Group Connectivity and Matchings

Abstract

Let $\alpha ^\prime (G)$ be the maximum number of independent edges in a graph $G$ and $A$ an abelian group with $|A|\ge 4$. If a graph $G^*$ is obtained by repeatedly contracting nontrivial $A$-connected subgraphs of $G$ until no such a subgraph left, then we say $G$ can be $A$-reduced to $G^*$. In this paper, we show that if $G$ is a 3-edge-connected simple graph and $\alpha ^\prime (G)\le 5$, then either $G$ is $A$-connected or $G$ can be $A$-reduced to the Petersen graph.

1 Introduction

The graphs considered in this paper are finite, loopless, and may have multiple edges. We follow [1] for terminology and notation not defined here.

Let $\alpha ^\prime (G)$ be the maximum number of independent edges in a graph $G$. Obviously $G$ has an $\alpha ^\prime (G)$-matching. For $k\ge 2$, a k-cycle is a cycle of length $k$. For two vertex-disjoint subsets $V_1, V_2\subseteq V(G)$, let $e_G(V_1,V_2)$ ($e(V_1, V_2)$ for short) denote the number of edges of $G$ with one endpoint in $V_1$ and the other endpoint in $V_2$. If $H_1$ and $H_2$ are two vertex-disjoint subgraphs of $G$, we write $e(H_1, H_2)$ instead of $e(V(H_1), V(H_2))$. If $V(H_1)=\{v\}$, we write $e(v, H_2)$ instead of $e(\{v\}, H_2)$. A graph $G$ is trivial if $|V(G)|=1$, and non-trivial otherwise.

Let $G$ be a graph, and $D$ an orientation of $G$. For a vertex $v \in V(G)$, denote by $E^+(v)$ the set of edges with tail at $v$ and $E^-(v)$ the set of edges with head at $v$. Let $A$ be a nontrivial additive abelian group and let $A^*=A-\{0\}$. Define $F(G,A)=\{f~|~f:E(G)\rightarrow A\}$ and $F^*(G, A)=\{f~|~f:E(G)\rightarrow A^*\}$.

For a given function $f\in F(G,A)$, let $\partial f: V(G)\rightarrow A$ be defined by, for all $v\in V(G)$,

$$\begin{aligned} \partial f (v) = \sum \limits _{e \in E^+(v)} f(e) - \sum \limits _{e \in E^-(v)} f(e), \end{aligned}$$

where “ $\sum$” refers to the addition in $A$.

A function $b: V(G)\rightarrow A$ is called an A-valued zero-sum function on $G$ if $\sum _ {v\in V(G)} b(v)=0$. Denoted by $Z(G,A)$ the set of all $A$-valued zero-sum functions on $G$. For a given $b\in Z(G,A)$, a function $f\in F^*(G,A)$ is called a nowhere-zero (A,b)-flow if $G$ has an orientation $D(G)$ such that $\partial f=b$. A graph $G$ is A-connected if for any $b\in Z(G,A), G$ has a nowhere-zero $(A, b)$-flow. A nowhere-zero A-flow is an $(A, 0)$-nowhere-zero flow. More specifically, a nowhere-zero k-flow is a nowhere-zero $Z_k$-flow where $Z_k$ is the cyclic group of order $k$.

For a subset $X \subseteq E(G)$, the contraction $G/X$ is the graph obtained from $G$ by identifying the two ends of each edge in $X$ and then deleting all loops generated in this process. For simplicity, we write $G/e$ for $G/\{e\}$, where $e \in E(G)$. If $H$ is a subgraph of $G$, then $G/H$ denotes $G/E(H)$.

Let $A$ be an abelian group. If a graph $G^*$ is obtained by repeatedly contracting nontrivial $A$-connected subgraphs of $G$ until no such a subgraph left, then we say $G$ can be A-reduced to $G^*$. We also say $G^*$ is the $A$-reduction of $G$.

The concept of group connectivity was introduced by Jaeger et al. [9] as a generalization of nowhere-zero flows. They posed the following longstanding conjecture.

Conjecture 1.1

Every 3-edge-connected graph is $Z_5$-connected.

Many authors were devoted themselves to investigating group connectivity for 3-edge-connected graphs. Jaeger et al. [9] proved that every 3-edge-connected graph is $A$-connected with $|A|\ge 6$. Lai in [11] studied that group connectivity of 3-edge-connected chordal graphs. Lai and Zhang in [14] proved that every 3-edge-connected planar graph is $Z_5$-connected. Yang and Li [19] proved that every 3-edge-connected graph with at most 11 vertices is $Z_5$-connected. For more results, the readers can be referred to [5, 10, 11, 13, 15, 16, 20] and others. However, Conjecture 1.1 is still open.

