Degree sum of 3 independent vertices and Z3 -connectivity

Discrete Mathematics 313 (2013) 2493–2505

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Degree sum of 3 independent vertices and Z3 -connectivity Fan Yang a,b , Xiangwen Li a,∗ a

Department of Mathematics, Huazhong Normal University, Wuhan 430079, China

b

Department of Mathematics and Physics, Jiangsu University of Science and Technology, Jiangsu 212003, China

article

abstract

info

Let G be a 2-edge-connected simple graph on n vertices and α(G) be the independent number of G. Denote by G5 the graph obtained from a K4 by adding a new vertex and two edges joining this new vertex to two distinct vertices of the K4 . It is proved in this paper that if when α(G) ≥ 3, d(x) + d(y) + d(z ) ≥ 3n/2 for every 3-independent set {x, y, z } of G and when α(G) ≤ 2, d(x) + d(y) ≥ n for every 2-independent set {x, y} of G, then G is not Z3 -connected if and only if G is one of the 12 specified graphs or G can be Z3 -contracted to one of the graphs {K3 , K4− , K4 , G5 }, which generalize the results of Luo et al. [R. Luo, R. Xu, J. Yin, G. Yu, Ore-condition and Z3 -connectivity, European J. Combin. 29 (2008) 1587–1595]. © 2013 Elsevier B.V. All rights reserved.

Article history: Received 22 October 2012 Received in revised form 12 July 2013 Accepted 13 July 2013 Available online 8 August 2013 Keywords: Nowhere-zero 3-flow Z3 -connectivity Independent number

1. Introduction Graphs considered here are finite and may have multiple edges without loops. We follow the notations and terminology in [1] except otherwise stated. For a graph G and k ≥ 2, define

σk (G) = min{d(v1 ) + d(v2 ) + · · · + d(vk ) : {v1 , v2 , . . . , vk } is an independent set of G}. For simplicity, we write σk for σk (G). The theory of k-flows was introduced by Tutte in [15] as a generalization of face k-coloring of planar graphs. A graph admits a nowhere-zero k-flow if its edges can be oriented and assigned numbers ±(k − 1), ±(k − 2), . . . , ±1 so that for every vertex, the sum of the values on incoming edges equals the sum of the outgoing edges. The concept of group connectivity was first introduced by Jaeger et al. in [7] as a generalization of nowhere-zero flow problems. Let D be an orientation of a graph G. If an edge e = uv is directed from a vertex u to a vertex v , then u is a tail of e, v is a head of e. For a vertex v ∈ V (G), let E + (v)(E − (v)) denote the set of all edges at v as a tail (a head). Let A be an abelian group with the additive identity 0, and let A∗ = A − {0}. For every mapping f : E (G) → A, the boundary of f is a function ∂ f : V (G) → A defined by, for each v ∈ V (G),

∂ f (v) =

 e∈E + (v)

f (e) −



f (e),

e∈E − (v)

where ‘‘ ’’ refers to the addition in A. If ∂ f (v) = 0 for each vertex v ∈ V (G), then f is called an A-flow of G. Moreover, if f (e) ̸= 0 for every e ∈ E (G), then f is a nowhere-zero A-flow of G.  A graph G is A-connected if for any mapping b : V (G) → A with v∈V (G) b(v) = 0, there exists an orientation of G and a mapping f : E (G) → A∗ such that ∂ f (v) = b(v) for each v ∈ V (G). Obviously, if G is A-connected, then G admits a nowhere-zero A-flow. The following two conjectures are long standing problems in this area.



∗

Corresponding author. E-mail addresses: [email protected], [email protected] (X. Li).

0012-365X/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.disc.2013.07.009

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Fig. 1. 12 specified graphs.

Conjecture 1.1 (Tutte [15]). Every 4-edge-connected graph admits a nowhere-zero 3-flow. Conjecture 1.2 (Jaeger et al. [7]). Every 5-edge-connected graph is Z3 -connected. Many authors are devoted themselves to work on these two conjectures. Recently, degree conditions have been used to guarantee the existence of nowhere-zero 3-flows and Z3 -connectivity of graphs. For the literature, some results can be seen in [5,6,10–13,16–19], a survey in [9] and others. A major progress in this area was made by Thomassen [14] who proved that every 8-edge-connected graph is Z3 -connected. However, Conjectures 1.1 and 1.2 are still open. Theorem 1.3 (Luo et al. [12]). Let G be a simple graph on n vertices. If σ2 ≥ n, then G is not Z3 -connected if and only if G is one of the special 12 graphs in Fig. 1. For X ⊆ E (G), the contraction G/X is obtained from G by contracting each edge of X and deleting the resulting loops. If H ⊆ G, we write G/H for G/E (H ). Let A be an abelian group with |A| ≥ 3. Denote by G′ the graph obtained by repeatedly contracting A-connected subgraphs of G until no such subgraph is left. We say that G can be A-contracted to G′ . Clearly, if a graph G can be contracted to K1 , then G is A-connected. Denote by α(G) the independent number of a graph G. For a simple graph G on n vertices, we define a family F of graphs as follows: G ∈ F if and only if G satisfies that when α(G) ≥ 3, σ3 ≥ 3n/2 and when α(G) ≤ 2, σ2 ≥ n. In this paper, we extend Theorem 1.3 by replacing σ2 ≥ n with σ3 ≥ 3n/2. The main theorem in this paper is presented as follows. Theorem 1.4. Let G ∈ F be a 2-edge-connected graph. The graph G is not Z3 -connected if and only if G is one of the 12 specified graphs shown in Fig. 1 or G can be Z3 -contracted to one of the graphs {K3 , K4 , K4− , G5 }. Proposition 1.5. Let G be a simple graph on n vertices. If σ2 ≥ n, then G cannot be Z3 -contracted to one of the graphs {K3 , K4 , K4− , G5 }. Proof. We prove here that G cannot be Z3 -contracted to K3 . The proofs for other cases are similar. Suppose otherwise that G can be Z3 -contracted to a K3 . Let V (K3 ) = {v1 , v2 , v3 }. Assume first that V (K3 ) contains precisely one vertex v1 into which a Z3 -connected subgraph H of G is contracted. Obviously, |V (H )| ≥ 5. This means that H contains a vertex v such that vv2 , vv3 ̸∈ E (G). On the other hand, d(v) ≤ n − 3. This implies that d(v) + d(v2 ) ≤ n − 3 + 2 = n − 1, contrary to the assumption that σ2 ≥ n. Thus, assume that V (K3 ) contains at least two vertices v1 , v2 such that two Z3 -connected subgraphs H1 , H2 of G can be contracted to v1 , v2 , respectively. Similarly, |V (Hi )| ≥ 5 for i = 1, 2. It follows that Hi contains a vertex vi′ which has all neighbors in Hi for i = 1, 2. Thus, d(vi′ ) ≤ |V (Hi )| − 1 for i = 1, 2. This implies that d(v1′ ) + d(v2′ ) ≤ |V (H1 )| − 1 + |V (H2 )| − 1 ≤ n − 3, a contradiction.  Remark. 1. Theorem 1.4 implies Theorem 1.3. Let G be a graph on n vertices. If α(G) = 2, then for any pair of nonadjacent vertices u and v , d(u) + d(v) ≥ n by each of Theorems 1.3 and 1.4. Thus, assume that α(G) ≥ 3. Let {u, v, w} be a set of 3 independent vertices of G. Since uv ̸∈ E (G), by Theorem 1.3, d(u) + d(v) ≥ n. We assume, without loss of generality, that d(u) ≥ d(v). It follows that d(u) ≥ 2n . Since vw ̸∈ E (G), Theorem 1.3 shows that d(v) + d(w) ≥ n. This implies that

d(u)+ d(v)+ d(w) ≥ 3n . By Proposition 1.5, G cannot be Z3 -contracted to one of the graphs {K3 , K4 , K4− , G5 }. By Theorem 1.4, 2 G is not Z3 -connected if and only if G is not one of the 12 specified graphs in Fig. 1. 2. Theorem 1.4 is stronger than Theorem 1.3 in the sense that there are a lot of graphs whose Z3 -connectedness is warranted by Theorem 1.4 but cannot be ensured by Theorem 1.3. As examples, for n ≥ 5, denote by Gn the graph obtained from a complete graph Kn with vertex set {u1 , u2 , . . . , un } by deleting one edge u1 u2 and adding a new vertex v and two edges v ui for i = 3, 4. Theorem 1.4 guarantees that Gn is Z3 -connected while Theorem 1.3 becomes invalid for Gn . The next theorem can be obtained immediately from Theorem 1.4.

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Theorem 1.6. Let G ∈ F be a 2-edge-connected graph. The graph G does not admit a nowhere-zero 3-flow if and only if one of the following holds: (i) G is one of the graphs {G4 , G6 , G7 , G9 , G10 , G12 } shown in Fig. 1; (ii) G can be Z3 -contracted to graph K4 . Proof. It is easy to verify that each graph of {G1 , G2 , G3 , G5 , G8 , G11 } admits a nowhere-zero 3-flow. If G can be contracted to one of the graphs {K1 , K3 , K4− , G5 }, by Lemma 2.1(7), G admits a nowhere-zero 3-flow. By Theorem 1.4, G does not admit a nowhere-zero 3-flow if and only if either G is one of the graphs {G4 , G6 , G7 , G9 , G10 , G12 } shown in Fig. 1 or G can be Z3 -contracted to graph K4 .  We end this section with some terminology and notation not defined in [1]. For V1 , V2 ⊆ V (G) and V1 ∩ V2 = ∅, denote by e(V1 , V2 ) the number of edges with one endpoint in V1 and the other endpoint in V2 . Let NG (v) denote the set of all vertices adjacent to vertex v ; set NG [v] = NG (v) ∪ {v}. We usually use N (v) and N [v] for NG (v) and NG [v] if there is no confusion. A k-vertex is a vertex of degree k. A k-cycle is a cycle of length k; a 3-cycle is also called a triangle. A wheel Wk is the graph obtained from a k-cycle by adding a new vertex and joining it to every vertex of the k-cycle. A wheel Wk is odd (even) if k is odd (even). For convenience, we define W1 as a triangle. An independent set S of G is a k-independent set of G if |S | = k. We organize this paper as follows. In Section 2, we state some known results and establish several lemmas. In Sections 3 and 4, we deal with the case for small graphs and the case for the minimum degree 3, respectively. In Section 5, we discuss the splitting operation and we will prove Theorem 1.4 in Section 6. 2. Preliminaries Some results in [2–4,7,8] on group connectivity are summarized as follows. Lemma 2.1. Let G be a graph and A be an abelian group with |A| ≥ 3. The following results are known. (1) (2) (3) (4) (5) (6) (7)

K1 is A-connected. Kn and Kn− are A-connected if n ≥ 5. An n-cycle is A-connected if and only if |A| ≥ n + 1. Km,n is A-connected if m ≥ n ≥ 4; neither K2,t (t ≥ 2) nor K3,s (s ≥ 3) is Z3 -connected. Each even wheel is Z3 -connected and each odd wheel is not. If H is a subgraph of G and H is A-connected, then G is A-connected if and only if G/H is A-connected. Let v is not a vertex of G. If G is A-connected and e(v, G) ≥ 2, then the graph induced by V (G) ∪ {v} is A-connected.

