How to Contract a Vertex Transitive 5-Connected Graph

<jats:p>M.Kriesell conjectured that there existed <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M1"><mml:mi>b</mml:mi></mml:math>, <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M2"><mml:mi>h</mml:mi></mml:math> such that every 5-connected graph <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M3"><mml:mi>G</mml:mi></mml:math> with at least <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M4"><mml:mi>b</mml:mi></mml:math> vertices can be contracted to a 5-connected graph <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M5"><mml:msub><mml:mrow><mml:mi>G</mml:mi></mml:mrow><mml:mrow><mml:mn>0</mml:mn></mml:mrow></mml:msub></mml:math> such that <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M6"><mml:mn>0</mml:mn><mml:mo><</mml:mo><mml:mfenced open="|" close="|" separators="|"><mml:mrow><mml:mi>V</mml:mi><mml:mfenced open="(" close=")" separators="|"><mml:mrow><mml:mi>G</mml:mi></mml:mrow></mml:mfenced></mml:mrow></mml:mfenced><mml:mo>−</mml:mo><mml:mfenced open="(" close=")" separators="|"><mml:mrow><mml:mi>V</mml:mi><mml:mfenced open="(" close=")" separators="|"><mml:mrow><mml:msub><mml:mrow><mml:mi>G</mml:mi></mml:mrow><mml:mrow><mml:mn>0</mml:mn></mml:mrow></mml:msub></mml:mrow></mml:mfenced></mml:mrow></mml:mfenced><mml:mo><</mml:mo><mml:mi>h</mml:mi></mml:math>. We show that this conjecture holds for vertex transitive 5-connected graphs.</jats:p>


Introduction
All graphs considered here are supposed to be simple, finite, and undirected graphs. For a connected graph G, a subset T ⊆ V(G) is called a smallest separator if |T| � κ(G) and G − T has at least two components. Let G be a k-connected graph, and let H be a subgraph of G. Let G/H stand for the graph obtained from G by contracting every component of H to a single vertex and replacing each resulting double edges by a single edge. A subgraph H of G is said to be k-contractible if G/H is still k-connected. An edge e is a k-contractible edge if G/e is k-connected; otherwise, we call it a noncontractible edge. Clearly, two end-vertices of a noncontractible edge are contained in some smallest separator. A k-connected graph without a k-contractible edge is said to be a contraction-critical k-connected graph.
Tutte's [1] wheel theorem showed that every 3-connected graph on more than four vertices contains a 3-contractible edge. For k ≥ 4, omassen and Toft [2] showed that there were infinitely many contraction-critical k-regular k-connected graphs. On the other hand, one can find that every 4connected graph can be reduced to a smaller 4-connected graph by contracting at most two edges. erefore, Kriesell [3] posted the following conjecture.
A smallest separator T of a k-connected graph is said to be trivial if G − T has exactly two components and one of them has exactly one vertex. A 5-connected graph G is essentially a 6-connected graph if every smallest separator of G is trivial. In ( [3]), Kriesell proved the following results.
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 TCC-5-connected graph. For a contraction-critical 5-connected graph, there are some results on the local structure of it [4][5][6][7][8][9][10].
To state our results, we need to introduce some further definitions. Let G be a 5-connected graph which is 5-regular. For any x ∈ V(G), we say that x has one of the following four types according the graph induced by the neighborhood of x(see Figures 1(a)-1(d)).
Furthermore, we need to introduce the graph G * (see Figure 1(e)). One can check that G * is vertex transitive, and G * can be reduced to K 6 by contracting yx 1 and xx 2 .
First, we have the following results on the local structure of TCC-5-connected graphs.
en, we will prove the following main result of the paper.
Theorem 5. Let G be a 5-connected vertex transitive graph which is neither K 6 nor icosahedron, and then, G can be contracted to e 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 TCC-graphs. In Section 4, we will prove eorem 5.

Terminology and Lemma
For terms not defined here, we refer the reader to [11]. Let G � (V(G), E(G)) be a graph, where V(G) denotes the vertex set of G and E(G) denotes the edge set of G. Let Aut(G) denote the automorphism group of G, and let κ(G) denote the vertex connectivity of G. Let P n denote a path on n vertices. An edge joining vertices x and y will be written as xy. Let [xy] stand for the new vertex obtained by contracting the edge xy. For x ∈ V(G), we define Let T be a smallest separator of a noncomplete connected G, and the union of at least one but not of all components of G − T is called a T-fragment. A fragment of G is a T-fragment for some smallest separator T. Let F be a T-fragment, and let F � V(G) − (F ∪ T). Clearly, F ≠ ∅, and F is also a T-fragment such that N G (F) � T � N G (F). A fragment with least cardinality is called an atom. For N G (x), d G (x), and N G (F), we often omit the index G if it is clear from the context. Furthermore, we need some special terminologies for 5connected graphs. Let A be a fragment of G, and let S � N(A). Let x ∈ S, and y ∈ N(x) ∩ A. A vertex z is said to be an admissible vertex of (x, y; A) if both of the following two conditions hold. (1) A vertex z is said to be an admissible vertex of (x; A), if z is an admissible vertex of (x, y; A) for some y ∈ N(x) ∩ A. Let Ad(x, y; A)(resp. Ad(x; A)) stand for the set of admissible vertices of (x, y; A)(resp. (x; A)). Let e be an edge of G, and a fragment A is said to be a fragment with respect to e if V(e) ⊆ N(A).
e following properties of fragment are well known (for the proof, see [12]), and we will use them without any further reference.