Jaeger et al. in [9] proved that every 4-edge-connected graph is $A$-connected with $|A|\ge 4$. What is next? Chen [6] proved that if $G$ is a 3-edge-connected graph on at most 11 vertices, then $G$ is collapsible or $G$ is the Petersen graph. Yang and Li [19] proved that if $G$ is a 3-edge-connected graph on at most 11 vertices, $G$ is $A$-connected for $|A|\ge 4$ or $G$ is the Petersen graph. Note that the Petersen graph has five independent edges. For at most five independent edges of a graph, one naturally ask whether this result can be extended without the restriction on the number of vertices. Motivated by Conjecture 1.1 and above observations, we present the following result in this paper.

Theorem 1.2

Let $A$ be an abelian group with $|A|\ge 4$. For any 3-edge-connected simple graph $G$, let $G^*$ be the $A$-reduction of $G$. If $\alpha '(G)\ge 5$, then $G$ is $A$-connected if and only if either $|A|\ge 5$ or $|A|=4$ and $G^*$ is not the Petersen graph.

Lai et al. in [15] proved that if $A$ is an abelian group with $|A|\ge 5$, then the Petersen graph is $A$-connected. The following corollary follows immediately from Theorem 1.2.

Corollary 1.3

Let $A$ be an abelian group with $|A|\ge 5$. If $G$ is a 3-edge-connected simple graph and $\alpha ^\prime (G)\le 5$, Then $G$ is $A$-connected.

The authors in [8] proved that every 3-edge-connected graph with at most 15 vertices is $Z_5$-connected; for $n\ge 17$, every 3-edge-connected graph $G$ with at least ${n-15\atopwithdelims ()2}+31$ edges is $Z_5$-connected; for an abelian group $A$ with $|A|\ge 4$ and $n\ge 13$, if $|E(G)|\ge {n-11\atopwithdelims ()2}+23$, then either $G$ is $A$-connected or $G$ can be $A$-reduced to the Petersen graph.

Such results does not imply Theorem 1.2 in the sense that there is an infinite family of graphs which are constructed as follows. For $k\ge 5$, let $n=5k$. Define a simple graph $G$ with vertex set $V(G)=X\cup Y\cup Z$, where $X=\{x_1, x_2,\ldots , x_5\}, Y=\{y_1, y_2, \ldots , y_5\}$ and $Z=\{z_1,z_2,\ldots ,z_{n-10}\}$, and edge set $E(G)=\cup _{i=1}^5 \{x_iy_i, x_iy_{i-1}, x_iy_{i+1}\}\cup _{i=1}^{n-10} \{v_iy_i, v_iy_{i+1}, v_iy_{i+2}\}$, where both subscripts of $y$ and $z$ are taken modular 5. It is easy to see that $\alpha ^\prime (G)=5$ and $G$ is 3-edge-connected. By Theorem 1.2, $G$ is $Z_5$-connected. However, $|E(G)|=3(n-5)<{n-15\atopwithdelims ()2}+31$ since $n\ge 25$. Therefore, the results in [8] are invalid to the graph $G$.

Corollary 1.3 tells us that Conjecture 1.1 holds for every 3-edge-connected graph with the independent edge number at most 5. This paper is organized as follows: In Sect. 2, we state several known results which are used in the proofs. Theorem 1.2 is proved in Sect. 3.

2 Preliminaries

We first state some known results in [9, 11] on group connectivity as follows.

Lemma 2.1

Let $G$ be a graph and $A$ be an abelian group with $|A|\ge 3$. Then each of the following holds.

• (i) $K_1$ is $A$-connected.

• (ii) If $G$ is $A$-connected and $e\in E(G)$, then $G/e$ is $A$-connected.

• (iii) Let $H$ be a subgraph of $G$. If $H$ is $A$-connected and $G/H$ is $A$-connected, then $G$ is $A$-connected.

• (iv) An $n$-cycle is $A$-connected if and only if $|A|\ge n+1$.

Let $\tau (G)$ be the maximum number of edge-disjoint spanning trees of graph $G$.

Lemma 2.2

([9]) Let $G$ be a graph and $A$ be an abelian group. If $\tau (G)\ge 2$, then $G$ is $A$-connected with $|A|\ge 4$.