Let G be a graph and u, v, w be three vertices of G with uv, uw ∈ E (G), and dG (u) ≥ 4. Let G[uv,uw] denote the graph obtained from G by deleting two edges uv and uw , and then adding edge wv , that is, G[uv,uw] = G ∪ {wv} − {uv, uw}. Lemma 2.2 ([2,8]). Let A be an abelian group with |A| ≥ 3. If G[uv,uw] is A-connected, then so is G. Lemma 2.3. Suppose that n ≥ 8 and G ∈ F is a 2-edge-connected graph. Then either G is Z3 -connected or G contains a K4− . Proof. Assume that G contains no K4− . Let u be a vertex of maximum degree in G, R = G − N [u]. It follows from the given degree condition that d(u) ≥ 2n and |V (R)| ≤ n/2 − 1. The proof here assumes that G has 3 independent vertices. Therefore, the discussion when α(G) = 2 is missing. This can be easily fixed. If there exists one vertex, say v , in N (u) such that dG[N (u)] (v)=2, then let v1 and v2 be such two vertices in G[N (u)] which are adjacent to v , respectively. Then G contains a K4− induced by {u, v, v1 , v2 }, a contradiction. Thus, we assume that dG[N (u)] (v) ≤ 1 for each vertex v ∈ N (u). Assume that G[N (u)] contains no edge. Then d(v) ≤ n/2 for each v ∈ N (u). In this case, α(G) ≥ 3. Since σ3 ≥ 3n/2, d(v) = n/2 for each v ∈ N (u). It follows that v is adjacent to all vertices of R for each v ∈ N (u) and |V (R)| = n/2 − 1, ∆(G) = n/2. If R contains one edge, then G contains a K4− , a contradiction. If R contains no edge, then G is a bipartite graph with one partition N (u) and the other V (R) ∪ {u}, where |N (u)| = |V (R) ∪ {u}| = n/2, where n ≥ 8. By Lemma 2.1(4), G is Z3 -connected. In the rest of our proof, we need a claim. Claim 1. If G[N (u)] contains an edge u1 u2 , then d(u1 ) + d(u2 ) ≤ n/2 + 3 < n. Proof of Claim 1. If there exists one vertex, say v , in R such that e(v, {u1 , u2 }) ≥ 2, then G contains a K4− induced by {u, u1 , u2 , v}, a contradiction. Thus, assume that e(v, {u1 , u2 }) ≤ 1 for each vertex v ∈ R. This implies that d(u1 ) + d(u2 ) ≤ |V (R)| + 4 ≤ n/2 + 3.  Now assume that G[N (u)] contains only one edge u1 u2 . Let u3 , u4 ∈ N (u) \ {u1 , u2 } since n ≥ 8. On the other hand, {u1 , u3 , u4 } and {u2 , u3 , u4 } are two 3-independent sets of G and hence d(u1 ) + d(u2 ) + 2d(u3 ) + 2d(u4 ) ≥ 3n. Since d(uj ) ≤ n/2 for j = 3, 4, d(u1 ) + d(u2 ) ≥ n, contrary to Claim 1. Next assume that G[N (u)] contains only two independent edges, say u1 u2 , u3 u4 . If d(u) ≥ 5, then there exists one vertex u5 ∈ N (u) \ {u1 , u2 , u3 , u4 }. By Claim 1, d(u1 ) + d(u2 ) < n and d(u3 ) + d(u4 ) < n. Note that {u1 , u3 , u5 } and {u2 , u4 , u5 }

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are two 3-independent sets of G. Since d(u5 ) ≤ n/2, 2 × 3n ≤ d(u1 ) + d(u2 ) + d(u3 ) + d(u4 ) + 2d(u5 ) < n + n + n ≤ 3n, 2 a contradiction. Thus, d(u) = 4. In this case, n = 8 and |V (R)| = 3. Set V (R) = {z1 , z2 , z3 }. If there exist two nonadjacent vertices in R, say z1 and z2 , then d(z1 ) + d(z2 ) ≥ 12 − 4 = 8 by the given degree condition. It follows that there exists one vertex in {z1 , z2 } with degree at least 4, which implies that G contains a K4− , a contradiction. Now we consider the case R ∼ = K3 . Since G does not contain a K4− , e(v, R) ≤ 1 for each v ∈ N (u). It follows that e(R, N (u)) ≤ 4, e(zi , N (u)) ≤ 1 for some zi ∈ R and zi u1 , zi u2 ̸∈ E (G). Thus, {u1 , uj , zi } for some j ∈ {3, 4} is a 3-independent set of G and d(u1 )+ d(uj )+ d(zi ) ≤ 3 + 3 + 3 = 9, contrary to the given degree condition that σ3 ≥ 12. Finally, assume that G[N (u)] contains at least three independent edges u1 u2 , u3 u4 , u5 u6 . By Claim 1, d(u1 ) + d(u2 ) < n, d(u3 )+ d(u4 ) < n, and d(u5 )+ d(u6 ) < n. It follows that d(u1 )+ d(u2 )+ d(u3 )+ d(u4 )+ d(u5 )+ d(u6 ) < 3n. On the other hand, note that {u1 , u3 , u5 } and {u2 , u4 , u6 } are two 3-independent sets of G and d(u1 )+ d(u2 )+ d(u3 )+ d(u4 )+ d(u5 )+ d(u6 ) ≥ 3n by the given degree condition. This contradiction proves our lemma.  3. Cases when n is small Lemma 3.1. Let G ∈ F be a simple 2-edge-connected graph on n vertices, where n ≤ 7. Then G is either Z3 -connected or one of the 12 graphs in Fig. 1. Proof. If n ≤ 4, then G is one of G1 , G2 , G3 and G4 . If G is a complete graph Kn (n ≥ 5), then G is Z3 -connected by Lemma 2.1(2). Thus, assume that 5 ≤ n ≤ 7 and G ∼ ̸ Kn . = Case 1. δ(G) ≥ 4. In this case, 6 ≤ n ≤ 7. Since δ(G) ≥ 4, G is not one of G1 , . . . , G12 in Fig. 1. Note that for each pair of nonadjacent vertices u and v , d(u) + d(v) ≥ 4 + 4 = 8. This implies that σ2 (G) ≥ n. Thus, G is Z3 -connected by Theorem 1.3. Next we assume that δ(G) = k ≤ 3. Let x be the vertex with dG (x) = δ(G). In the rest of our proof in this lemma, define N (x) = {x1 , . . . , xk } and R = V (G) − N [x] = {z1 , z2 , . . . , zl }. Case 2. δ(G) = 2. We claim that G[R] is a complete graph. Suppose otherwise that z1 z2 ̸∈ E (G[R]). Then {x, z1 , z2 } is a 3-independent set of G. Then d(zi ) ≤ n − 3 for each i ∈ {1, 2}. In this case, d(x) + d(z1 ) + d(z2 ) ≤ 2 + 2n − 6 = 2n − 4. Since σ3 ≥ 3n/2, 2n − 4 ≥ 3n/2, which implies that n ≥ 8, contrary to that n ≤ 7. So far, we have proved that G[R] is a complete graph. Now we claim that α(G) ≤ 2 or G is Z3 -connected. If n = 5, then G[R] is a K2 , that is, z1 z2 . Since G ∈ F , G contains a 5-cycle. Thus, α(G) ≤ 2. If n = 6, then G[R] is K3 . If α(G) ≥ 3, then the 3-independent set of G must be {x1 , x2 , zi } for some i ∈ {1, 2, 3}. Without loss of generality, we assume that i = 1. Then d(xi ) ≤ 3 for each i ∈ {1, 2} and d(z1 ) = 2. Thus d(x1 ) + d(x2 ) + d(z1 ) ≤ 8, contrary to σ3 ≥ 3n/2. Thus α(G) ≤ 2. If n = 7, then G[R] is K4 . If α(G) ≥ 3, then the 3-independent set of G must be {x1 , x2 , zi } for some i ∈ {1, 2, 3, 4}. Without loss of generality, we assume that i = 1. Then d(xi ) ≤ 4 for each i ∈ {1, 2} and d(z1 ) = 3. Thus d(x1 ) + d(x2 ) + d(z1 ) ≤ 11. Since σ3 ≥ 11 by the given degree condition, d(xi ) = 4 for each i ∈ {1, 2} and d(z1 ) = 3. In this case G contains a K5− induced by R ∪ {x1 }, and by Lemma 2.1(2), (7), G is Z3 -connected. Thus, α(G) ≤ 2. We now assume that α(G) ≤ 2. Since G ∈ F , G satisfies σ (G) ≥ n. By Theorem 1.3, G is either Z3 -connected or one of {G5 , G6 } if n = 5, and G is Z3 -connected if n = 6, 7. Case 3. δ(G) = 3. If n = 5, then R = {z }. Since δ(G) = 3, zxi ∈ E (G) for each i ∈ {1, 2, 3} and G[N (x)] contains at least two edges. Thus, G contains an even wheel W4 . Contracting this W4 and repeatedly contracting 2-cycles generated in the process, we finally get a K1 . By Lemma 2.1(1), (5), (6), G is Z3 -connected. If n = 6, then R = {z1 , z2 }. If z1 z2 ̸∈ E (G), then zi is adjacent to all the neighbors of x for each i ∈ {1, 2}. When G[N (x)] contains no edge, G is K3,3 , that is, G ∼ = G8 ; when G[N (x)] contains one edge, G is K3+,3 , that is, G ∼ = G9 ; when G[N (x)] contains at least two edges, say x1 x2 , x2 x3 , then G contains an even wheel W4 induced by N [x] ∪ {z1 } with its center at x2 and e(z2 , W4 ) ≥ 2. Thus, G is Z3 -connected by Lemma 2.1(5), (7). Thus, assume that z1 z2 ∈ E (G). If G has a 3-independent set T , then R ∩ T = ∅ since δ(G) = 3. Thus, T must be {x1 , x2 , x3 }. In this case, G is K3+,3 , that is, G9 . Thus, α(G) ≤ 2. By Theorem 1.3, either G is Z3 -connected or G is one of the graphs {G7 , G8 , G9 , G10 , G11 , G12 }. If n = 7, then R = {z1 , z2 , z3 }. By the given degree condition, G[R] contains at least two edges. If G[R] contains two edges, say z1 z2 and z2 z3 , then d(z1 ) ≤ 4, d(z3 ) ≤ 4. Note that {x, z1 , z3 } is a 3-independent set of G and d(x) + d(z1 ) + d(z3 ) ≤ 3 + 4 + 4 = 11. By the given degree condition, d(z1 ) = d(z3 ) = 4. Then zi xj ∈ E (G) for each i ∈ {1, 3}, j ∈ {1, 2, 3}. Since δ(G) = 3, e(z2 , N (x)) ≥ 1. We assume, without loss of generality, that x1 z2 ∈ E (G). Since σ3 ≥ 11, G[N (x)] contains at least one edge, say xi xj for some i, j ∈ {1, 2, 3} or e(z2 , N (x)) ≥ 2. In the former case, by repeatedly contracting 2-cycles, G[z1 xi ,z1 xj ] can be Z3 -contracted to K1 . In the latter case, set xi z2 ∈ E (G) (i ∈ {1, 2}), by repeatedly contracting 2-cycles, G[x1 z1 ,x1 z2 ] can be Z3 -contracted to K1 . In each case, by Lemma 2.2, G is Z3 -connected. Thus, we assume that G[R] contains three edges. If T is a 3-independent set of G, then T = {x1 , x2 , x3 } or {xi , xj , zk } for some i, j, k ∈ {1, 2, 3} and i ̸= j. In the former case, we may assume that d(x1 ) = d(x2 ) = 4 and 3 ≤ d(x3 ) ≤ 4 by the given degree condition and by symmetry. This means that e(x1 , R) = 3, e(x2 , R) = 3 and 2 ≤ e(x3 , R) ≤ 3. The subgraph induced by {x1 , x2 , z1 , z2 , z3 } is a K5− , which is Z3 -connected. Contracting this K5− and repeatedly contracting 2-cycles generated in the process, we get a K1 . In the latter case, we may assume, without loss of generality, that T = {x1 , x3 , z2 }. By the given

