#### Abstract

M.Kriesell conjectured that there existed , such that every 5-connected graph with at least vertices can be contracted to a 5-connected graph such that . We show that this conjecture holds for vertex transitive 5-connected graphs.

#### 1. Introduction

All graphs considered here are supposed to be simple, finite, and undirected graphs. For a connected graph , a subset is called a smallest separator if and has at least two components. Let be a -connected graph, and let be a subgraph of . Let stand for the graph obtained from by contracting every component of to a single vertex and replacing each resulting double edges by a single edge. A subgraph of is said to be -contractible if is still -connected. An edge is a -contractible edge if is -connected; otherwise, we call it a noncontractible edge. Clearly, two end-vertices of a noncontractible edge are contained in some smallest separator. A -connected graph without a -contractible edge is said to be a contraction-critical -connected graph.

Tutteâ€™s [1] wheel theorem showed that every 3-connected graph on more than four vertices contains a 3-contractible edge. For , Thomassen and Toft [2] showed that there were infinitely many contraction-critical -regular -connected graphs. On the other hand, one can find that every 4-connected graph can be reduced to a smaller 4-connected graph by contracting at most two edges. Therefore, Kriesell [3] posted the following conjecture.

Conjecture 1. *(see [3]). There exists , such that every -connected graph with at least vertices can be contracted to a k-connected graph such that .*

Clearly, Conjecture 1 is true for . By Kriesellâ€™s examples [3], Conjecture 1 fails for . Hence, it is still open for .

A smallest separator of a -connected graph is said to be trivial if has exactly two components and one of them has exactly one vertex. A 5-connected graph is essentially a 6-connected graph if every smallest separator of is trivial. In ([3]), Kriesell proved the following results.

**(a)**

**(b)**

**(c)**

**(d)**

**(e)**

Theorem 1. *(see [3]). Every essentially 6-connected graph with at least 13 vertices can be contracted to a 5-connected graph such that .*

In this paper, we will show that Conjecture 1 is true for vertex transitive 5-connected graphs. Clearly, Conjecture 1 holds for 5-connected graphs which contain a contractible edge. Hence, in order to show that Conjecture 1 holds for vertex transitive 5-connected graphs, we have to show that all vertex transitive contraction-critical 5-connected graphs have a small contractible subgraph. So, the key point of this paper is to characterize the local structure of a vertex transitive contraction-critical 5-connected graph and, then, to find the contractible subgraph of it. In the following, for convenience, a vertex transitive contraction-critical 5-connected graph will be called a -5-connected graph. For a contraction-critical 5-connected graph, there are some results on the local structure of it [4â€“10].

To state our results, we need to introduce some further definitions. Let be a 5-connected graph which is 5-regular. For any , we say that has one of the following four types according the graph induced by the neighborhood of (see Figures 1(a)â€“1(d)).(i)Type 1: (ii)Type 2: (iii)Type 3: (iv)Type 4:

Moreover, for , has if every vertex of has type .

Furthermore, we need to introduce the graph (see Figure 1(e)). One can check that is vertex transitive, and can be reduced to by contracting and .

First, we have the following results on the local structure of -5-connected graphs.

Theorem 2. *Let be a -5-connected graph. If , then either or .*

Theorem 3. *Let be a -5-connected graph. If , then has type 1, type 2, type 3, or type 4.*

Theorem 4. *Let be a -5-connected graph. If has type 2, then is isomorphic to icosahedron.*

Then, we will prove the following main result of the paper.

Theorem 5. *Let be a 5-connected vertex transitive graph which is neither nor icosahedron, and then, can be contracted to a 5-connected such that .*

The organization of the paper is as follows. Section 2 contains some preliminary results. In Section 3, we will characterize the local structure of 5-connected -graphs. In Section 4, we will prove Theorem 5.

#### 2. Terminology and Lemma

For terms not defined here, we refer the reader to [11]. Let be a graph, where denotes the vertex set of and denotes the edge set of . Let denote the automorphism group of , and let denote the vertex connectivity of . Let denote a path on vertices. An edge joining vertices and will be written as . Let stand for the new vertex obtained by contracting the edge . For , we define . For , we define . Furthermore, let denote the subgraph induced by , and let denote the graph obtained from by deleting all the vertices of together with the edges incident with them. Let stands for the set of edge with one end in and the other end in .