Let ${\fancyscript{T}}$ denote a family of graphs such that a graph $G\in {\fancyscript{T}}$ if and only if $\tau (G)\ge 2$ or $G$ is a cycle of length $3$. A graph $G^*$ is called the T-reduction of $G$ if it is obtained from $G$ by repeatedly contracting nontrivial subgraphs of $G$ in ${\fancyscript{T}}$ until no subgraph in ${\fancyscript{T}}$ left. Let $F(G)$ denote the minimum number of additional edges that must be added to $G$ so that the resulting graph has two edge-disjoint spanning trees.

Lemma 2.3

([3, 18]) Let $G$ be a $T$-reduced graph. If $G$ is nontrivial, then each of the following holds.

• (i) $2|V(G)|-|E(G)|\ge 3$ and $\delta (G)\le 3$.

• (ii) $F(G)=2|V(G)|-|E(G)|-2$.

Denote by $O(G)$ the set of all vertices of odd degree in a graph $G$. A graph $G$ is collapsible if for every even set $R\subseteq V(G)$, there is a spanning connected subgraph $H_R$ of $G$ such that $O(H_R)=R$. A graph $G$ is contractible to a graph $H$ if $H$ can be obtained from $G$ by contracting all maximal collapsible subgraphs of $G$.

Theorem 2.4

([4]) Let $G$ be a connected graph. If $F(G)\le 2$, then $G$ is collapsible or $G$ is contractible to a $K_2$ or a $K_{2,t}$ for some integer $t\ge 1$.

Theorem 2.5

([12]) Let $A$ be an abelian group with $|A|=4$. If $G$ is a collapsible, then $G$ is A-connected.

Theorem 2.6

([19]) Let $G$ be a 3-edge-connected simple graph on $n$ vertices. If $n\le 11$ and $A$ is an abelian group with $|A|\ge 4$, then $G$ is $A$-connected or $G$ is the Petersen graph.

Let $M$ be a matching in a graph $G$. A vertex $v$ of $G$ is said to be M-saturated if some edge of $M$ is incident with $v$; otherwise, $v$ is M-unsaturated. A matching $M$ is a maximum matching if $G$ has no matching $M'$ with $|M'|>|M|$. An M-alternating path in $G$ is a path whose edges are alternately in $E(G)\setminus M$ and $M$. An M-augmenting path is an $M$-alternating whose origin and terminus are $M$-unsaturated. Berge characterized the property of a maximum matching in $G$ as follows.

Theorem 2.7

A matching $M$ in $G$ is a maximum matching if and only if $G$ contains no $M$-augmenting path.

An odd component of $G$ is one that has an odd number of vertices. Denote by $q(G)$ the number of odd components of $G$.

Theorem 2.8

(Berge [2] and Tutte [17]) Let $G$ be a graph on $n$ vertices. If

$$\begin{aligned} t={\underset{S\subset V(G)}{max}}\{q(G-S)-|S|\}, \end{aligned}$$

then $|\alpha ^\prime (G)|=\frac{n-t}{2}$.

3 Proof of Theorem  1.2

Following the idea in [7], we can prove next lemma.

Lemma 3.1

Let $G$ be a graph on $n$ vertices. Suppose that $G$ is a $T$-reduced connected graph with $\kappa '(G)\ge 3$. Then $\alpha ^\prime (G)\ge min \{\frac{n-1}{2},\frac{n+3}{3}\}$.

Let $G$ be a graph. For an integer $i\ge 1$, define $D_i(G)=\{v\in V(G): d_G(v)=i\}$.

Lemma 3.2

Suppose that $G$ is a $T$-reduced graph on $n=12$ vertices with $\kappa '(G)\ge 3$. If one of the following holds:

• (i) $|D_3|=12$, or

• (ii) $|D_3|=11$ and $|D_5|=1$, or

• (iii) $|D_3|=10$ and $|D_4|=2$, then $\alpha ^\prime (G)=6$.

Proof

By contradiction, suppose that $\alpha ^\prime (G)\le 5$. In order to complete our proof, we need to establish the following claims.

Claim 1. $\alpha ^\prime (G)=5$.

Proof of Claim 1. Suppose otherwise that $\alpha ^\prime (G)\le 4$. Since $\frac{n-1}{2}\ge \frac{n+3}{3}$ for $n=12$, by Lemma 3.1, $\alpha ^\prime (G)\ge \frac{n+3}{3}$. It follows that $n\le 9$, contrary to that $n=12$. This proves Claim 1.