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degree condition, d(z2 ) = 3 and x1 x2 , x1 z1 , x1 z3 , x3 x2 , x3 z1 , x3 z3 ∈ E (G). In this case, G∗ = G[x3 z3 ,x3 x2 ] contains an even wheel W4 induced by {x1 , x2 , z1 , z2 , z3 } with the center at z3 . Contracting this wheel and all 2-cycles generated in the process, the resulting graph is a K1 . In each case, by Lemmas 2.1 and 2.2, G is Z3 -connected. Thus α(G) ≤ 2, and G is Z3 -connected by Theorem 1.3.  4. The case when δ(G ) = 3 Lemma 4.1. Suppose that G ∈ F is a 2-edge-connected graph on n ≥ 8 vertices and that G has two nonadjacent 3-vertices and δ(G) ≥ 3. Then G is Z3 -connected. Proof. Suppose that x and y are two 3-vertices in G. We claim that for each v ∈ V (G) − {x, y}, v x ∈ E (G) or v y ∈ E (G). Suppose otherwise that there exists a vertex z such that {x, y, z } is a 3-independent set of G. By the given degree condition, d(z ) ≥ 3n/2 − 6. However, d(z ) ≤ n − 3. Thus, n − 3 ≥ 3n/2 − 6, which implies that n ≤ 6, contrary to n ≥ 8. From the claim and the fact that xy ̸∈ E (G), n = 8. Thus, we may assume that V (G) = {x, y, x1 , x2 , x3 , y1 , y2 , y3 }, where N (x) = {x1 , x2 , x3 } and N (y) = {y1 , y2 , y3 }. We claim that α(G) ≥ 3. Suppose otherwise that α(G) = 2. Since xy ̸∈ E (G), d(x) + d(y) = 6 < σ2 (G) = 8, a contradiction. Thus, let S be a 3-independent set of G. We claim that |S ∩{x, y}| = 1. By the claim above, |S ∩{x, y}| ≤ 1. Suppose that |S ∩{x, y}| = 0. If S ⊆ N (x), then {x1 , x2 , y} is an independent set and d(x1 ) + d(x2 ) + d(y) ≤ 11, contrary to the given degree condition. Thus, |S ∩ N (x)| ≤ 2. By symmetry, |S ∩ N (y)| ≤ 2. Thus, we assume, without loss of generality, that S = {x1 , x2 , y3 }. It follows that d(x1 ) + d(x2 ) + d(y3 ) ≤ 11, contrary to the given degree condition. We have proved that |S ∩ {x, y}| = 1. By the claims above and symmetry, assume that S = {x1 , x2 , y} is a 3-independent set of G. Since d(xi ) ≤ 5 for each i ∈ {1, 2}, d(x1 ) = 5, d(x2 ) ≥ 4 by the given degree condition. It follows that x1 x3 , x1 yi ∈ E (G) for each i ∈ {1, 2, 3}. If x2 x3 ̸∈ E (G), then {x2 , x3 , y} is a 3-independent set. By the given degree condition, d(x3 ) = 5 and d(x2 ) = 4. It follows that x2 yi , x3 yj ∈ E (G) for each i ∈ {1, 2, 3} and each j ∈ {1, 2, 3}. Note that d(y1 ) ≥ 4. By repeatedly contracting 2-cycles, G[y1 x1 ,y1 x3 ] can be Z3 -contracted to a K1 which is Z3 -connected. By Lemmas 2.1 and 2.2, G is Z3 -connected. Thus, assume that x2 x3 ∈ E (G). If G[N (y)] is not a complete graph, then by symmetry, let y1 y2 ̸∈ E (G). Similarly, assume that d(y1 ) = 5, d(y2 ) ≥ 4, y1 y3 , y1 xj ∈ E (G) for each j ∈ {1, 2, 3}, y2 y3 ∈ E (G) and e({x2 , x3 }, {y2 , y3 }) ≥ 1 since d(x2 ) ≥ 4 and d(y2 ) ≥ 4. By repeatedly contracting 2-cycles, G[y1 x1 ,y1 x3 ] can be Z3 -contracted to a K1 . By Lemmas 2.1(1), (6) and 2.2, G is Z3 -connected. If G[N (y)] is a complete graph, then G[N [y] ∪ {x1 }] is K5− . By the given degree condition, e(x2 , {y1 , y2 , y3 }) ≥ 2 and e(x3 , {y1 , y2 , y3 }) ≥ 1. Since K5− is Z3 -connected, by Lemma 2.1(7), G[N (y) ∪ {x1 , x2 }] is Z3 -connected. Contracting this Z3 -connected subgraph and contracting all 2-cycles generated in the process, we finally get a K1 . By Lemma 2.1(1), (6), G is Z3 -connected.  Suppose that x and y are two 3-vertices in G such that xy ∈ E (G). Define R = N (x)∪ N (y)−{x, y}, S = V (G)− N (x)∪ N (y). Let |R| = a. Since x and y are 3-vertices, a ∈ {2, 3, 4}. Lemma 4.2. Suppose that α(G[S ]) ≥ 2. If a = 2, 3, then G[S ] is Z3 -connected; if a = 4, then G[S ] is Z3 -connected or one of the graphs {G2 , G3 }. Proof. Suppose that z1 and z2 are a pair of nonadjacent vertices in S. It follows that {x, z1 , z2 } is a 3-independent set of G. By the given degree condition, d(z1 ) + d(z2 ) ≥ 3n/2 − 3. Thus dG[S ] (z1 ) + dG[S ] (z2 ) ≥ 3n/2 − 3 − 2a. If a = 2, 3, then dG[S ] (z1 ) + dG[S ] (z2 ) ≥ 3n/2 − 3 − 2a ≥ n − 2 − a = |S |. Thus σ2 (G[S ]) ≥ |S |. By Theorem 1.3, G[S ] is either Z3 -connected or one of the graphs {G2 , G3 , G5 , G6 , G7 , G8 , G9 , G10 , G11 , G12 }. In the former case, we are done. In the latter case, if G[S ] ∼ = G2 or G3 , then n = a + 6; if G[S ] ∼ = G5 or G6 , then n = a + 7; if G[S ] is one of the graphs {G7 , G8 , G9 , G10 , G11 , G12 }, then n = a + 8. Recall that z1 and z2 are two nonadjacent vertices of G[S ] and {x, z1 , z2 } is a 3-independent set of G. If n = a + 6, then d(zi ) ≤ a + 2 for i ∈ {1, 2}. Thus, d(x) + d(z1 ) + d(z2 ) ≤ 2a + 7, contrary to that σ3 ≥ 3(a + 6)/2. If n = a + 7, then d(z1 ) ≤ a + 2, d(z2 ) ≤ a + 3. Thus, d(x) + d(z1 ) + d(z2 ) ≤ 2a + 8, contrary to that σ3 ≥ 3(a + 7)/2. If n = a + 8, then d(zi ) ≤ a + 3 for i ∈ {1, 2}. It follows that d(x) + d(z1 ) + d(z2 ) ≤ 2a + 9, contrary to that σ3 ≥ 3(a + 8)/2. Thus, assume that a = 4. In this case, dG[S ] (z1 ) + dG[S ] (z2 ) ≥ 3n/2 − 3 − 2a = 3n/2 − 11. If n = 8, let S = {z1 , z2 }. Since σ3 (G) ≥ 3n/2, d(z1 ) + d(z2 ) ≥ 12 − 3 = 9. On the other hand, d(zi ) ≤ 4 for each i ∈ {1, 2} since z1 z2 ̸∈ E (G), a contradiction. If n ≥ 9, then dG[S ] (z1 ) + dG[S ] (z2 ) ≥ n − 6 = |S |, which implies that σ2 (G[S ]) ≥ |S |. By Theorem 1.3, G[S ] is either Z3 -connected or one of the graphs {G2 , G3 , G5 , G6 , G7 , G8 , G9 , G10 , G11 , G12 }. By the argument above, we can prove that G[S ] is not one of the graphs {G5 , G6 , G7 , G8 , G9 , G10 , G11 , G12 }.  Lemma 4.3. (i) If a = 3 and N (x) = {y, x1 , u}, N (y) = {x, y1 , u}, then for each vertex v ∈ {x1 , y1 }, e(v, S ) ≥ n/2 − 2. (ii) If a = 4 and N (x) = {y, x1 , x2 }, N (y) = {x, y1 , y2 }, then for each vertex v ∈ {x1 , x2 , y1 , y2 }, e(v, S ) ≥ n/2 − 3. Proof. (i) By symmetry, let v = x1 . If there exists one vertex z ∈ S such that z v ̸∈ E (G), then {v, y, z } is a 3-independent set . Since d(z ) ≤ n − 4 and n ≥ 8, d(v) ≥ n/2 + 1 and so e(v, S ) ≥ n/2 + 1 − 3 = n/2 − 2. of G. Thus d(v) + d(y) + d(z ) ≥ 3n 2 Thus, assume that for each vertex z ∈ S, z v ∈ E (G). In this case, e(v, S ) = n − 5 ≥ 3. Since n ≥ 8, e(v, S ) ≥ n/2 − 2. (ii) The proof is similar. 

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Lemma 4.4. Suppose that G ∈ F is a 2-edge-connected graph on n ≥ 8 vertices and that G has at least two 3-vertices and δ(G) ≥ 3. Then G is not Z3 -connected if and only if G can be Z3 -contracted to one of K4 and G5 . Proof. Suppose that x and y are two 3-vertices in G. By Lemma 4.1, we are left to prove the case xy ∈ E (G). R, S and a have been defined above. We consider the following three cases. Case 1. |N (x) ∩ N (y)| = 2. In this case, a = 2 and define N (x) = {y, x1 , y1 } and N (y) = {x, x1 , y1 }. Then |S | = n − 4 ≥ 4 for n ≥ 8. If α(G[S ]) ≥ 2, then by Lemma 4.2, G[S ] is Z3 -connected; if α(G[S ]) = 1, then G[S ] ∼ = Kn−4 . This implies that G[S ] is Z3 -connected or G[S ] ∼ = K4 by Lemma 2.1(2). Thus, G[S ] is Z3 -connected or G[S ] ∼ = K4 . Assume that G[S ] is Z3 -connected. If e(x1 , S ) ≥ 2 and e(y1 , S ) ≥ 2, then G is Z3 -connected by Lemma 2.1(7). Thus, assume that e(x1 , S ) ≤ 1 or e(y1 , S ) ≤ 1. Suppose first that x1 y1 ∈ E (G). Since G is 2-edge-connected, e({x1 , y1 }, S ) ≥ 2. If either e(x1 , S ) ≥ 1 and e(y1 , S ) ≥ 2 or e(x1 , S ) ≥ 2 and e(y1 , S ) ≥ 1, then G is Z3 -connected by Lemma 2.1(7); if e(x1 , S ) = 1 and e(y1 , S ) = 1, G′ ∼ = G5 ; if either e(y1 , S ) ≥ 2 and e(x1 , S ) = 0 or e(x1 , S ) ≥ 2 and e(y1 , S ) = 0, then G′ ∼ = K4 . Thus x1 y1 ̸∈ E (G). By symmetry, assume that e(x1 , S ) ≤ 1. Since δ(G) ≥ 3, e(x1 , S ) = 1. If there exists a vertex in S, say z, such that x1 z , y1 z ̸∈ E (G), then by the given degree condition, d(x1 ) + d(y1 ) ≥ 3n/2 − n + 5 = n/2 + 5 since d(z ) ≤ n − 5, which implies that e({x1 , y1 }, S ) = d(x1 ) + d(y1 ) − 2 − 2 ≥ n/2 + 1, that is, e(y1 , S ) ≥ n/2. Otherwise, each vertex of S is adjacent to x1 or y1 . Thus, e(y1 , S ) = n − 5. In either case, G[S ] ∪ {y1 } is Z3 -connected by Lemma 2.1(7). Thus G′ ∼ = K4 . Assume next that G[S ] ∼ = K4 . If α(G) ≤ 2, then G is Z3 -connected by Theorem 1.3. Thus, assume that α(G) ≥ 3. In this case, x1 y1 ̸∈ E (G) and there is a vertex z ∈ S such that T = {x1 , y1 , z } is a 3-independent set with d(z ) = 3. By the given degree condition, d(x1 ) + d(y1 ) ≥ 12 − d(z ) = 9 and d(x1 ), d(y1 ) ≤ 5. We assume, without loss of generality, that d(x1 ) = 5 and d(y1 ) ≥ 4. In this case, G contains a K5− and G is Z3 -connected by Lemma 2.1(2), (7). Case 2. |N (x) ∩ N (y)| = 1. In this case, a = 3, and define N (x) = {y, x1 , u} and N (y) = {x, y1 , u}. Then |S | = n − 5. If α(G[S ]) = 1, then G[S ] ∼ = Kn−5 . Thus either G[S ] is Z3 -connected or G[S ] ∼ = Km , where m = 3, 4. Thus, assume α(G[S ]) ≥ 2. By Lemma 4.2, G[S ] is Z3 -connected. Thus, G[S ] is Z3 -connected or G[S ] ∼ = Km , where m = 3, 4. Suppose first that G[S ] is Z3 -connected. By Lemma 4.3, e(x1 , S ) ≥ n/2 − 2 ≥ 2 and e(y1 , S ) ≥ n/2 − 2 ≥ 2. Since δ(G) ≥ 3, e(u, {x1 , y1 } ∪ S ) ≥ 1. If e(u, {x1 , y1 } ∪ S ) ≥ 2, G is Z3 -connected by Lemma 2.1(7). If e(u, {x1 , y1 } ∪ S ) = 1, then G′ ∼ = K4 . Next, suppose that G[S ] ∼ = K3 . In this case, n = 8 and let S = {z1 , z2 , z3 }. By Lemma 4.3, e(v, S ) ≥ n/2 − 2 ≥ 2 for each v ∈ {x1 , y1 }. Let e(x1 , S ) = e(y1 , S ) = 3. It follows that the subgraph induced by S ∪ {x1 , y1 } is a K5− , which is Z3 -connected by Lemma 2.1. In this case, α(G) = 2. By the given degree condition, e(u, S ∪ {x1 , y1 }) ≥ 2. Contracting the K5− and contracting all 2-cycles generated in the process, we get a K1 . By Lemma 2.1, G is Z3 -connected. Let e(x1 , S ) = 2 and e(y1 , S ) = 3. We assume, without loss of generality, that x1 z1 , x1 z2 ∈ E (G). Since {y, x1 , z3 } is a 3-independent set, the condition that d(y) + d(x1 ) + d(z3 ) ≥ 12 implies that x1 u, x1 y1 , uz3 ∈ E (G). In this case, the subgraph induced by S ∪ {x1 , y1 } is an even wheel W4 with the center at y1 . Contracting this wheel and contracting all 2-cycles generated in the process, we get a K1 . By Lemma 2.1, G is Z3 -connected. Let e(x1 , S ) = 2 and e(y1 , S ) = 2. We assume, without loss of generality, that x1 z1 , x1 z2 , y1 z2 , y1 z3 ∈ E (G). For two 3-independent sets {x, y1 , z1 } and {y, x1 , z3 }, by the given degree condition, x1 y1 , x1 u, y1 u, uz1 , uz3 ∈ E (G). In this case, the subgraph induced by S ∪ {x1 , y1 } is an even wheel W4 with the center at z2 . Contracting this wheel and contracting all 2-cycles generated in the process, we get a K1 . By Lemma 2.1, G is Z3 -connected. Finally, suppose that G[S ] ∼ = K4 . In this case, n = 9 and let S = {z1 , z2 , z3 , z4 }. By Lemma 4.3, e(v, S ) ≥ 3 for each v ∈ {x1 , y1 }. If one of {x1 , y1 }, say x1 , has precisely three neighbors in S, that is, e(x1 , S ) = 3, let x1 z1 , x1 z2 , x1 z3 ∈ E (G). Then {y, x1 , z4 } is a 3-independent set. Applying the given degree condition to this 3-independent set, z4 u ∈ E (G) and e(u, S ∪ {x1 }) ≥ 2. Note that the subgraph induced by S ∪ {x1 } contains a K5− , which is Z3 -connected. Contracting this K5− and contracting all 2-cycles generated in the process, we get a K1 which is Z3 -connected. By Lemma 2.1, G is Z3 -connected. Thus, we assume that each of {x1 , y1 } has four neighbors in S. In this case, by the given degree condition, e(u, S ∪{x1 , y1 }) ≥ 2. The subgraph induced by S ∪ {y1 } contains a K5 , which is Z3 -connected. Contracting this K5− and contracting all 2-cycles generated in the process, we get a K1 which is Z3 -connected. By Lemma 2.1, G is Z3 -connected. Case 3. |N (x) ∩ N (y)| = 0. In this case, a = 4 and we define N (x) = {y, x1 , x2 } and N (y) = {x, y1 , y2 }. Then |S | = n − 6. If n ≥ 9, then by Lemma 4.3, e(v, S ) ≥ 2 for each v ∈ {x1 , x2 , y1 , y2 }. If α(G[S ]) = 1, then G[S ] ∼ = Kn−6 . Since n ≥ 8, either G[S ] is Z3 -connected or G[S ] ∼ = Km , where m = 2, 3, 4. If α(G[S ]) ≥ 2, then by Lemma 4.2, G[S ] is either Z3 -connected or one of the graphs {G2 , G3 }. Suppose that G[S ] is Z3 -connected. By Lemma 2.1, |S | ≥ 5. Thus, n ≥ 11. By Lemma 4.3, for each vertex v ∈ {x1 , x2 , y1 , y2 }, e(v, S ) ≥ 2. By Lemma 2.1(7), G is Z3 -connected. Suppose that G[S ] ∼ = K2 ; let S = {z1 , z2 }. If there exists one vertex, say x1 ∈ R such that z1 x1 ̸∈ E (G), then {x1 , y, z1 } is a 3-independent set of G. Thus d(x1 ) + d(y) + d(z1 ) ≥ 12. Since d(z1 ) ≤ n − 4 = 4, d(x1 ) ≥ 12 − 4 − 3 = 5. Thus d(x1 ) = 5