Lemma 6. Let G be a vertex transitive connected graph, and then, for any two vertices x and y, G[N(x)] � G[N(y)].
Proof. Since G is a vertex transitive graph, there exist g ∈ Aut(G) such that x g � y. It follows that (N(x)) g � N(y).  Proof. To the contrary, we may assume that δ(G) > p since δ(G) ≥ κ(G). It follows that every atom of G has at least two vertices. Since G is a vertex transitive graph, then every vertex of G is contained in some atom. First, we show that any two atoms of G are disjoint. Otherwise, let A and B be two distinguished atoms of G such is implies that |A| > |B|, a contradiction. us, any two atoms of G are disjoint.
Let A and C be two erefore, N(A) is the disjoint union of some atom, since any two atoms of G are disjoint and every vertex of G is contained in some atom.
is means that |A| is a subdivision of |N(A)|, and hence, |A| � p. It follows that N|A| � C. By symmetry, we see that Proof. Clearly, Lemma 7 assures us that G is 5-regular, which implies that G has an even order. Suppose that then G has six vertices. It follows that G � K 6 , which implies G contains K 4 , a contradiction.
Hence, we may assume that Proof. Since G is a vertex transitive graph, we only show that N(x 1 ) has a cycle of length 4. By Claim 1, we see that for Discrete Dynamics in Nature and Society By symmetry, we may assume that has a cycle of length 4. Now, we are ready to complete the proof of Lemma 8. By Claim 2, we see that If |B| � 2, we can observe that B has a vertex with a degree of at most 4, a contradiction. Hence, we may assume that |B| � 3. Now, Lemma 4 shows that Ad( Proof. Since G has type 4, we see that for any Hence, we may assume that |A| ≤ 2. If |A| � 1, then we see that d(x 1 ) < 5, a contradiction. So, we may assume |A| � 2. It follows that |V(G)| � 9, which contradicts the fact that G has an even order. Hence, Claim 3 holds. □ Claim 4. If A is a fragment of G, then |A| ≠ 3.

Proof. We first show that G[A] is a connected graph. Otherwise, let A 1 be a component of G[A]
such that A 1 has exactly one vertex. It follows that A 2 � A − A 1 is a fragment of cardinality 2, a contradiction. Next, we show that G[A] is a path. Suppose G[A] is a cycle, then a simple calculation shows that |z(A)| � 9. is implies that one vertex of N(A), say w, has exactly one neighbor in A. Now, we find that A − N(w) is a fragment of cardinality 2, a contradiction.
Let xyz be the path of G[A], and let N(A) � x 1 , . . . , x 5 . Without the loss of generality, let Proof. Notice that G has type 4; we find that |N(x) ∩ Without the loss of generality, we may assume that erefore, then N(y) has a triangle, a contradiction. It follows that then we find that there is a cycle of length four in N(y), a contradiction. us, Now, we are ready to complete the proof of Claim 4.
Focusing on x 2 , we find that N( ] is connected. By Subclaim 2, we may assume that x 4 ∈ N(x 2 ). Now, we find that there is a cycle of length four in N(x 2 ), a contradiction. Proof. Otherwise, assume that G − T has at least three components. Let A 1 , A 2 , and A 3 be three connected components of G − T.
Proof. Let y ∈ T, let N(y) � y 1 , . . . , y 5 . Without the loss of generality, we may assume that y i ∈ A i , i ∈ 1, 2, 3 { }. Now, we find that N(y) ∩ T ≠ ∅, since G has type 4. Suppose y 4 ∈ N(y) ∩ T. If N(y) ∩ T � y 4 , then the fact that G[N(y)] is connected shows that y 4 has three neighbors in G[N(y)], which contradicts the fact that G has type 4. So, we have N(y) ∩ T � y 4 , y 5 .
Next, we assume that G is not essentially 6-connected. It follows that there is a fragment B such that |B| ≥ 2 and { is a fragment such that |B| ≥ 2 and |B| ≥ 2}, and let t � min |B | |B ∈ B { }. By Claims 3 and 4, we and |B| � t}. Let A ∈ B 1 , and let y ∈ N(A). Now, since G is vertex transitive, every vertex of G is contained in some member of B 1 . erefore, there exist B ∈ B 1 such that y ∈ B. Next, we will analyse the local structure of A and B.