Let be a smallest separator of a noncomplete connected , and the union of at least one but not of all components of is called a -fragment. A fragment of is a -fragment for some smallest separator . Let be a -fragment, and let . Clearly, , and is also a -fragment such that . A fragment with least cardinality is called an atom. For , and , we often omit the index if it is clear from the context.

Furthermore, we need some special terminologies for 5-connected graphs. Let be a fragment of , and let . Let , and . A vertex is said to be an of if both of the following two conditions hold.

A vertex is said to be an of , if is an of for some . Let (resp. ) stand for the set of admissible vertices of (resp. ). Let be an edge of , and a fragment is said to be a fragment with respect to if .

The following properties of fragment are well known (for the proof, see [12]), and we will use them without any further reference.

Lemma 1. *(see [12]). Let and be two distinct fragments of ; , . Then, the following statements hold.*(1)*If , then *(2)*If and is not a fragment of G, then and *(3)

*If , then both and are fragments of*

*G*, andLemma 2. *(see [4]). Let be a -connected graph, and is a fragment of . Let . If , then .*

Lemma 3. *(see [5]). Let be a contraction-critical 5-connected graph, and then, contains a vertex such that every edge incident with is contained in some triangle.*

Lemma 4. *(see [6]). Let be a contraction-critical 5-connected graph. Let , and be a fragment such that , , and . If , then .*

Lemma 5. *(see [7]). Let be a fragment of a contraction-critical 5-connected graph such that , and let be two vertices of such that . Then, either or .*

Lemma 6. *Let be a vertex transitive connected graph, and then, for any two vertices and , .*

*Proof. *Since is a vertex transitive graph, there exist such that . It follows that . Hence, is an isomorphic of and , where is the restriction of on .

Lemma 7. *Let be a prime integer, and let be a vertex transitive graph with ; then, is a -regular graph.*

*Proof. *To the contrary, we may assume that since . It follows that every atom of has at least two vertices. Since is a vertex transitive graph, then every vertex of is contained in some atom.

First, we show that any two atoms of are disjoint. Otherwise, let and be two distinguished atoms of such that . By the definition of atom, is not a fragment. Lemma 1 assures us that and . This implies that , a contradiction. Thus, any two atoms of are disjoint.

Let and be two atoms of such that . It follows that . We show that . Otherwise, suppose . If is a fragment of , then we see that , since . This contradicts the definition of atom. So, is not a fragment of . Lemma 1 assures us that and . It follows that , a contradiction. Hence, . Therefore, is the disjoint union of some atom, since any two atoms of are disjoint and every vertex of is contained in some atom. This means that is a subdivision of , and hence, . It follows that . By symmetry, we see that , which implies that , a contradiction.

Lemma 8. *Let be a -5-connected graph. If does not contain as subgraph, then for any , .*

*Proof. *Clearly, Lemma 7 assures us that is 5-regular, which implies that has an even order. Suppose that with such that adjacent to at least three vertices of . Let , and it follows . If , then has six vertices. It follows that , which implies contains , a contradiction. Hence, we may assume that . It follows that is a fragment of . By symmetry, we assume that . Now, we observe that and . Let . Let , and then . Now, the fact that is even assures us that . It follows that is a fragment of . Furthermore, we see that and .Now, Lemma 5 assures us that either or . If , then without loss of generality, assume . Therefore, is a connected graph. If , then, similarly, we have that is a connected graph. Now, since is vertex transitive, the following claims hold.

*Claim 1. *For any , is a connected graph.

*Claim 2. *For any , contains a cycle of length 4.

*Proof. *Since is a vertex transitive graph, we only show that has a cycle of length 4. By Claim 1, we see that for . On the other hand, we observe that , since does not contain . This implies that every member of is either adjacent to or . It follows that or . By symmetry, we may assume that . It follows that contains a cycle of length 4. Hence, for any , has a cycle of length 4.

Now, we are ready to complete the proof of Lemma 8. By Claim 2, we see that for . This implies that . Now, we see that every vertex of is adjacent to exactly one vertex of . If , say , then and , a contradiction. If , we can observe that has a vertex with a degree of at most 4, a contradiction. Hence, we may assume that . Now, Lemma 4 shows that , which implies for some , a contradiction.

Lemma 9. *Let be a -5-connected graph. If has type 4, then is essentially 6-connected.*

*Proof. *Since has type 4, we see that for any , .