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and d(z1 ) = 4. That is N (x1 ) = {x, x2 , y1 , y2 , z2 } and N (z1 ) = {z2 , x2 , y1 , y2 }. We can deduce that e(zi , R) ≥ 3 for each i ∈ {1, 2}, and there is a vertex, say x2 , in R such that e(x2 , {z1 , z2 }) = 2. If e(zi , R) = 4, then e(x2 , {z1 , z2 }) = 2. In each case, we can get a K1 from G[x2 z1 ,x2 z2 ] by contracting 2-cycles generated in the process. By Lemma 2.2, G is Z3 -connected. Suppose that G[S ] ∼ = Km , where m = 3, 4. As the proof in the case G[S ] ∼ = K2 , we can prove that there is a vertex in t ∈ R such that d(t ) ≥ 4 and e(t , S ) = 2, say x1 z1 , x1 z2 ∈ E (G). We can get a K1 K1 from G[x1 z1 ,x1 z2 ] by contracting 2-cycles generated in the process. Then G is Z3 -connected by Lemma 2.2. Suppose that G[S ] is one of the graphs {G2 , G3 }. If G[S ] ∼ = G2 , then n = 10 and 3n/2 = 15. Let z1 , z2 be a pair of two nonadjacent 2-vertices of G[S ]. Then {x, z1 , z2 } is a 3-independent set of G. Clearly, d(z1 ) ≤ 6, d(z2 ) ≤ 6. Since d(x)+ d(z1 )+ d(z2 ) ≥ 15, d(z1 ) = d(z2 ) = 6. Similarly, d(z ) = 6 for each z ∈ S. Then G contains an even wheel W4 induced by S ∪{x1 } and e(xi , S ) = 4, e(yi , S ) = 4 for each i ∈ {1, 2}, by Lemma 2.1(5), (7), G is Z3 -connected. If G[S ] ∼ = G3 , then n = 10 and 3n/2 = 15. Let z1 , z2 be a pair of two nonadjacent 2-vertices of G[S ]. By the argument above, d(z1 ) = d(z2 ) = 6. We claim that there exists one vertex, say u in R such that uz3 ∈ E (G) or uz4 ∈ E (G). Suppose otherwise that for each vertex u ∈ R, e(u, {z3 , z4 }) = 0. In this case, {y, x1 , z3 } is a 3-independent set of G. By the given degree condition, d(x1 ) + d(z3 ) ≥ 12. However, d(x1 ) ≤ 6 and d(z3 ) = 3, a contradiction. Thus, we assume that x1 z3 ∈ E (G). We can get a K1 K1 from G[uz1 ,uz3 ] by contracting 2-cycles generated in the process. Then G is Z3 -connected by Lemma 2.2.  5. Splitting operation Throughout this section, we assume that G ∈ F is a 2-edge-connected graph on n ≥ 8 vertices and G contains at most one 3-vertex with δ(G) ≥ 3. Let xyz and xyw be two triangles in G with d(z ) ≥ 4, and H be a maximal Z3 -connected subgraph of G[zx,zy] containing 2-cycles xyx. In the rest of this section, when the maximality of H is mentioned, we mean that H is a maximal Z3 -connected subgraph of G[zx,zy] . Let G∗ = G[zx,zy] /H , R = G − V (H ), u∗ be the vertex into which H is contracted, and |V (G∗ )| = n∗ . Then G∗ is simple and n∗ = n − (|V (H )| − 1) ≤ n − 2. Lemma 5.1. If α(G∗ ) ≥ 3, then σ3 (G∗ ) ≥ 3n∗ /2. Proof. Let {u1 , u2 , u3 } be a 3-independent set of G∗ . If ui ̸∈ {z , u∗ } for each i ∈ {1, 2, 3}, then dG∗ (ui ) = d(ui ) for each i ∈ {1, 2, 3} and {u1 , u2 , u3 } is also a 3-independent set of G. Thus, dG∗ (u1 ) + dG∗ (u2 ) + dG∗ (u3 ) = d(u1 ) + d(u2 ) + d(u3 ) ≥ 3n/2 ≥ 3n∗ /2. If u3 = z and ui ̸∈ {z , u∗ } for each i ∈ {1, 2}, then dG∗ (ui ) = d(ui ) for each i ∈ {1, 2} and dG∗ (u3 ) = d(u3 ) − 2. It follows that dG∗ (u1 ) + dG∗ (u2 ) + dG∗ (u3 ) = d(u1 ) + d(u2 ) + d(u3 ) − 2 ≥ 3n/2 − 2 ≥ 3n∗ /2. If u3 = u∗ and ui ̸= z for each i ∈ {1, 2}, then dG∗ (ui ) = d(ui ) for each i ∈ {1, 2} and dG∗ (u3 ) ≥ d(a) − (|V (H )| − 1) + eG[zx,zy] (H − a, R) for each a ∈ V (H ) − {x, y} (in the rest section, we use e(H − a, R) for eG[zx,zy] (H − a, R) in no confusion). This implies that dG∗ (u1 ) + dG∗ (u2 ) + dG∗ (u3 ) ≥ d(u1 ) + d(u2 ) + d(a) − (|V (H )| − 1) + e(H − a, R) ≥ 3n/2 − (|V (H )| − 1) + e(H − a, R) = n/2 + n∗ + e(H − a, R) ≥ 3n∗ /2. If u2 = z, u3 = u∗ and u1 ̸∈ {z , u∗ }, then dG∗ (u1 ) = d(u1 ), dG∗ (u2 ) = d(u2 )− 2 and dG∗ (u3 ) ≥ d(a)−(|V (H )|− 1)+ e(H − a, R) for each a ∈ V (H ) − {x, y}. Then dG∗ (u1 ) + dG∗ (u2 ) + dG∗ (u3 ) ≥ d(u1 ) + d(u2 ) − 2 + d(a) − (|V (H )| − 1) + e(H − a, R) ≥ 3n/2 − 2 − (|V (H )| − 1) + e(H − a, R) = n/2 + n∗ − 2 + e(H − a, R) = 3n∗ /2 + (|V (H )| − 5)/2 + e(H − a, R). If |V (H )| ≥ 5, then dG∗ (u1 ) + dG∗ (u2 ) + dG∗ (u3 ) ≥ 3n∗ /2. If |V (H )| = 4, then dH (v) ≤ 3 for each v ∈ V (H ) − {x, y}. Since G has at most one vertex with degree 3 and δ(G) ≥ 3, there exists one vertex a ∈ V (H ) − {x, y} such that e(H − a, R) ≥ 1. Then dG∗ (u1 ) + dG∗ (u2 ) + dG∗ (u3 ) ≥ 3n∗ /2 − 1/2 + 1 > 3n∗ /2. If |V (H )| = 3, then dH (v) = 3 for each v ∈ {x, y} and V (H ) = {x, y, w}. Since G has at most one vertex with degree 3 and δ(G) ≥ 3, e(H − w, R) ≥ 1. Thus, dG∗ (u1 ) + dG∗ (u2 ) + dG∗ (u3 ) ≥ 3n∗ /2 − 1 + 1 ≥ 3n∗ /2.  Lemma 5.2. If α(G∗ ) ≤ 2 and α(G) ≤ 2, then either G∗ is Z3 -connected or σ2 (G∗ ) ≥ n∗ /2. Proof. Let {u1 , u2 } be a 2-independent set of G∗ . If ui ̸∈ {z , u∗ } for each i ∈ {1, 2}, then dG∗ (ui ) = d(ui ) for each i ∈ {1, 2} and {u1 , u2 } is a 2-independent set of G. Then dG∗ (u1 ) + dG∗ (u2 ) = d(u1 ) + d(u2 ) ≥ n ≥ n∗ . If u1 = z and u2 ̸∈ {z , u∗ }, then dG∗ (u1 ) = d(u1 ) − 2, dG∗ (u2 ) = d(u2 ) and {u1 , u2 } is a 2-independent set of G. Then dG∗ (u1 ) + dG∗ (u2 ) = d(u1 ) − 2 + d(u2 ) ≥ n − 2 ≥ n∗ . If u1 = u∗ and u2 ̸∈ {z , u∗ }, then dG∗ (u1 ) ≥ d(a) − (|V (H )| − 1) + e(H − a, R) for any a ∈ V (H ) − {x, y}, dG∗ (u2 ) = d(u2 ) and {a, u2 } is a 2-independent set of G. Then dG∗ (u1 ) + dG∗ (u2 ) ≥ d(a) − (|V (H )| − 1) + e(H − a, R) + d(u2 ) ≥ n − (|V (H )| − 1) + e(H − a, R) = n∗ + e(H − a, R) ≥ n∗ . If u1 = u∗ and u2 = z, then dG∗ (u1 ) ≥ d(a) − (|V (H )| − 1) + e(H − a, R) for any a ∈ V (H ) − {x, y}, dG∗ (u2 ) = d(u2 ) − 2 and {a, u2 } is a 2-independent set of G. Then dG∗ (u1 ) + dG∗ (u2 ) ≥ d(a) − (|V (H )| − 1) + e(H − a, R) + d(u2 ) − 2 ≥ n − (|V (H )| − 1) + e(H − a, R) − 2 = n∗ + e(H − a, R) − 2. If there exists a ∈ V (H ) − {x, y} such that e(H − a, R) ≥ 2, then dG∗ (u1 ) + dG∗ (u2 ) ≥ n∗ . Thus, we may assume that e(H − a, R) ≤ 1 for each a ∈ V (H ) − {x, y}. If |V (H )| ≥ 5, then there exists at most one vertex v1 ∈ V (H ) − {x, y} such that e(v1 , R) = 1, pv1 ∈ E (G), p ∈ V (R) and for each vertex v ∈ V (H ) − {v1 }, e(v, R) = 0. It implies that for each vertex v ∈ V (H )−{v1 , x, y}, dG (v) ≤ |V (H )|− 1. By the given degree condition, for each u ∈ V (G)\(V (H )∪{z }), d(u) ≥ n −(|V (H )|− 1). On the other hand, for each u ∈ V (R)\{p, z }, d(u) ≤ |V (R)|− 1 ≤ n −|V (H )|− 1. This contradiction shows that V (R) = {p, z }. Since δ(G) ≥ 3, e(p, z ) ≥ 2, contrary to that G is a simple graph.