By Claim 1, let $M=\{x_1y_1,x_2y_2,x_3y_3,x_4y_4,x_5y_5\}$ be a maximum matching of graph $G$. By Theorem 2.7, $G$ contains no $M$-augmenting path. Since $n=12$, let $\{u, v\}=V(G)\setminus V(M)$. By Claim 1, $u$ is nonadjacent to $v$. We claim that if $u$ is adjacent to $x_i$ (or $y_i$), then $v$ is nonadjacent to $y_i$ (or $x_i$). Otherwise, let $P=ux_iy_iv$. Then $P$ is an $M$-augmenting path of $G$, contrary to Theorem 2.7. Since $G$ is a $T$-reduced graph, $G$ contains no 3-cycle. We assume, without loss of generality, that $N(u)\cup N(v)\subseteq \{x_1,x_2,x_3,x_4,x_5\}$. Let $X=N(u)\cup N(v)$ and $Y=\{y_i|x_iy_i\in M, x_i\in X\}$ and $N_u=\{y_i|x_iy_i\in M, x_i\in N(u)\setminus N(v)\}$ and $N_v=\{y_i|x_iy_i\in M, x_i\in N(v)\setminus N(u)\}$. Let $H_X$ and $H_Y$ denote the subgraphs of $G$ induced by $X$ and $Y$, respectively.

Claim 2. If $x_i\in N(u)\cap N(v)$, then $e(y_i,Y\setminus \{y_i\})=0$.

Proof of Claim 2. By way of contradiction, we assume that there exists a vertex $y_j\in Y$ such that $y_iy_j\in E(G)$. By the definition of $Y$, it implies that $x_j\in N(u)\cup N(v)$, where $i\not =j$. Then either $P=ux_iy_iy_jx_jv$ or $P=vx_iy_iy_jx_ju$ is an $M$-augmenting path of $G$, contrary to Theorem 2.7. This proves Claim 2.

Claim 3. $e(N_u,N_v)=0$.

Proof of Claim 3. Suppose otherwise that there exist two vertices $y_i\in N_u$ and $y_j\in N_v$ such that $y_iy_j\in E(G)$. Define $P=ux_iy_iy_jx_jv$. Then $P$ is an $M$-augmenting path of $G$, contrary to Theorem 2.7. Claim 3 is proved.

Claim 4. Suppose that $x_i\notin N(u)\cup N(v)$ and $x_j\in N(u)\cap N(v)$. If $y_iy_j\in E(G)$, then $e(x_i,Y)=0$.

Proof of Claim 4. Suppose otherwise that there exists a vertex $y_k\in Y$ such that $x_iy_k\in E(G)$. Since $G$ contains no 3-cycle, $x_i$ is nonadjacent to $y_j$. It follows that $y_k\ne y_j$. Then either $P=ux_jy_jy_ix_iy_kx_kv$ or $P=vx_jy_jy_ix_iy_kx_ku$ is an $M$-augmenting path of $G$, contrary to Theorem 2.7. This proves Claim 4.

Claim 5. $|N(u)\cup N(v)|\ge 4$.

Proof of Claim 5. Suppose otherwise that $|N(u)\cup N(v)|\le 3$. Since $G$ is 3-edge-connected, $d(u)=d(v)=3$ and $|N(u)\cup N(v)|=3$. Without loss of generality, we assume that $N(u)=N(v)=\{x_1, x_2, x_3\}$. Then $Y=\{y_1,y_2,y_3\}$. By Claim 2, $e(y_i,Y\setminus \{y_i\})=0$ for all $i=1,2,3$.

Suppose that $\{x_1,x_2,x_3\}$ has at least two vertices, each of which is degree of $3$. We assume, without loss of generality, that $d(x_2)=d(x_3)=3$. In this case, all neighbors of $y_1$ are in $\{x_1, x_4, y_4, x_5, y_5\}$. Since $G$ is a $T$-reduced graph, $y_1$ is adjacent to exactly one of $x_4$ and $y_4$ and exactly one of $x_5$ and $y_5$. We assume, without loss of generality, that $y_1y_4, y_1y_5\in E(G)$. By Claim 4, $e(x_4,Y)=0$ and $e(x_5,Y)=0$.