Proof. Suppose A ∩ B ≠ ∅ and A ∩ B � ∅. Now, Lemma 1 assures us that |A ∩ N(B)| ≥ |B ∩ N(A)|. It follows that
is contradicts the choice of A.
is contradicts the choice of A. Hence, we see that If A ∩ B � ∅, then Claim 7 assures us that A ∩ B � ∅. Since A ∩ B ≠ ∅, Lemma 1 assure us that has at least two components, a contradiction. Hence, we may assume that A ∩ B � ∅.
en, by the choice of B, we know that |B| ≥ |B|. It follows that |B ∩ N(A)| ≥ 3. Hence, we find that |N(A)| ≥ |B ∩ N(A)|+ |B ∩ N(A)| ≥ 6, a contradiction. Now, we are ready to complete the proof of the Lemma. By Claims 8 and 9, we find that A � A ∩ N(B) and

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 G be a TCC-5-connected graph. If
Proof. Recall that G has an even order. It follows that either |V(G)| � 6 or |V(G)| � 8. If |V(G)| � 6, then G � K 6 . So, we may assume |V(G)| � 8. It follows that G has a fragment of Proof. Otherwise, we find that G[N(A)] � C 5 . It follows that Let N(x) � x 1 , x 2 , x 3 , x 4 , y . By symmetry, we may assume that N(y) � x 1 , x 2 , x 3 , x 5 , x . We find that y has at least three neighbors in G[N(x)]. Hence, Lemma 8 implies that G contains K 4 . It follows that G[N(x)] contains a triangle.

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Proof. Since G is a vertex transitive graph, we know that every vertex of G is contained in a K 4 . Let x be a vertex of G, and let N(x) � x 1 , . . . , x 5 . Without the loss of generality, Proof. We only show that |N(x 4 ) ∩ x 1 , x 2 , x 3 | ≤ 1, and the other one can be handled similarly. Otherwise, by symmetry, we may assume then N(A) is a separator of order 4, a contradiction. us, t 1 ≠ x 5 . erefore, A is a fragment of G. Furthermore, since |V(G)| ≥ 10, we see that |A| ≥ 3. Let B � x, x 2 , and let N( Notice that t 2 ∈ A and t 1 ∈ N(A), we see that 4 . It follows that either x 3 or x 4 has two neighbors in N(C). By symmetry, let x 3 have two neighbors in N(C). It follows that d(x 3 ) ≥ 6, a contradiction. Hence, Claim 12 holds.
Proof. We only show that N(x 4 ) ∩ x 1 , x 2 , x 3 � ∅, and the other one can be handled similarly. Otherwise, by symmetry, we may assume that x 4 x 1 ∈ E(G). Now, by Claim 12, Proof. Suppose x 2 x 5 ∈ E(G), and then, x, x 2 is fragments of G. Furthermore, we see that G[N(x)] is a connected graph, and this implies that for any is connected, and we see that for every vertex t of G, G[N(t)] is connected.
has at least two components, a contradiction. erefore, |N( On the other hand, by Claim 12, is fact implies that either By symmetry, we may assume has only one vertex of degree 3. On the other hand, we find that G[N(x)] has two vertex of degree 3, and this implies that , a contradiction. is contradiction shows that x 2 x 5 ∉ E(G). By symmetry, x 3 x 5 ∉ E(G). Hence, Subclaim 4 holds.
Proof. Suppose x 4 x 5 ∉ E(G). Let P ′ be a graph which is got from the path x 3 x 2 x 1 x 4 x 5 by adding the edge x 1 x 3 . us, Claim 13 holds. By Claim 13 and Lemma 3, x 4 x 5 ∈ E(G), and hence, x has type 1. erefore, G has type 1. □ Theorem 7. Let G be a TCC-5-connected graph. If |V(G)| ≥ 10, then G has type 1, type 2, type 3, or type 4.
Proof. If G contains K 4 , then Lemma 10 assures us that G has type 1. So, we may assume that G does not contain K 4 . Hence, Lemma 8 assures us that for any x ∈ V(G), Δ(G[N(x)]) ≤ 2. Now, Lemma 3 assures us that G has either type 2 or type 3 or type 4. □ Theorem 8. Let G be a TCC-5-connected graph. If G has type 2, then G is isomorphic to icosahedron.

Proof of Theorem 5
Since G is 5-regular, we see that |V(G)| is even. If G has a contractible edge, then we are done. erefore, in the rest of the paper, we may assume that G is a contraction-critical 5connected graph. Hence, by eorem 1, we can assume that |V(G)| ≥ 10. By eorem 2, we see that G has type 1, type 2, type 3, or type 4. Next, we complete the proof of eorem 5 by showing that the following lemmas are true.  Proof. By the definition of type 1, we know that G contains K 4 as a subgraph. Since G is vertex transitive graph, every vertex of G is contained in some K 4 . Let x be a vertex of G, and let N(x) � x 1 , . . . , x 5 . Furthermore, without the loss of generality, suppose G[ x, x 1 , x 2 , x 3 ] � K 4 .
Furthermore, we can observe that