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If |V (H )| = 4, then let V (H ) = {x, y, w, v}. Since e(H − a, R) ≤ 1 for each a ∈ V (H ) − {x, y}, H contains at most two vertices w, v such that e(w, R) = e(v, R) = 1. This means that R contains at most two vertices v1 , v2 ∈ V (R) such that e(vi , H ) = 1 for i = 1, 2. In this case, H contains at least one vertex u such that d(u) ≤ 4 and e(u, R) ≤ 1. Since α(G) ≤ 2, σ2 ≥ n. Since n ≥ 8, there is a vertex v ∈ V (R) \ {v1 , v2 , z } such that d(v) ≥ n − 4 and e(v, H ) = 0, contrary to that G is a simple graph. If |V (H )| = 3, then let V (H ) = {x, y, w}. Since G has at most one degree 3-vertex, e({x, y}, R) = 1. Without loss of generality, we assume e(x, R) = 1 and e(y, R) = 0. Since n ≥ 8, |V (R) − {z }| ≥ 4. Since dG (z ) ≥ 4, dG∗ (z ) ≥ 2. Since yv ̸∈ E (G) for v ∈ V (R) − {z }, by the given degree condition, d(v) ≥ n − 3. By the maximality of H, for each vertex v ∈ V (R), e(v, H ) ≤ 1. It implies that R is a complete graph. If |V (R)| ≥ 5, then G∗ is Z3 -connected by Lemma 2.1(2), (7).  Lemmas 5.3–5.7 are preparations for the proof of Lemma 5.8. The following lemma is straightforward. Lemma 5.3. Let G be a graph with α(G) ≤ 2. If for a pair of nonadjacent vertices u1 and u2 of G, d(u1 ) + d(u2 ) ≥ n − 2, then each of the following holds. (i) If |NG (u1 ) ∩ NG (u2 )| = 0, then G[NG [ui ]] is a complete graph for each i ∈ {1, 2}. (ii) If |NG (u1 ) ∩ NG (u2 )| = 1, say {b} = NG (u1 ) ∩ NG (u2 ), then G[NG [ui ] − {b}] is a complete graph for each i ∈ {1, 2} and there is at least one i ∈ {1, 2} such that G[NG [ui ]] is a complete graph. Lemma 5.4. Let α(G∗ ) ≤ 2 and α(G) ≥ 3. If there exists v ∈ V (G∗ ) such that v ̸∈ {u∗ , z } and z v ̸∈ E (G∗ ), then one of the following holds. (i) dG∗ (z ) + dG∗ (v) ≥ n∗ , or (ii) dG∗ (u∗ ) ≥ n/2, or (iii) |V (H )| = 3, 4. Furthermore, if V (H ) = {x, y, w}, then z w ∈ E (G) if zu∗ ∈ E (G∗ ); vw ∈ E (G) if v u∗ ∈ E (G∗ ). If V (H ) = {x, y, w, t }, then z w, v t ∈ E (G) or zt , vw ∈ E (G). Proof. If there exists one vertex a ∈ V (H ) − {x, y} such that {z , v, a} is a 3-independent set of G, then d(z ) + d(v) + d(a) ≥ 3n/2. Since d(a) ≤ (V (H ) − 1) + e(a, R), dG∗ (z ) + dG∗ (v) = d(z ) − 2 + d(v) ≥ 3n/2 − d(a) − 2 ≥ 3n/2 − (|V (H )| − 1) − e(a, R) − 2 = n∗ + n/2 − e(a, R) − 2. If n/2 ≥ e(a, R) + 2, then dG∗ (z ) + dG∗ (v) ≥ n∗ and (i) holds. If n/2 < e(a, R) + 2, then e(a, R) ≥ n/2−1. Since α(G∗ ) ≤ 2 and v z ̸∈ E (G∗ ), zu∗ or v u∗ ∈ E (G∗ ). This means that dG∗ (u∗ ) = e(a, R)+e(H −a, R) ≥ n/2 and (ii) holds. If there exists no vertex a ∈ V (H )−{x, y} such that {z , v, a} is a 3-independent set of G, then each vertex a ∈ V (H )−{x, y} is adjacent to z or v . If |V (H )| ≥ 5, there are two vertices of V (H ) − {x, y} both adjacent to z or v . By Lemma 2.1(7), z ∈ V (H ) or v ∈ V (H ), contrary to the maximality of H. If |V (H )| = 3 and V (H ) = {x, y, w}, then z w ∈ E (G) if zu∗ ∈ E (G∗ ); vw ∈ E (G) if v u∗ ∈ E (G∗ ). If |V (H )| = 4 and V (H ) = {x, y, w, t }, then z w, v t ∈ E (G) or zt , vw ∈ E (G) and (iii) holds.  Lemma 5.5. Suppose that α(G) ≥ 3 and α(G∗ ) ≤ 2. If u1 and u2 are a pair of nonadjacent vertices of G∗ such that ui ̸∈ {z , u∗ }, then dG∗ (u1 ) + dG∗ (u2 ) ≥ n∗ or G∗ is Z3 -connected. Proof. In this case, dG∗ (ui ) = d(ui ) for each i ∈ {1, 2}. By the maximality of H and Lemma 2.1(7), e(v, H ) ≤ 1 for each v ∈ V (R). Thus, H contains at least one vertex which is not adjacent to u1 nor u2 if |V (H )| ≥ 3. If H contains two vertices, say a1 and a2 , such that e(ai , {u1 , u2 }) = 0, then d(u1 ) + d(u2 ) + d(ai ) ≥ 3n/2, and d(ai ) ≤ dG∗ (u∗ )+(V (H )− 1)− e(H − ai , R)+ 1 for each i ∈ {1, 2}. Then dG∗ (u1 )+ dG∗ (u2 ) = d(u1 )+ d(u2 ) ≥ 3n/2 − d(ai ) ≥ 3n/2 − dG∗ (u∗ )−(|V (H )|− 1)+ e(H − ai , R)− 1 = n∗ + n/2 − e(ai , R)− 1 for each i ∈ {1, 2}. If n/2 ≥ e(ai , R)+ 1 for some i ∈ {1, 2}, then dG∗ (u1 ) + dG∗ (u2 ) ≥ n∗ ; if n/2 < e(ai , R) + 1 for each i ∈ {1, 2}, then e({a1 , a2 }, R) = e(a1 , R) + e(a2 , R) > n − 2. Since |V (R)| ≤ n − 3, a1 and a2 have at least one common adjacent vertex in R, contrary to e(v, H ) ≤ 1 for each v ∈ V (R). Thus, assume that H contains one vertex, say a, such that e(a, {u1 , u2 }) = 0. By Lemma 2.1(7), V (H ) = {x, y, w}, e(ui , H ) = 1 for each i ∈ {1, 2} and N (u1 ) ∩ H ̸= N (u2 ) ∩ H. That is, n = n∗ + 2. In this case, dG∗ (u1 ) + dG∗ (u2 ) = n∗ − 1. By Lemma 5.3, we assume, without loss of generality, that G∗ [NG∗ [u1 ]] is a complete graph and b = u∗ . Since u∗ u1 ∈ E (G), by the maximality of H, G∗ [NG∗ [u1 ]] ∼ = K4 for δ(G) ≥ 3. Note that {u1 , u2 , a} is a 3-independent set of G and d(a) ≤ dG∗ (u∗ ) + (V (H ) − 1) − e(H − a, R) + 1. This implies that d(u1 ) + d(u2 ) ≥ 3n/2 − d(a) ≥ 3n/2 − dG∗ (u∗ ) − (|V (H )| − 1) + e(H − a, R) − 1 = n∗ + n/2 − e(a, R) − 1. If n/2 ≥ e(a, R) + 1, then dG∗ (u1 ) + dG∗ (u2 ) ≥ n∗ . If n/2 < e(a, R) + 1, then dG∗ (u∗ ) = e(H − a, R) + e(a, R) ≥ n/2 + e(H − a, R). Since e(H − a, R) ≥ 2, dG∗ (u∗ ) ≥ n/2 + 2 ≥ 6. Since G contains at most one 3-vertex, G∗ [NG∗ [u2 ] − {u∗ }] ∼ = Km (m ≥ 4). Then G∗ contains a K5− as a subgraph on vertex set NG∗ [u2 ], which is Z3 -connected by Lemma 2.1(2), contrary to the maximality of H.  If G∗ is Z3 -connected, by Lemmas 2.1 and 2.2, G is Z3 -connected, which is the desired result of Lemma 5.8. Thus, in the rest of this section, when Lemma 5.5 is applied to a pair of nonadjacent vertices u1 , u2 of G∗ , where u1 , u2 ̸∈ {z , u∗ }, we have dG∗ (u1 ) + dG∗ (u2 ) ≥ n∗ . Application of Lemma 5.6 is similar. Lemma 5.6. Suppose that α(G) ≥ 3 and α(G∗ ) ≤ 2. If u1 and u2 are a pair of nonadjacent vertices of G∗ such that u1 = z , u2 ̸∈ {z , u∗ }, then dG∗ (u1 ) + dG∗ (u2 ) ≥ n∗ or G∗ is Z3 -connected.

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Proof. In this case, dG∗ (z ) = d(z ) − 2 and dG∗ (u2 ) = d(u2 ). By Lemma 5.3, we assume that G∗ [NG∗ [z ]] is a complete graph Kt . Since dG (z ) ≥ 4, t ≥ 3 and dG∗ (z ) ≥ 2. Case 1. dG∗ (z ) ≥ 4. In this case, G∗ [NG∗ [z ]] is Z3 -connected by Lemma 2.1(2). Let NG∗ (z ) = {z1 , z2 , . . . , zt −1 }. By the maximality of H, u∗ ̸∈ V (Kt ) and so u∗ z ̸∈ E (G∗ ). Since α(G∗ ) ≤ 2 and zu2 ̸∈ E (G∗ ), u∗ u2 ∈ E (G∗ ). By Lemma 5.3, u2 has at most one neighbor in {z1 , z2 , . . . , zt −1 }. Thus, we may assume that none of {z1 , . . . , zt −2 } is a neighbor of u2 . For a pair of nonadjacent vertices zi and u2 for i ∈ {1, 2, . . . , t − 2}, by Lemma 5.5, dG∗ (u2 ) + dG∗ (zi ) ≥ n∗ . Note that u∗ ∈ NG∗ (u2 ) \ NG∗ (z ) and e({z1 , . . . , zt −2 }, NG∗ (u2 )) ≥ t − 2. By contracting G∗ [NG∗ [z ]] and all 2-cycles generated in the process, we finally get a K1 . By Lemma 2.1, G∗ is Z3 -connected. Case 2. dG∗ (z ) = 3. In this case, G∗ [NG∗ [z ]] ∼ = K4 . Let NG∗ (z ) = {z1 , z2 , z3 }. Let |NG∗ (z ) ∩ NG∗ (u2 )| = ε . If ε = 0, then by Lemma 5.3, G∗ [NG∗ [u2 ]] ∼ = Km and m ≥ 4 since δ(G) ≥ 3. Define NG∗ (u2 ) = {u21 , u22 , . . . , u2(m−1) }. For two nonadjacent vertices u2 and zi , where zi ̸= u∗ , by Lemma 5.5, dG∗ (u2 ) + dG∗ (zi ) ≥ n∗ . We may assume that e(zi , NG∗ (u2 )) ≥ 2 for i = 1, 2. If m ≥ 5, then by Lemma 2.1(2), (7), G∗ is Z3 -connected. Thus assume m = 4. This implies that dG∗ (u2 ) = 3. Since G contains at most one 3-vertex, d(u2i ) ≥ 4 for i ∈ {1, 2, 3}. Assume that d(u21 ) ≤ d(u2i ). Contracting 2-cycles generated in the process in the graph G∗[u21 u2 ,u21 u22 ] , we finally get a K1 . By Lemma 2.2, G∗ is Z3 -connected. Thus, assume that ε = 1. By Lemma 5.3, we assume that G∗ [NG∗ [u2 ] − {z3 }] ∼ = Km , where m = n∗ − 4 and z3 ∈ NG∗ (z ) ∩ NG∗ (u2 ). Set NG∗ (u2 ) − {z3 } = {u21 , u22 , . . . , u2(m−1) }. In this case, m ≥ 3. Consider first the case when m = 3. Since G contains at most one 3-vertex, d(u2i ) ≥ 4 for each i ∈ {1, 2}. This means that e(u2i , {z1 , z2 }) ≥ 2 for i = 1, 2 and we may assume that u21 z1 , u21 z2 ∈ E (G∗ ). Contracting 2-cycles generated in the process in the graph G∗[u21 z1 ,u21 z2 ] , we finally get a K1 . By Lemma 2.2, G∗ is Z3 -connected. Consider the case that m = 4. In this case, n∗ = 8 and n ≥ 10. Assume first, without loss of generality, that u∗ = z3 . For a pair of nonadjacent vertices u2 and zi for each i ∈ {1, 2}, by Lemma 5.5, e(zi , NG∗ (u2 ) − {z3 }) ≥ 1. If there exists some zi , say z1 , such that e(zi , NG∗ (u2 ) − {z3 }) ≥ 2, then we can get a K1 from G∗[u2 u21 ,u2 u22 ] by contracting all 2-cycles generated in the process. By Lemma 2.2, G∗ is Z3 -connected. Thus we assume e(zi , NG∗ (u2 ) − {z3 }) = 1 for each i ∈ {1, 2}. Without loss of generality, we assume that z1 u21 ∈ E (G∗ ). For vertex z2 , z2 u21 ∈ E (G∗ ) or z2 u21 ̸∈ E (G∗ ). In the former case, since G has at most one vertex of degree 3, we may assume that e(u22 , {z1 , z2 , z3 }) ≥ 1. In this case, we can get a K1 from G∗[u2 u21 ,u2 u22 ]