If $d(x_1)=3$, then $N(x_j)\subseteq \{y_4, y_5\}$ for $j=4, 5$. Thus, $d(x_4)\le 2$ and $d(x_5)\le 2$, contrary to that $\kappa '(G)\ge 3$. Therefore, $d(x_1)>3$. If $d(x_1)=4$, then $x_1$ is adjacent to at most one of $x_4$ and $x_5$. Thus, $N(x_4)\subseteq \{y_4, y_5\}$ or $N(x_5)\subseteq \{y_4, y_5\}$. It follows that either $d(x_4)\le 2$ or $d(x_5)\le 2$, contrary to that $\kappa '(G)\ge 3$. Thus, we assume that $d(x_1)=5$. In this case, by our hypothesis (ii), $d(x_4)=d(x_5)=3$. We conclude that $x_4x_1,x_4y_5, x_5x_1, x_5y_4\in E(G)$. By $\kappa '(G)\ge 3$ and Claim 2, $y_2y_4, y_2y_5, y_3y_4,y_3y_5\in E(G)$. Hence, $d(y_4)=d(y_5)=5$, contrary to our assumption (ii).

Therefore, suppose that $\{x_1,x_2,x_3\}$ contains exactly one vertex of degree 3. By the hypothesis (iii), we may assume that $d(x_1)=d(x_2)=4$ and $d(x_3)=3$. It follows that there exists a vertex $y_i\in Y$ such that $e(y_i,X)=1$ for $i\in \{1, 2, 3\}$. By Claim 2, $e(y_i, Y\setminus \{y_i\})=0$. Since $\kappa '(G)\ge 3$, $e(y_i, \{x_4, y_4, x_5, y_5\})\ge 2$. Since $G$ is $T$-reduced, $y_i$ is adjacent to exactly one of $x_4$ and $y_4$ and adjacent to exactly one of $x_5$ and $y_5$. Without loss of generality, we assume that $y_iy_4, y_iy_5\in E(G)$. By Claim 4, $e(x_4,Y)=0$ and $e(x_5,Y)=0$. Since $d(x_4)=d(x_5)=3$, $x_4$ is adjacent to $x_1$ or $x_2$ and $x_5$ is adjacent to $x_2$ or $x_1$. It follows that $x_4y_5,x_5y_4\in E(G)$. By Claim 2 and $\kappa '(G)\ge 3$, $y_1y_4,y_1y_5,y_2y_4, y_2y_5, y_3y_4,y_3y_5\in E(G)$. Therefore, $d(y_4)=d(y_5)=5$, contrary to the hypothesis of our lemma. This completes the proof of Claim 5.

Claim 6. $|N(u)\cap N(v)|\ge 3$.

Proof of Claim 6. Suppose otherwise that $|N(u)\cap N(v)|\le 2$. On the other hand, $|N(u)\cap N(v)|\ge 1$ since $N(u)\cup N(v)\subseteq \{x_1, \ldots , x_5\}$ and $G$ is 3-edge-connected. Assume first that $|N(u)\cap N(v)|=1$. Since $G$ is 3-edge-connected, $d(u)=d(v)=3$. Without loss of generality, assume that $N(u)=\{x_1,x_2,x_3\}$ and $N(v)=\{x_1,x_4,x_5\}$. Recall that $X=\{x_1,x_2,x_3,x_4,x_5\}$ and $Y=\{y_1,y_2,y_3,y_4,y_5\}$. In this case, $N_u=\{y_2,y_3\}$ and $N_v=\{y_4,y_5\}$. By Claims 2 and 3, $H_Y$ contains at most two edges $y_2y_3$ and $y_4y_5$. Since $\kappa '(G)\ge 3$ and $G$ is $T$-reduced, there are at least $11$ edges from $Y$ to $X$. By our assumption, we just have three cases: $|D_3|=12$; $|D_3|=11$ and $|D_5|=1; |D_3|=10$ and $|D_4|=2$. If $|D_3|=12$, then there are at most 9 edges from $X$ to $Y$, a contradiction. If $|D_3|=11$ and $|D_5|=1$ or $|D_3|=10$ and $|D_4|=2$, then there are at most 11 edges from $X$ to $Y$. Thus, $e(X,Y)=11$. It follows that $H_X$ contains no edges and $H_Y$ contains exactly two edges $y_2y_3$, $y_4y_5$ and $d(y_i)=3$ where $i=1,\ldots ,5$. Since $d(y_1)=3$, there are two vertices in $\{x_2,x_3,x_4,x_5\}$ adjacent to $y_1$. If $y_1x_2\in E(G)$, then $P=vx_1y_1x_2u$ is an $M$-augmenting path of $G$, contrary to Theorem 2.7. Similarly, we can prove that $y_1$ is not adjacent to one of $\{x_3, x_4, x_5\}$. Thus, we conclude that $|N(u)\cap N(v)\not =1$.