*Claim 3. *If is a fragment of , then .

*Proof. *Suppose . If , then has three neighbors in , a contradiction. So, we may assume . It follows that . Let be the path of . It follows that . If , then Lemma 4 implies that . Hence, either or . This is a contradiction, since . Hence, we may assume that . If , then we see that , a contradiction. So, we may assume . It follows that , which contradicts the fact that has an even order. Hence, Claim 3 holds.

*Claim 4. *If is a fragment of , then .

*Proof. *We first show that is a connected graph. Otherwise, let be a component of such that has exactly one vertex. It follows that is a fragment of cardinality 2, a contradiction. Next, we show that is a path. Suppose is a cycle, then a simple calculation shows that . This implies that one vertex of , say , has exactly one neighbor in . Now, we find that is a fragment of cardinality 2, a contradiction.

Let be the path of , and let . Without the loss of generality, let .â€‰Subclaim 1. and .â€‰Proof. Notice that has type 4; we find that . If , then we find that , a contradiction. Hence, . By symmetry, .â€‰Without the loss of generality, we may assume that . Now, if , then is a cycle of , a contradiction. Therefore, , which implies that .â€‰Subclaim 2. .â€‰Proof. If , then has a triangle, a contradiction. It follows that . Similarly, we have . Furthermore, if , then we find that there is a cycle of length four in , a contradiction. Thus, . It follows that .Now, we are ready to complete the proof of Claim 4. Focusing on , we find that since is connected. By Subclaim 2, we may assume that . Now, we find that there is a cycle of length four in , a contradiction.

*Claim 5. *For every smallest separator , has exactly two components.

*Proof. *Otherwise, assume that has at least three components. Let , , and be three connected components of .â€‰Subclaim 3. For any , , and .â€‰Proof. Let , let . Without the loss of generality, we may assume that . Now, we find that , since has type 4. Suppose . If , then the fact that is connected shows that has three neighbors in , which contradicts the fact that has type 4. So, we have . It follows that and .By Subclaim 3, , which implies that is a cycle of length 5. Hence, we see that . Furthermore, by Subclaims 3 and 4, for each . Focusing on , we find that , which implies that . Recall that , and Lemma 4 shows that . Without the loss of generality, assume that . This implies that , which contradicts Subclaim 3. Hence, Claim 5 holds.

Next, we assume that is not essentially 6-connected. It follows that there is a fragment such that and . Let is a fragment such that and , and let . By Claims 3 and 4, we see that . Let and . Let , and let . Now, since is vertex transitive, every vertex of is contained in some member of . Therefore, there exist such that . Next, we will analyse the local structure of and .

*Claim 6. *If , then .

*Proof. *Suppose and . Now, Lemma 1 assures us that . It follows thatThis contradicts the choice of .

*Claim 7. * if and only if .

*Proof. *Suppose . Now, Lemma 1 assures us that . If , then we see thatThis contradicts the choice of . Hence, we see that . By symmetry, we see that implies .

*Claim 8. *.

*Proof. *Suppose . By Claim 6, we know that . Hence, is a fragment of . By the choice of , we know that . Furthermore, since , Lemma 1 assures us that .

If , then Claim 7 assures us that . Furthermore, implies that . Hence, we find that , a contradiction.

So, we may assume . Then, Claim 7 assures us that . Hence, both and are fragments of . By the choice of and , we know that . It follows that .

Since , Lemma 1 assure us that . If , then , a contradiction.

Therefore, let . It follows that . Since and , Lemma 1 assures us that . This implies that . Let . Now, , since is a fragment. Similarly, we find that , , and . Now, we find that has at least two components, a contradiction.

*Claim 9. * and .

*Proof. *Suppose . By Claim 7, we see that . Hence, both and are fragments of . By the choice of , we see that . It follows that .

If , then . Hence, we see that , a contradiction.

Hence, we may assume that . Then, by the choice of , we know that . It follows that . Hence, we find that , a contradiction.

Now, we are ready to complete the proof of the Lemma. By Claims 8 and 9, we find that and . Now, we find that , since . It follows that . Now, Lemma 1 implies that . It follows that , a contradiction.

#### 3. The Local Structure of TCC-5-Connected Graphs

In this section, since Lemma 7 holds, all TCC-5-connected graphs were supposed to be 5-regular and have an even order.