by contracting all 2-cycles and K5− generated in the process. By Lemma 2.2, G∗ is Z3 -connected. In the latter case, we may assume that z2 u22 ∈ E (G∗ ). For a pair of nonadjacent vertices u23 and z2 , by Lemma 5.5, dG∗ (u23 ) + dG∗ (z2 ) ≥ n∗ = 8. Thus, u23 u∗ ∈ E (G∗ ). In this case, we finally get a K1 from G∗[u21 u2 ,u21 u22 ] by contracting 2-cycles and K5− generated in the process. By Lemma 2.2, G∗ is Z3 -connected. Next, we assume that z3 ̸= u∗ . This leads to that u∗ = zi for some i ∈ {1, 2} or u∗ = u2j for some j ∈ {1, 2, 3}. In the former case, by symmetry, let u∗ = z1 . In this case, dG∗ (z3 ) ≥ 4. For a pair of nonadjacent vertices z2 and u2 , by Lemma 5.5, dG∗ (z2 ) ≥ 4. If d(z3 ) = 4, then e({z2 , u∗ }, NG∗ (u2 ) − {z3 }) ≥ 3 by Lemma 5.5. Then we can get a K1 from G∗[u2 u21 ,u2 u22 ] by contracting 2-cycles generated in the process. By Lemma 2.2, G∗ is Z3 -connected. If d(z3 ) ≥ 5, say z3 u21 ∈ E (G∗ ), then e({z2 , u∗ }, NG∗ (u2 ) − {z3 }) ≥ 2 by Lemma 5.5. Then we can get a K1 from G∗[u23 u21 ,u23 u22 ] by contracting 2-cycles generated in the process. In the latter case, the proof is similar. Finally, consider the case when m ≥ 5. In this case, G∗ [NG∗ [u2 ] − {z3 }] is Z3 -connected by Lemma 2.1(2). If e({z1 , z2 , z3 }, NG∗ [u2 ] − {z3 }) ≥ 3, then G∗ is Z3 -connected by Lemma 2.1(2), (7). Thus, assume that e({z1 , z2 , z3 }, NG∗ [u2 ] − {z3 }) ≤ 2. If e(z3 , NG∗ [u2 ] − {z3 }) = 2, then dG∗ (zi ) = 3 for each i = 1, 2. Thus u∗ = zi since G contains at most one 3-vertex, say u∗ = z2 . In this case, dG∗ (z1 ) + dG∗ (u2 ) < n∗ , contrary to Lemma 5.5. If e(b, NG∗ [u2 ] − {z3 }) = 1 and e(z2 , NG∗ [u2 ] − {z3 }) = 1, then we can find some u2i ̸= u∗ satisfying z2 u2i ̸∈ E (G∗ ). Then dG∗ (z2 ) + dG∗ (u2i ) = 4 + m − 1 = m − 3 < n∗ , contrary to Lemma 5.5. Case 3. dG∗ (z ) = 2. In this case, let NG∗ (z ) = {z1 , z2 } and |NG∗ (z ) ∩ NG∗ (u2 )| = ε . If ε = 0, then by Lemma 5.3, G∗ [NG∗ [u2 ]] ∼ = Km and m ≥ 4 since δ(G) ≥ 3. Suppose that u∗ = z1 . For a pair of nonadjacent vertices z2 and u2 , by Lemma 5.5, dG∗ (z2 ) + dG∗ (u2 ) ≥ n∗ , which implies that e(z2 , NG∗ (u2 )) ≥ 2. For a pair of nonadjacent vertices z and u2 , dG∗ (z ) + dG∗ (u2 ) < n∗ , by Lemma 5.4, either dG∗ (u∗ ) ≥ n/2 or |V (H )| = 3, 4. In the former case, |V (H )| ≥ 5, which implies that n ≥ 11. Thus, dG∗ (u∗ ) ≥ 6. Then G∗ contains a K5− induced by NG∗ [u2 ] ∪ {u∗ }. By Lemma 2.1(2), (7), G∗ is Z3 -connected. In the latter case, if |V (H )| = 3, then dG∗ (u∗ ) ≥ 4 since G contains at most one 3-vertex. If m ≥ 5, then by Lemma 2.1(2), (7) G∗ is Z3 -connected. If m = 4, then dG∗ (z2 ) ≥ 4 and e(u2j , {u∗ , z2 }) ≥ 1 since G contains at most one 3-vertex. Let u∗ u21 , u∗ u22 ∈ E (G∗ ). Then we can get a K1 from G∗[u23 u21 ,u23 u22 ] by contracting all 2-cycles generated in the process. By Lemma 2.2, G∗ is Z3 -connected. If |V (H )| = 4, then u∗ u2 ∈ E (G∗ ) for otherwise H contains a , where n = 10. This contradicts vertex x such that {x, z , u2 } is a 3-independent set and d(x) + d(u2 ) + d(z ) ≤ 4 + 4 + 3 ≤ 3n 2 that |NG∗ (z ) ∩ NG∗ (u2 )| = 0. Suppose that u∗ = u21 . By Lemma 5.5 and the argument above, e(zi , NG∗ (u2 )) ≥ 2 for each i ∈ {1, 2}. If m ≥ 5, then by Lemma 2.1(2), (7) G∗ is Z3 -connected. If m = 4, then let dG∗ (u23 ) ≤ dG∗ (u2i ) for i = 1, 2. We can get a K1 from G∗[u23 u21 ,u23 u22 ] by contracting all 2-cycles generated in the process. By Lemma 2.2, G∗ is Z3 -connected. If ε = 1, then by Lemma 5.3, we assume that G∗ [NG∗ [u2 ] − {z3 }] ∼ = Km , where m = n∗ − 3 and b ∈ NG∗ (u2 ) ∩ NG∗ (z ). Thus, m ≥ 3 and let

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NG∗ (z ) = {b, z1 }. If m ≥ 5, then by Lemma 2.1(2), G∗ [NG∗ [u2 ] − {b}] is Z3 -connected. Since δ(G) ≥ 3 and G∗ contains at most one 3-vertex, e(z1 , NG∗ [u2 ] − {b}) ≥ 1. If e({b, z1 }, NG∗ [u2 ] − {b}) ≥ 3, then by Lemma 2.1(7), G∗ is Z3 -connected. Thus, assume e({b, z1 }, NG∗ [u2 ] − {b}) ≤ 2. This implies that e({v}, NG∗ [u2 ] − {b}) = 1 for each v ∈ {b, z1 }. In this case, there exists a vertex u2i such that {u2i , z1 } is a 2-independent set of G∗ such that dG∗ (z1 ) + dG∗ (u2i ) = 7 < n∗ . By Lemma 5.5, G∗ is Z3 -connected. If m = 4 and b ̸= u∗ , then dG∗ (b) ≥ 3 and u∗ = z1 or u∗ = u2i for some i ∈ {1, 2, 3}. If dG∗ (b) ≥ 5, then G∗ contains a K5− induced by NG∗ [u2 ]. If z1 = u∗ , then for a pair of nonadjacent vertices z , u2i , dG∗ (z ) + dG∗ (u2i ) < n∗ , by Lemma 5.4, dG∗ (z1 ) ≥ 3 or |V (H )| = 3, 4. If |V (H )| = 3, dG∗ (z1 ) ≥ 3 since G has at most one vertex of degree 3. If |V (H )| = 4, then by the given degree condition, dG∗ (z1 ) ≥ 3; if z1 ̸= u∗ , then dG∗ (z1 ) ≥ 3 since δ(G) ≥ 3. In either case, by Lemma 2.1(2), (7), G∗ is Z3 -connected. Suppose that dG∗ (b) = 4. If dG∗ (z1 ) ≥ 4, let z1 u21 , z1 u22 ∈ E (G∗ ). In this case, we can get a K1 from G[u2 u21 ,u2 u22 ] by contracting 2-cycles generated in the process. By Lemma 2.2, G is Z3 -connected. Thus, let dG∗ (z1 ) = 3. In this case, z1 = u∗ or u2i = u∗ for some i ∈ {1, 2, 3} and in either case dG∗ (u∗ ) = 3 since G contains at most one 3-vertex. Then dG∗ (u∗ ) = 3 < n/2. By Lemma 5.4, |V (H )| = 3, 4. Since u∗ u2 ̸∈ E (G∗ ) or u∗ z ̸∈ E (G∗ ), |V (H )| = 3. This leads to that G contains at least two 3-vertices, a contradiction. If dG∗ (b) = 3, then u2i z1 ∈ E (G∗ ) for each i ∈ {1, 2, 3} since G contains at most one vertex of degree 3. Thus G∗ contains a K5− induced by (NG∗ [u2 ] − {b}) ∪ {z1 }. By Lemma 2.1(2), (7), G∗ is Z3 -connected. If b = u∗ , then n∗ = 7. This implies that n ≥ 9. If for each vertex u2i , dG∗ (u2i ) ≥ 5, then G∗ contains a K5 . By Lemma 2.1, G∗ is Z3 -connected. Thus, assume that there is a vertex u2i with dG∗ (u2i ) ≤ 4. Thus, G∗ contains a pair of nonadjacent vertices z and u2i with dG∗ (z ) + dG∗ (u2i ) < n∗ . By Lemma 5.4, either dG∗ (u∗ ) ≥ n/2 or |V (H )| = 3, 4. In the former case, dG∗ (u∗ ) ≥ 5. Thus e(u∗ , NG∗ [u2 ] − {b}) ≥ 3. So G∗ contains a K5− induced by NG∗ [u2 ]. Then by Lemma 2.1(2), (7), G∗ is Z3 -connected. In the latter case, since u∗ u2 , u∗ z ∈ E (G∗ ), by Lemma 5.4, |V (H )| = 4 and dG∗ (u∗ ) < n/2 = 5. In this case, n = 10. Note that dG∗ (z1 ) ≥ 3 and dG∗ (u∗ ) ≥ 3. If dG∗ (z1 ) = 3 and for a pair of two nonadjacent vertices z1 and u2i , then by Lemma 5.5, dG∗ (u2i ) ≥ 5, contrary to dG∗ (u2i ) < 5. Thus, assume that dG∗ (z1 ) ≥ 4. If dG∗ (u∗ ) ≥ 4, then we can get a K1 from G[u2 u21 ,u2 u22 ] by contracting 2-cycles generated in the process. By Lemma 2.2, G∗ is Z3 -connected. If dG∗ (z1 ) = 5, then G∗ contains a K5− induced by (NG∗ [u2 ] − {u∗ }) ∪ {z1 }. By Lemma 2.1(2), (7), G∗ is Z3 -connected. Thus, we are left to discuss the case when dG∗ (z1 ) = 4 and dG∗ (u∗ ) = 3. There is a vertex a ∈ V (H ) such that {a, z1 , u2 } is a 3-independent set of G and d(a) ≤ 4 while ≤ d(a) + d(u2 ) + d(z1 ) ≤ 4 + 4 + 4 = 12, a contradiction. 3 10 2 If m = 3, then n∗ = 6. Assume that u∗ ̸∈ {u21 , u22 }. Since G contains at most one 3-vertex, dG∗ (z1 ) = 4, dG∗ (b) = 5. In this case, G∗ contains a K5− induced by NG∗ [u2 ] ∪ {z1 }. By Lemma 2.1(2), (7), G∗ is Z3 -connected. By symmetry, assume that u∗ = u21 . Since G contains at most one 3-vertex, dG∗ (z1 ) = 4, dG∗ (b) = 4, dG∗ (u∗ ) = 3. In this case, |V (H )| ≥ 4 since G contains at most one 3-vertex. Then n ≥ 9. Note that dG∗ (u∗ ) = 3 < n/2, by Lemma 5.4, |V (H )| = 4 and zu∗ , u2 u∗ ∈ E (G∗ ), contrary to that u∗ = u21 .  Lemma 5.7. Suppose that α(G) ≥ 3 and α(G∗ ) ≤ 2. If u1 , u2 are a pair of nonadjacent vertices of G∗ such that u1 = u∗ , u2 ̸∈ {z , u∗ }, then dG∗ (u1 ) + dG∗ (u2 ) ≥ n∗ or G∗ is Z3 -connected. Proof. In this case, dG∗ (u2 ) = d(u2 ) and dG∗ (u∗ ) ≥ 2. Let NG∗ (u∗ ) = {v1 , v2 , . . . , vm1 }, NG∗ (u2 ) = {u21 , u22 , . . . , u2m2 } and |NG∗ (u∗ ) ∩ NG∗ (u2 )| = ε, where ε = 0, 1. By the maximality of H, if ε = 0, dG∗ (u∗ ) = 2, 3; if ε = 1, dG∗ (u∗ ) = 2, 3, 4. Since δ(G) ≥ 3, dG∗ (u2 ) ≥ 3. If ε = 0, then by Lemma 5.3, G∗ [NG∗ [u∗ ]] and G∗ [NG∗ [u2 ]] are two complete graphs. Note that dG∗ (u∗ ) = 2, 3 and m1 = 2, 3. In this case, G∗ [NG∗ [u2 ]] ∼ = Km , where m = n∗ − 2 or n∗ − 3, m ≥ 4. For a pair of nonadjacent vertices u2 and vi , by Lemma 5.5, e(vi , NG∗ (u2 )) ≥ 2 for each i = 1, 2, 3. If m ≥ 5, then G∗ [NG∗ [u2 ]] contains a K5 . Contracting this K5 and contracting all 2-cycles generated in the process, we finally get a K1 . By Lemma 2.1, G∗ is Z3 -connected. Thus, m = 4. We assume that v1 u11 , v1 u22 , v2 u22 , v2 u23 ∈ E (G∗ ). Since G contains at most one vertex of degree 3, d(u23 ) ≥ 4. We can get a