Next, we assume that $|N(u)\cap N(v)|=2$. It follows that either $3\le d(u)\le 4$ and $d(v)=3$ or $d(u)=3$ and $3\le d(v)\le 4$. By symmetry, we consider two cases: $d(u)=d(v)=3; d(u)=4$ and $d(v)=3$.

In the case that $d(u)=d(v)=3$. Without loss of generality, assume that $N(u)=\{x_1, x_2, x_3\}$ and $N(v)=\{x_1, x_2, x_4\}$. Then $Y=\{y_1,y_2,y_3,y_4\}$. By Claims 2 and 3, $H_Y$ contains no edges.

We claim that for each $j\in \{1, 2, \ldots , 5\}, d(y_j)=3$. Suppose otherwise that there is some $j_0\in \{1, 2, \ldots , 5\}$ such that $d(j_0)\ge 4$. In the case that $j_0\in \{1, 2, 3, 4\}$. Since $G$ is $T$-reduced, for each $j\in \{1, 2, 3, 4\}, e(y_j, \{x_5, y_5\})\le 1$. Since $|D_3|\ge 10, \{x_1, x_2, \ldots , x_4\}$ contains at most one vertex, say $x_{i_0}$, of degree at least 4. Thus there are at least $\sum _{j=0}^4d(y_j)-4=9$ edges from $\{y_1, y_2, y_3, y_4\}$ to $\{x_1, x_2, x_3, x_4\}$. On the other hand, since $\Delta (G)\le 5$, there are at most $\sum _{i=0}^4d(x_i)-6=3+3+3+4-6=7$ edges from $\{x_1, x_2, x_3, x_4\}$ to $\{y_1, y_2, y_3, y_4\}$. This is a contradiction. In the case that $j_0=5$. Let $e(y_5, \{y_1, y_2, y_3, y_4\})=l$. Since $G$ is $T$-reduced, $e(x_5, \{y_1, y_2, y_3, y_4\})\le 4-l$. Since $3\le d(x_5)\le 4$, $e(y_5, \{y_1, y_2, y_3, y_4\})\in \{0, 1, 2\}$ and $l\le 2$. It follows that $e(y_5, \{x_1, x_2, x_3, x_4\})\ge 1$. Thus there are at least $\sum _{j=0}^4d(y_j)-4=8$ edges from $\{y_1, y_2, y_3, y_4\}$ to $\{x_1, x_2, x_3, x_4\}$. On the other hand, since $e(y_5, \{x_1, x_2, x_3, x_4\})\ge 1$, there are at most $\sum _{i=0}^4d(x_i)-6=3+3+3+4-5=7$ edges from $\{x_1, x_2, x_3, x_4\}$ to $\{y_1, y_2, y_3, y_4\}$. This is a contradiction.

By assumption (i), (ii) and (iii), $d(y_5)=3, e(y_5,Y)=2, e(\{x_1,x_2,x_3,x_4,x_5\},\{y_1,y_2,y_3,y_4,y_5\}) =11$ and $H_X$ contains no edges and $e(x_5,X)=0$. If $y_5$ is adjacent to either $y_1$ or $y_2$, then by Claim 4, $e(x_5,Y)=0$. This implies $d(x_5)=1$, a contradiction. Thus, $y_5$ is adjacent to $y_3$ and $y_4$. Since $d(x_5)\ge 3$ and $G$ is a $T$-reduced graph, $x_5y_1,x_5y_2\in E(G)$. Then $P=ux_1y_1x_5y_5y_4x_4v$ is an $M$-augmenting path of $G$, a contradiction.

In the case that $d(u)=4$ and $d(v)=3$. By our assumption (iii), $|D_4|=2$. In this case, let $N(u)=\{x_1, x_2, x_3, x_4\}$ and $N(v)=\{x_1, x_2, x_5\}$. By Claims 2 and 3, $H_Y$ contains at most one edge $y_3y_4$. It follows that there are at least $\sum _{j=0}^5d(y_j)-2\ge 3\times 5-2=13$ edges from $Y$ to $X$. On the other hand, there are at most $\sum _{i=0}^5d(x_i)-7\le 3\times 4+4-7=9$ edges from $X$ to $Y$. This is a contradiction. This completes the proof of Claim 6.