Theorem 6. *Let be a TCC-5-connected . If , then either or .*

*Proof. *Recall that has an even order. It follows that either or . If , then . So, we may assume . It follows that has a fragment of cardinality 2. Let be a fragment of . Let , and let .

*Claim 10. *.

*Proof. *Otherwise, we find that . It follows that . On the other hand, . It follows that , a contradiction.

Let . By symmetry, we may assume that . We find that has at least three neighbors in . Hence, Lemma 8 implies that contains . It follows that contains a triangle.

*Claim 11. *.

*Proof. *Suppose . Notice that for , , we see that . Therefore, , which implies that does not contain a triangle, a contradiction.

Now, we observe that is 2-regular. Hence, is a cycle of length 5. Now, by symmetry, we may assume that and . It follows that since is a cycle of length 5. Therefore, we have .

Lemma 10. *Let be a -5-connected with . If contains as a subgraph, then has type 1.*

*Proof. *Since is a vertex transitive graph, we know that every vertex of is contained in a . Let be a vertex of , and let . Without the loss of generality, let .

*Claim 12. *.

*Proof. *We only show that , and the other one can be handled similarly. Otherwise, by symmetry, we may assume . Let . Let , and it follows that . Furthermore, recall that , . If , then is a separator of order 4, a contradiction. Thus, . Therefore, is a fragment of . Furthermore, since , we see that . Let , and let , where . Similarly, is a fragment of such that . Furthermore, we have .

Notice that and , we see that . Now, we see that , , , , , and .

Now, since , we find that . Let . Clearly, is a fragment with . Notice that , , and . Now, Lemma 4 implies that . It follows that either or has two neighbors in . By symmetry, let have two neighbors in . It follows that , a contradiction. Hence, Claim 12 holds.

*Claim 13. *.

*Proof. *We only show that , and the other one can be handled similarly. Otherwise, by symmetry, we may assume that . Now, by Claim 12, . Let and . Clearly, and . Now, since is 5-connected, we observe that . Therefore, is a fragment of .â€‰Subclaim 4. .â€‰Proof. Suppose , and then, is fragments of . Furthermore, we see that is a connected graph, and this implies that for any , is a connected graph. Let , and it follows , where . Now, since is 5-connected, we see that . We observe that , , , , , and . Furthermore, we see that and . Notice that is connected, and we see that for every vertex of , is connected.Now, since and , the fact shows that . Furthermore, is a fragment.

If , then has at least two components, a contradiction. Therefore, and .

On the other hand, by Claim 12, , and . This fact implies that either or .

By symmetry, we may assume . Now, since , we see that has only one vertex of degree 3. On the other hand, we find that has two vertex of degree 3, and this implies that , a contradiction. This contradiction shows that . By symmetry, . Hence, Subclaim 4 holds.â€‰Subclaim 5. .â€‰Proof. Suppose . Let be a graph which is got from the path by adding the edge . Clearly, . Now, since , we find that . This implies that . Let . Furthermore, has a triangle, since has a triangle. Therefore, is a triangle. It follows that has a Hamilton cycle. This implies that , a contradiction. Thus, Subclaim 5 holds.By Subclaims 4 and 5, has two components, and one of them has exactly one vertex. If , then , a contradiction. So, assume that , which implies that . By Claim 12 and Subclaim 5, . Now, we see that which contains is contained in . Hence, is a connected graph, a contradiction. Thus, Claim 13 holds.

By Claim 13 and Lemma 3, , and hence, has type 1. Therefore, has type 1.

Theorem 7. *Let be a TCC-5-connected . If , then has type 1, type 2, type 3, or type 4.*

*Proof. *If contains , then Lemma 10 assures us that has type 1. So, we may assume that does not contain . Hence, Lemma 8 assures us that for any , . Now, Lemma 3 assures us that has either type 2 or type 3 or type 4.

Theorem 8. *Let be a TCC-5-connected . If has type 2, then is isomorphic to icosahedron.*

*Proof. *Let , and let be the cycle of . Furthermore, let . Since has type 2, we may assume that is a cycle of . Let , and then and . Now, is a cycle of . Let . If , then, since has type 2, . This implies that , a contradiction. Hence, we have . Since does not contain , we see that . Now, we observe that is the cycle of . Let