K1 from G∗[u23 u21 ,u23 u22 ] by contracting 2-cycles generated in the process. By Lemma 2.1, G∗ is Z3 -connected. If ε = 1, then by Lemma 5.3, we may assume that G∗ [NG∗ [u∗ ] − {vm1 }] and G∗ [NG∗ [u2 ]] = Km , where m ≥ 4, are two complete graphs. Note that dG∗ (u∗ ) = 2, 3, 4. For a pair of nonadjacent vertices u2 and vi , by Lemma 5.5, e(vi , NG∗ ) ≥ 2 for each vi ∈ NG∗ (u∗ ) − {vm1 }. If m ≥ 5, then by the argument above, G∗ is Z3 -connected by Lemma 2.1. Thus, assume m = 4. If dG∗ (u∗ ) = 3, 4, then m1 = 3, 4. Note that e(vi , NG∗ ) ≥ 2 for each i ∈ {1, . . . , m1 }. We assume, without loss of generality, that v1 u21 , v1 u22 , v2 u22 , v2 u23 ∈ E (G∗ ). Since G contains at most one vertex of degree 3, d(u23 ) ≥ 4. Then we can get a K1 from G∗[u23 u21 ,u23 u22 ] by contracting 2-cycles generated in the process; by Lemma 2.2, G∗ is Z3 -connected. We are left to consider the case when dG∗ (u∗ ) = 2. Since G contains one 3-vertex, |V (H )| ≥ 4. Thus H contains a vertex v ∈ V (H ) \ {x, y} such that {z , v1 , u2 } is an independent 3-set of G while d(z ) + d(v) + d(v1 ) ≤ |V (H )| − 1 + 3 + 4 ≤ n − 5 + 3 + 4 = n + 2, which implies that 3n/2 ≤ n + 2 and n ≤ 4, a contradiction.  Lemma 5.8. Let G ∈ F be a graph on n ≥ 8 vertices with δ(G) ≥ 3. If G contains at most one 3-vertex, then either G is Z3 -connected or G∗ ∈ F . Proof. By Lemmas 5.1 and 5.2, we only consider the case when α(G∗ ) ≤ 2 and α(G) ≥ 3. Let {u1 , u2 } be a 2-independent set of G∗ . Since α(G∗ ) ≤ 2, v u1 ∈ E (G∗ ) or v u2 ∈ E (G∗ ) for each v ∈ V (G∗ )−{u1 , u2 }. This implies that dG∗ (u1 )+dG∗ (u2 ) ≥ n∗ −2. If dG∗ (u1 ) + dG∗ (u2 ) ≥ n∗ , then we are done. Thus, we may assume that dG∗ (u1 ) + dG∗ (u2 ) = n∗ − 2 or n∗ − 1. By Lemma 2.2, we are left to show that G∗ is Z3 -connected. By Lemmas 5.5–5.7, we are left to consider the case when u1 = u∗ and u2 = z.

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Let NG∗ (u∗ ) = {v1 , v2 , . . . , vm1 }, NG∗ (z ) = {z1 , z2 , . . . , zm2 } and |NG∗ (u∗ ) ∩ NG∗ (z )| = ε , where ε = 0, 1. By the maximality of H, if ε = 0, then m1 = 2, 3; if ε = 1, then m1 = 2, 3, 4. Since d(z ) ≥ 4, m2 ≥ 2. Assume first that ε = 0. In this case, G∗ [NG∗ [u∗ ]] and G∗ [NG∗ [z ]] are two complete graphs and n∗ = m1 + m2 + 2. For a pair of nonadjacent vertices z and vi , by Lemma 5.6, then e(vi , NG∗ (z )) ≥ 2 for each i ∈ {1, 2, . . . , m1 }. If m2 ≥ 4, then G∗ is Z3 -connected by Lemma 2.1(2), (7). Thus, m2 ≤ 3. Let m2 = 3 and m1 = 3. We assume, without loss of generality, that v1 zi ∈ E (G∗ ) for each i ∈ {1, 2}. If m2 = 3 and m1 = 2, then by Lemma 5.7, e(zi , NG∗ (u∗ )) ≥ 2 for each i ∈ {1, 2, 3}. Thus e(vi , NG∗ (z )) = 3 for each i ∈ {1, 2}. In either case, we can get a K1 from G∗[v1 z1 ,v1 z2 ] by contracting 2-cycles generated in the process; by Lemma 2.2, G∗ is Z3 -connected. Thus we only need to consider the case m2 = 2. If m1 = 3, then by Lemma 5.6 e(vi , NG∗ (z )) ≥ 2 for each i ∈ {1, 2, 3}. The proof is similar to the case that m2 = 3 and we are done. If m1 = 2, then vi zj ∈ E (G∗ ) for each i, j ∈ {1, 2} by Lemmas 5.6 and 5.7. Since α(G) ≥ 3, we can find v ∈ V (H ) − {x, y} such that {v, vi , z } is a 3-independent set of G. Thus 3n/2 ≤ d(v) + d(vi ) + d(z ) ≤ |V (H )| − 1 + 4 + 4 = n − 6 + 8 = n + 2; we can deduce that n ≤ 4, a contradiction. Next, assume that ε = 1. By Lemma 5.3, we assume that G∗ [NG∗ [z ]] and G∗ [NG∗ [u∗ ] − {vm1 }] are complete graphs. For a pair of nonadjacent vertices z and vi , by Lemma 5.6, then e(vi , NG∗ (z )) ≥ 2 for each i ∈ {1, 2, . . . , m1 − 1}. If m2 ≥ 4, then G∗ is Z3 -connected by Lemma 2.1(2), (7). If m2 = 3 and m1 = 3, 4, then we assume that say v1 zi ∈ E (G∗ ) for i = 1, 2. We can get a K1 from G∗[v1 z1 ,v1 z2 ] by contracting 2-cycles. By Lemma 2.2, G∗ is Z3 -connected. Thus we only need to consider the case m1 = 2. In this case, n∗ = 6 and v2 = z3 . Then v1 zi ∈ E (G∗ ) for each i = 1, 2. If v2 z1 ∈ E (G∗ ), then the subgraph induced by NG∗ [z ] ∪ {v1 } is an even wheel W4 . Then G∗ is Z3 -connected by Lemma 2.1(3), (5), (6). Thus assume v1 z3 ̸∈ E (G∗ ). Since α(G) ≥ 3, there is a vertex v ∈ V (H ) such that {v, v1 , z } is a 3-independent set of G. Thus 3n/2 ≤ d(v) + d(v1 ) + d(z ) ≤ |V (H )| − 1 + 5 + 4 = n − 6 + 8 = n + 2, which implies that n ≤ 4, a contradiction. If m2 = 2 and m1 = 3, 4, then n∗ = 5, 6, 7. Since α(G) ≥ 3, there is a vertex v ∈ V (H ) such that {v, v1 , z } is a 3-independent set of G. In this case, 3n/2 ≤ d(v) + d(v1 ) + d(z ) ≤ |V (H )| − 1 + 4 + 4 = n − 5 + 8 = n + 3, which implies that n ≤ 6, a contradiction. Thus we only need to consider the case when m1 = 2. Then n∗ = 5 and G contains at least two 3-vertices, a contradiction.  6. Proof of Theorem 1.4 By Lemma 3.1, it is sufficient for us to show that when n ≥ 8, G can be contracted to one of the graphs {K1 , K3 , K4 , K4− , G5 }. Clearly, δ(G) ≥ 2. Assume that n ≥ 8 and the theorem holds for any graph with the number of vertices less than n. Case 1. δ(G) = 2. In this case, let x ∈ V (G) with d(x) = 2, N (x) = {x1 , x2 }, S = V (G) − N [x], and |S | ≥ 5. If G[S ] is a complete graph, then G[S ] = Kt and t ≥ 5 since n ≥ 8. By Lemma 2.1, G[S ] is Z3 -connected. If S contains two nonadjacent vertices u and v , then by the given degree condition, d(u) + d(v) ≥ 3n/2 − 2. It follows that dG[S ] (u) + dG[S ] (v) ≥ 3n/2 − 6 ≥ n − 3 = |S | since n ≥ 8. By Theorem 1.3, either G[S ] is Z3 -connected or G[S ] is one of the graphs {G5 , G6 , G7 , G8 , G9 , G10 , G11 , G12 }. Assume first that G[S ] is Z3 -connected. If e(xi , S ) ≥ 2 for each i = 1, 2, then G is Z3 -connected by Lemma 2.1(7). Thus, we assume that e(x1 , S ) ≤ 1. If e(x1 , S ) = 0, then x1 x2 ∈ E (G). Since G is 2-edge-connected, e(x2 , S ) ≥ 2. By Lemma 2.1(7), G′ ∼ = K3 . If e(x1 , S ) = 1, then e(x2 , S ) ≥ 1. In this case, we consider x1 x2 ̸∈ E (G) or x1 x2 ∈ E (G). In the former case, if G[S ] is a complete graph, then α(G) = 2. By the given degree condition, d(x2 ) ≥ n − d(x1 ) = n − 2. This implies that e(x2 , S ) ≥ 2. Assume that G[S ] is not a complete graph. If each vertex of S is adjacent to x1 or x2 , then e(x2 , S ) ≥ |S |− 1 ≥ 4; otherwise, S contains a vertex y such that e(y, {x1 , x2 }) = 0. In this case, {x1 , x2 , y} is a 3-independent set of G and by the given degree condition, d(x2 ) ≥ 3n/2 − 2 − (n − 4) = n/2 + 2. This implies that e(x2 , S ) ≥ 2. By Lemma 2.1(7), G′ ∼ = K3 . In the latter case, by Lemma 2.1(7), G is Z3 -connected if e(x2 , S ) ≥ 2; G′ ∼ = K4− if e(x2 , S ) = 1. Next, assume that G[S ] is one of the graphs {G5 , G6 }. In both cases, n = 8. Let y be the vertex of degree 2 and z be the vertex of degree 3 in G[S ] such that yz ̸∈ E (G[S ]). It follows that {x, y, z } is a 3-independent set and d(x) + d(y) + d(z ) ≤ 2 + 4 + 5 = 11, contrary to σ3 ≥ 12. Finally, assume that G[S ] is one of the graphs {G7 , G8 , G9 , G10 , G11 , G12 }. In these cases, n = 9. Let y and z be the two vertices of degree 3 in G[S ] such that yz ̸∈ E (G[S ]). It follows that {x, y, z } is a 3-independent set and d(x) + d(y) + d(z ) ≤ 2 + 5 + 5 = 12, contrary to σ3 ≥ 14. Case 2. δ(G) ≥ 3. By Lemma 4.4, we may assume that G has at most one 3-vertex. By Lemma 2.3, either G is Z3 -connected or G contains a K4− as a subgraph. In the former case, we are done. In the latter case, let K4− be the union of two triangles xyz and xyw with xy in common and d(z ) ≥ 4. Note that G[zx,zy] contains a 2-cycle xyx, which is Z3 -connected by Lemma 2.1. Let H be the maximal Z3 -connected subgraph of G[zx,zy] containing xyx and let G∗ = G[zx,zy] /H. By Lemma 5.8, either G is Z3 -connected or G∗ ∈ F . Subcase 2.1. G∗ is 2-edge-connected. In this case, by the induction hypothesis, either G∗ is isomorphic to one of the 12 graphs shown in Fig. 1 or G∗ can be contracted to one of the graphs {K1 , K3 , K4− , K4 , G5 }. Claim. If G∗ is isomorphic to one of the 12 graphs shown in Fig. 1, then G is Z3 -connected.