We are ready to complete our proof. By Claim 6, $|N(u)\cap N(v)|\ge 3$. Assume first that $|N(u)\cap N(v)|=4$. In this case, by our assumption (iii), $d(u)=d(v)=4$. Without loss of generality, assume that $N(u)=N(v)=\{x_1, x_2, x_3,x_4\}$. Then $Y=\{y_1,y_2,y_3,y_4\}$. By Claim 2, $H_Y$ contains no edges. There are at most 7 edges from $\{x_1,x_2,x_3,x_4,x_5\}$ to $\{y_1,y_2,y_3,y_4,y_5\}$. Since $d(y)=3$, $e(y_5, \{y_1, y_2, y_3, y_4\})\le 2$. Therefore, there are at least $3\times 5-4=11$ edges from $\{y_1, \ldots , y_5\}$ to $\{x_1, \ldots , x_5\}$. This is a contradiction.

Therefore, $|N(u)\cap N(v)|=3$. If $d(u)=d(v)=3$, then $|N(u)\cup N(v)|=3$, contrary to Claim 5. Therefore, we assume, without loss of generality, that $d(u)\ge 4$. If $d(u)=5$, then by our assumption (ii), $|D_3|=11$. By Claim 2, $H_Y$ contains at most one edge. It follows that there are at least 13 edges from $Y$ to $X$. However, there are at most 7 edges from $X$ to $Y$, a contradiction. If $d(u)=4$ and $d(v)=3$, then by our assumption (iii), $|D_4|=2$. By Claim 2, $H_Y$ contains no edges. Without loss of generality, assume that $x_5\notin X$. If $d(x_{i_0})=4$ for some $i_0\in \{1,\ldots ,5\}$, then there are at most 9 edges from $\{x_1,x_2,x_3,x_4,x_5\}$ to $\{y_1,y_2,y_3,y_4,y_5\}$. In this case, since $d(y_5)=3, e(y_5, \{y_1, \ldots , y_4\})\le 2$. It follows that there are at least 11 edges from $\{y_1, \ldots , y_5\}$ to $\{x_1, \ldots , x_5\}$. This is a contradiction. If $d(y_{j_0})=4$ for some $j_0\in \{1,\ldots ,5\}$, then there are at most 8 edges from $\{x_1,x_2,x_3,x_4,x_5\}$ to $\{y_1,y_2,y_3,y_4,y_5\}$. Similarly, there are at least 9 edges from $\{y_1, \ldots , y_5\}$ to $\{x_1, \ldots , x_5\}$. This is a contradiction. If $d(u)=d(v)=4$, then by Claim 2, $H_Y$ contains no edges. Similarly, there are $15$ edges from $Y$ to $X$, yet there are at most 7 edges from $X$ to $Y$, a contradiction. $\square$

Lemma 3.3

Let $A$ be an abelian group with $|A|\ge 4$. Suppose that $G$ is a $T$-reduced graph with $n=12$ and $\kappa '(G)\ge 3$. If $|F(G)|\le 2$, then $G$ is A-connected.

Proof

Suppose otherwise that $G$ is not A-connected. Since $G$ is 3-edge-connected, $|D_1|=|D_2|=0$. We first claim that $|D_3|\le 11$. Suppose otherwise that $|D_3|=12$. Then by Lemma 2.3(ii), $F(G)=2|V(G)|-|E(G)|-2=4$, contrary to that $|F(G)|\le 2$. Since $|D_3|\le 11$, it implies that there exists a vertex $u$ with degree at least $4$. Let $N(u)=\{u_1,u_2,\ldots ,u_k\}$ where $k\ge 4$. Since $d(u_i)\ge 3$ and $G$ contains no 3-cycles, there exist $u_i$ and $u_j$ such that $|N(u_i)\cap N(u_j)\setminus \{u\}|\ge 1$. Let $v\in N(u_i)\cap N(u_j)\setminus \{u\}$. It follows that $G$ contains a 4-cycle: $uu_ivu_ju$. Contracting this 4-cycle and repeatedly contracting all cycles of length at most 4 generated in process, we obtain the resulting graph, denoted by $G_1$, which is also 3-edge-connected. Then $|V(G_1)|\le 12-3=9$. By Theorem 2.6, $G_1$ is A-connected for $|A|\ge 5$. By Lemma 2.1, $G$ is $A$-connected for $|A|\ge 5$. On the other hand, since $F(G)\le 2$ and $G$ is 3-edge-connected, $G$ is collapsible by Theorem 2.4. By Theorem 2.5, $G$ is A-connected for $|A|=4$. We thus conclude that $G$ is $A$-connected for $|A|\ge 4$. $\square$

Lemma 3.4

Let $A$ be an abelian group with $|A|\ge 4$. Suppose that $G$ is a $T$-reduced graph with $n=12$ and $\kappa '(G)\ge 3$. If $\alpha ^\prime (G)\le 5$, then $G$ is A-connected.

