path in . Let . Then is a cycle. If does not meet , then intersects in at least two vertices giving a cycle which is disjoint from , a contradiction. Hence contains a vertex of . If this path contains another vertex of , then contains two cycles, where is an path in . This is a contradiction. Hence . This implies that the path meets in a vertex . Let be an path in . Then contains two cycles, which is a contradiction. This proves the claim. *Claim 2.* If and are cycles in , then .

Assume that the claim is false. Let and be cycles in such that . Then is edge-disjoint from . By Claim 1, each component of contains or . Thus has exactly two components. We may assume that is contained in a component of for . Obviously, is contained in or . We may assume that is contained in . By the maximality of , it follows that . By Claim 1, has only two components and , with each containing or . We may assume that is contained in for . Since is connected, it is contained in . Let be the set of all edges in , having one end vertex in and the other end vertex in . Then . By a hypothesis, . Let be the set of all edges in , having one end vertex in and the other end vertex in . Then and .

If is connected, then is connected, a contradiction. Therefore we may assume that is disconnected. Obviously, is a component of . Let . Note that , , , and . Hence . Let . Then is vertex disjoint from both and . If contains a cycle then contains two cycles and hence we get two cycles in . By the maximality of , we get a cycle that is edge-disjoint from both and such that is connected, which is a contradiction. Thus is a forest. Let . Then is a forest. Every edge of with one end vertex in and the other end vertex in is either in or in and hence or . As for all and , at most one vertex of is isolated. Hence contains an edge. Let be a nontrivial tree in . Let and be two pendant vertices of . Since , for . We may assume that . Suppose that . Let and let be the path in . Then contains at least two cycles. These cycles are edge-disjoint from and . Now, a contradiction follows from the maximality of . Hence . Similarly, . This implies that for . If has a third pendant vertex , then is in or , which is a contradiction. Hence has exactly two pendant vertices and therefore it is a path. As for all , . This implies that has no internal vertex. Hence is isomorphic to . Thus every component of is isomorphic to either or . Let be an edge of with and . Suppose that has an another edge . Then for each . We may assume that and . Since , we may assume by suitable labeling that at least one edge of is not in . It follows that the subgraph is edge-disjoint from the subgraph and further, it contains at least two cycles. Obviously, is connected and contains more edges than . A contradiction follows from the maximality of . Hence contains exactly one edge. Since , . As at most one vertex of is isolated, . Let . We may assume that , , , and . Obviously, or . We may assume that . The subgraph contains two cycles and further, it is edge-disjoint from the subgraphs and . A contradiction follows from the maximality of . The proof of Claim 2 is completed.

By Claim 2, any two cycles in has at least two vertices in common. This implies that there exist two vertices and and three mutually internally disjoint , paths , , and in . We may assume that . Let . Then is a cycle in . By Claim 1, has exactly two components each containing or . Let be the component of containing for . We may assume that the path is contained in the component . Hence the vertices and are in . It follows from the maximality of that . Let be the set of all edges in that have one end vertex in and the other end vertex in . Then . By a hypothesis, . In fact, . By Lemma 5, no edge of has an end vertex in as . Obviously, or . Let if ; otherwise . Thus is a cycle in that is edge-disjoint from both and and further, and for some . By Claim 1, has exactly two components and . We may assume that is contained in and is contained in . As , is connected and hence contains . So the vertices and do not belong to . Thus has at least two components, one of which is .

By a hypothesis, . Further, every component of contains at least one vertex of and hence contains at least one vertex of . Suppose that there is a component of containing two vertices and of with . Let be an path in . Obviously, contains at least two cycles. Thus there are at least two cycles in . Further, . Since and are connected and are vertex disjoint subgraphs, by the maximality of , there exists a cycle in that is edge-disjoint from and hence edge-disjoint from such that is connected, which is a contradiction. Thus has exactly one vertex in common with every component of that is different from . Since and are common vertices of and , there are two components and of , both different from such that and .

Suppose that , , and are the only components of . Let . Then . Suppose that does not meet . Then, as meets each of and only once, the corresponding two edges of form a 2-edge-cut of , which is a contradiction. Therefore meets . Since is a path and for . If meets , then is connected, which is a contradiction as is a cycle in . Thus does not meet . Hence . This shows that . Traverse the path beginning at and suppose that meets at the first time in the vertex . Let be the last vertex of on the subpath of . Then the subpath of contains at most two edges of . Extend the path to a path by adding a path in and an path in . Thus . As contains at least two cycles, there are at least two cycles that are edge-disjoint from the connected subgraphs and of . Since is contained in , is contained in , and is connected, by the maximality of , there exists a cycle that is edge-disjoint from such that is connected, which is a contradiction. Thus has at least four components.

Since and are the only components of , the path meets every component of except . Also, . It follows that we can choose two components and of and two vertices , such that for the subpath of we have . Let and let and be paths in and , respectively, from to and to . Then contains at least two cycles and further, it is edge-disjoint from , and . Due to the maximality of , there exists a cycle that is edge-disjoint from both and such that is connected, which is a contradiction. This completes the proof of the theorem.

*Proof of Corollary 4. *Let , , and be mutually edge-disjoint cycles in . By Theorem 3, there exists a cycle in such that is connected. Since is eulerian, every component of is eulerian. Hence has a cycle decomposition containing the cycle . Since and are edge-disjoint from and contains at least one cycle other than . As is edge-disjoint from , by Theorem 3, there exists a cycle in such that is connected. Obviously, each component of is eulerian. Further, is a cycle in . There exists a cycle decomposition of containing . It is easy to see that contains at least one edge. Hence contains at least one edge which is not in . This implies that contains a cycle with . Hence, by Theorem 3, there exists a cycle in such that is connected. Obviously, , , and are mutually edge-disjoint cycles.