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Proof of Claim. By our assumption, δ(G) ≥ 3. Thus, G∗ has at most two 2-vertices. This implies that G∗ cannot be G1 or G2 . If G∗ ∼ = G3 , then the two 2-vertices must be u∗ and z which implies that G contains at least two 3-vertices, a contradiction. If G∗ is one of the graphs {G4 , G6 , G7 , G8 , G9 , G10 , G11 , G12 }, we can deduce that G contains at least two 3-vertices, a contradiction. Assume that G∗ ∼ = G5 . Let V (G5 ) = {v1 , v2 , v3 , v4 , v5 } with d(v1 ) = 2, d(v2 ) = d(v3 ) = 3, d(v4 ) = d(v5 ) = 4. Since δ(G) ≥ 3 and G has at most one 3-vertex, u∗ and z must be a 2-vertex and a 3-vertex in G∗ , respectively. Consider first that u∗ = v1 is a 2-vertex and z = v2 is a 3-vertex in G∗ . If G[V (H )] contains two nonadjacent vertices, say w2 and w3 , then {v2 , w2 , w3 } is a 3-independent set. By the given degree condition, d(w2 ) + d(w3 ) ≥ 3n/2 − 3 ≥ |V (H )| + 1 since n ≥ 8 and n = |V (H )| + 4. Note that H contains 2 parallel edges between x and y. Denote by H ′ the graph from H by removing an edge from x to y. It implies that dH ′ (w2 ) + dH ′ (w3 ) ≥ |V (H ′ )|. Since n ≥ 8, by Theorem 1.3, either G[V (H ′ )] is Z3 -connected or G[V (H ′ )] is one of the 10 graphs {G2 , G3 , G5 , G6 , G7 , . . . , G12 }. In the former case, since e(z , H ′ ) = 2, by Lemma 2.1(8), G is Z3 -connected. In the latter case, when G[V (H ′ )] is one of the graphs {G2 , G3 , G6 , G7 , G8 , G12 }, we can deduce that G contains at least two 3-vertices or δ(G) = 2, a contradiction. If G[V (H )] ∼ = G5 , let w2 and w3 be 2-vertex and 3-vertex in G[V (H )], respectively. Clearly, d(w2 ) = d(w3 ) = 4. Then {v3 , w2 , w3 } is a 3-independent set of G. Then d(v3 )+ d(w2 )+ d(w3 ) = 3 + 4 + 4 = 11 < 3n/2, where n = 9, a contradiction. If G[V (H )] ∈ {G9 , G10 , G11 }, we choose two nonadjacent vertices w2 and w3 with degree 3 in G10 (or G11 ). Clearly, d(w2 ) = d(w3 ) = 4. Then {v3 , w2 , w3 } is a 3-independent set of G. Then d(v3 ) + d(w2 ) + d(w3 ) = 3 + 4 + 4 = 11 < 3n/2, where n = 10, a contradiction. Thus, assume that G[V (H )] is a complete graph. If n ≥ 9, G[V (H )] is a complete graph on more than 5 vertices. By Lemma 2.1(2), (7), G is Z3 -connected. If n = 8, G[V (H )] = K4 , then let V (H ) = {x, y, w, a}. Clearly, d(v) = 4 for each v ∈ V (H ). Then we can get a trivial graph from G[ax,ay] by contracting 2-cycles. By Lemmas 2.1(1) and 2.2, G is Z3 -connected. Therefore, we are left with the case when u∗ = v2 is a 3-vertex and z = v1 is a 2-vertex in G∗ . Define H ′ as in the above paragraph. Since n ≥ 8, there exists one vertex w2 in V (H ′ ) − {x, y} such that {z , v3 , w2 } is a 3-independent set of G. By the given degree condition, d(w2 ) ≥ 3n/2 − 3 − 4 = 3n/2 − 7. On the other hand, d(w2 ) ≤ |V (H ′ )| − 1 + 2 = |V (H ′ )| + 1. This means that n = |V (H ′ )| + 4 ≥ d(w2 ) + 3 ≥ 3n/2 − 4, n ≤ 8. Thus, n = 8, |V (H ′ )| = 4, d(w2 ) = |V (H ′ )| + 1 = 5. Then G[V (H ′ )] ∼ = K4 . Let V (H ′ ) = {x, y, w2 , a}. Then we can get a K1 from G[ax,ay] by contracting 2-cycles. By Lemma 2.2, G is Z3 -connected.  By the claim, we assume that G∗ can be contracted to one of the graphs {K1 , K3 , K4− , K4 , G5 }. If G∗ can be contracted to K1 , then G is Z3 -connected by Lemmas 2.1(1), (6) and 2.2. ¯ where G¯ ∈ {K3 , K4− , K4 , G5 }. Let V (G¯ ) = {u∗ , u1 , u2 , . . . , ut }, From now on, we assume that G∗ can be contracted to G, where t ≥ 2, and Hi be the maximal Z3 -connected subgraph in G∗ which is contracted to ui for each i ∈ {1, 2, . . . , t }. We assume, without loss of generality, that z ∈ V (H1 ). Consider the case when G∗ can be contracted to K3 . Then t = 2. Note that |V (H )| ≥ 3. Since δ(G) ≥ 3 and G is a simple graph, |V (H2 )| ≥ 5 and |V (H1 )| = 1 or ≥ 5. If |V (H )| = 3, then we can deduce that G contains at least two 3-vertices, a contradiction. Thus, assume that |V (H )| ≥ 4. In this case, we can pick xi ∈ V (Hi ) for each i ∈ {1, 2} and x3 ∈ V (H ) such that {x1 , x2 , x3 } is a 3-independent set of G and such that for i ∈ {1, 2}, d(xi ) ≤ |V (Hi )| − 1 if |V (Hi )| ≥ 5. By the given degree condition, d(x1 ) + d(x2 ) + d(x3 ) ≥ 3n/2. If |V (H1 )| = 1, then x1 = z and d(x1 ) = 4 = |V (H1 )| + 3, d(x2 ) ≤ |V (H2 )| − 1, d(x3 ) ≤ |V (H )|. If |V (H1 )| ≥ 5, then d(xi ) ≤ |V (Hi )| − 1 for each i ∈ {1, 2}, d(x3 ) ≤ |V (H )|. In each case, n = |V (H1 )| + |V (H2 )| + |V (H )| ≥ d(x1 ) − 3 + d(x2 ) + 1 + d(x3 ) ≥ 3n/2 − 2. Then n ≤ 4, a contradiction. ¯ where G¯ ∈ {K4− , K4 }. In this case, t = 3. By the argument above, Consider the case when G∗ can be contracted to G, there exists one i such that |V (Hi )| ≥ 5, where i ∈ {2, 3}. We assume, without loss of generality, that |V (H2 )| ≥ 5. We can pick xi ∈ V (Hi ) for each i ∈ {1, 2} and x3 ∈ V (H ) such that {x1 , x2 , x3 } is a 3-independent set of G. This implies that d(x1 ) ≤ 5 = |V (H1 )| + 4 or d(x1 ) ≤ |V (H1 )| − 1, d(x2 ) ≤ |V (H2 )| − 1, d(x3 ) ≤ |V (H )| − 1 + 2 = |V (H )| + 1. It follows that n = |V (H1 )| + |V (H2 )| + |V (H3 )| + |V (H )| ≥ d(x1 ) − 4 + d(x2 ) + 1 + 1 + d(x3 ) − 1 ≥ 3n/2 − 3, which implies that n ≤ 6, a contradiction. It remains for us to consider the case when G∗ can be contracted to G5 . In this case, t = 4. Suppose that u∗ and u1 are 2-vertex and 3-vertex in G5 , respectively. Let u2 be the other 3-vertex and u3 , u4 be two 4-vertices in G5 . If |V (Hi )| = 1 for each i ∈ {1, 2, 3, 4}, then G∗ ∼ = G5 ; this case we have discussed. Then |V (Hi )| ≥ 5 for some i ∈ {1, 2, 3, 4}. Without loss of generality, we assume that |V (H1 )| ≥ 5. Then we can find x1 ∈ V (H1 ) − {z } and x2 ∈ V (H ) − {x, y} such that {x1 , x2 , u2 } is a 3-independent set of G and d(x1 ) ≤ |V (H1 )| − 1 + 2 = |V (H1 )| + 1, d(x2 ) ≤ |V (H )| − 1 + 2 = |V (H )| + 1, d(u2 ) = 3 = |V (H2 )|+ 2. Then n = |V (H1 )|+|V (H2 )|+|V (H3 )|+|V (H4 )|+|V (H )| ≥ d(x1 )− 1 + d(u2 )− 2 + 1 + 1 + d(x2 )− 1 ≥ 3n/2 − 2. Then n ≤ 4, a contradiction. The proofs of other cases are similar. Subcase 2.2. G∗ is not 2-edge-connected. Assume that G∗ has an edge cut X ∗ with |X ∗ | ≤ 1. This means that X = X ∗ ∪ {zx, zy} is an edge cut of G. Let G1 and G2 be the connected component of G − X , where z ∈ V (G1 ) and x, y, w ∈ V (G2 ). Let |V (Gi )| = ni for each i ∈ {1, 2}. Then n = n1 + n2 . Since G contains at most one 3-vertex and δ(G) ≥ 3, n1 ≥ 5 and n2 ≥ 4. Thus, n ≥ 9. If G1 contains a pair of nonadjacent vertices, say z1 and z2 , then there exists one vertex, say a, in G2 − {x, y} such that {z1 , z2 , a} is a 3-independent set of G. If zi ̸= z for each i ∈ {1, 2}, then d(z1 ) + d(z2 ) ≤ n1 − 2 + n1 − 2 + 1 = 2n1 − 3. If z1 = z, then d(z1 ) + d(z2 ) ≤ n1 − 2 + 2 + n1 − 2 + 1 = 2n1 − 1. Thus, in both cases, d(z1 ) + d(z2 ) ≤ 2n1 − 1. Since d(a) ≤ n2 − 1 + 1 = n2 , d(z1 ) + d(z2 ) + d(a) ≤ 2n1 + n2 − 1 = n + n1 − 1. This implies that n + n1 − 1 ≥ 3n/2, n1 ≥ n/2 + 1. If G2 also contains a pair of nonadjacent vertices, similarly, n2 ≥ n/2. Thus, there is at least one subgraph Gi such that Gi is a complete graph. If

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G1 is a complete graph, then by Lemma 2.1 G1 is Z3 -connected since n1 ≥ 5. Let G2 be a complete graph. Note that n2 ≥ 4. When n2 = 4, the subgraph of G induced by V (G2 ) ∪ {z } is a K5− which is Z3 -connected; when n2 ≥ 5, G2 is Z3 -connected and hence the subgraph induced by V (G2 ) ∪ {z } is Z3 -connected by Lemma 2.1(7). Thus, if G1 and G2 are both a complete graph, then G is Z3 -connected. Thus, we assume that only one of G1 and G2 is a complete graph. We first assume that G1 is a complete graph and G2 is not a complete graph. In this case, n2 ≥ n/2 ≥ 4.5. Then n2 ≥ 5. By the induction hypothesis, G/G1 is Z3 -connected or G/G1 can be contracted to {K3 , K4 , K4− , G5 } or G/G1 is one of the graphs {G7 , G8 , G9 , G10 , G11 , G12 }. In the first case, G is Z3 -connected by Lemma 2.1; in the second case, G can be contracted to {K3 , K4 , K4− , G5 }. Since G contains at most one 3-vertex, the third case does not happen. Next, assume that G2 is a complete graph and G1 is not a complete graph. Then n1 ≥ n/2 + 1 ≥ 5.5, which implies that n1 ≥ 6. If n2 ≥ 5, then by the induction hypothesis, G/{G2 ∪ {z }} is Z3 -connected or G/{G2 ∪ {z }} can be contracted to {K3 , K4 , K4− , G5 } or G/{G2 ∪ {z }} is one of the graphs {G7 , G8 , G9 , G10 , G11 , G12 }. In the first case, G is Z3 -connected by Lemma 2.1; in the second case, G can be contracted to {K3 , K4 , K4− , G5 }. Since G contains at most one 3-vertex, the third case does not happen. If n2 = 4, then let V (G2 ) = {x, y, w, v}. Without loss of generality, we assume that d(w) = 4, d(v) = 3. In this case, n = n1 + 4 ≥ n/2 + 1 + 4. Then n ≥ 10, n1 ≥ 6. For every pair of nonadjacent vertices z1 and z2 of G1 , {z1 , z2 , v} is a 3-independent set. By the given degree condition, dG1 (z1 ) + dG1 (z2 ) ≥ d(z1 ) + d(z2 ) − 3 ≥ 3n/2 − 6 ≥ n − 4 = n1 . By Theorem 1.3, either G1 is Z3 -connected or G1 is one of the graphs {G7 , G8 , G9 , G10 , G11 , G12 }. In the former case, we can deduce that G is Z3 -connected. Since G contains at most one 3-vertex, the latter case does not happen. Acknowledgment The second author was supported by National Science Foundation of China (11171129). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19]

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