Proof

Since $\alpha ^\prime (G)\le 5, |D_3|\le 11$ by Lemma 3.2(i). Since $G$ is 3-edge-connected, $|D_1|=|D_2|=0$.

In the case that $|D_3|\le 9$. Let $|D_3|=m\le 9$. Then $2|E(G)|\ge 3m+4(12-m)=48-m\ge 39$, which implies that $|E(G)|\ge 20$. By Lemma 2.3(ii), $F(G)=2|V(G)|-|E(G)|-2\le 2$. By Lemma 3.3, $G$ is A-connected.

In the case that $|D_3|=10$. Since $\alpha ^\prime (G)\le 5$, $|D_4|=t\le 1$ by Lemma 3.2(ii). Therefore, $2|E(G)|\ge 30+4t+5(12-10-t)=40-t\ge 39$, which implies that $|E(G)|\ge 20$. By Lemma 2.3(ii), $F(G)=2|V(G)|-|E(G)|-2\le 2$. By Lemma 3.3, $G$ is A-connected.

We are left to the case that $|D_3|=11$. Since $\alpha ^\prime (G)\le 5$, $|D_5|=0$ by Lemma 3.2(iii). Since every graph contains even number of vertices with odd degree, $G$ contains a vertex with odd degree at least 7. Thus, $2|E(G)|\ge 33+7=40$. Thus, $|E(G)|\ge 20$. By Lemma 2.3(ii), $F(G)=2|V(G)|-|E(G)|-2\le 2$. By Lemma 3.3, $G$ is A-connected. $\square$

Lemma 3.5

Let $A$ be an abelian group with $|A|\ge 4$. Suppose that $G$ is a $T$-reduced graph with $n\le 12$ and $\kappa '(G)\ge 3$. If $\alpha ^\prime (G)\le 5$, then either $G$ is $A$-connected or $G$ is the Petersen graph.

Proof

If $n=12$, by Lemma 3.4, then $G$ is $A$-connected. If $n\le 11$, by Theorem 2.6, then either $G$ is $A$-connected or $G$ is the Petersen graph. $\square$

Note that if for an abelian group $A$ with $|A|\ge 4$, a $T$-reduction of $G$ is $A$-connected, then $G$ is $A$-connected. By Lemma 2.2, if $G$ can be $T$-reduced to $G^*$, then $G$ can be $A$-reduced to $G^*$. We obtain next lemma immediately from Lemma 3.5.

Lemma 3.6

Let $A$ be an abelian group with $|A|\ge 4$. Suppose that $G$ is an $A$-reduced graph with $n\le 12$ and $\kappa '(G)\ge 3$. If $\alpha ^\prime (G)\le 5$, then either $G$ is $K_1$ or $G$ is the Petersen graph.

Lemma 3.7

Let $A$ be an abelian group with $|A|\ge 4$. Suppose that $G$ is a nontrivial $A$-reduced graph with $\kappa '(G)\ge 3$. Then $\alpha ^\prime (G)\ge \frac{n+3}{3}$.

Proof

Since $G$ is nontrivial, $G$ is the Petersen graph when $n\le 11$ by Theorem 2.6. Therefore, either $G$ is the Petersen graph or $n\ge 12$. Since $\frac{n+3}{3}\le \frac{n-1}{2}$ for $n\ge 9$, we are done by Lemma 3.1. $\square$

Proof of Theorem 1.2

Let $G^*$ be the $A$-reduced graph of $G$. Since the independence edges number and edge connectivity will not increase under reduction, $G^*$ is also 3-edge-connected with $\alpha ^\prime (G^*)\le 5$. If $G^*$ is trivial, then we are done. Therefore, we assume that $G^*$ is nontrivial. By Lemma 3.7, $\alpha ^\prime (G^*)\ge \frac{|V(G^*)|+3}{3}$. Thus, $|V(G^*)|\le 12$. By Lemma 3.6, $G^*$ is the Petersen graph. Therefore, $G$ can be $A$-reduced to the Petersen graph. $\square$