We show that if $G$ is a simple triangle-free graph with $n\geq 3$ vertices, without a perfect matching, and having a minimum degree at least $\frac{n-1}{2}$, then $G$ is isomorphic either to $C_5$ or to $K_{\frac{n-1}{2},\frac{n+1}{2}}$.
We consider finite undirected graphs that do not contain loops or multiple edges. Let V (G) and E(G) denote the sets of vertices and edges of G, respectively. The degree of a vertex v ∈ V (G) is denoted by d G (v) and the diameter of G by diam(G). For a graph G, let δ(G) and ∆(G) denote the minimum and maximum degree of G, respectively. For n ≥ 3, let C n denote the cycle of length n. The cycle C 3 is called a triangle. For m, n ≥ 1, let K m,n denote the complete bipartite graph one part of which has m vertices and the other part n vertices. Terms and concepts that we do not define can be found in [6].
It is well-known that triangle-free graphs play an important role in graph theory. One of the first results concerns triangle-free graphs is the following theorem of Mantel [4].
Note that the upper bound in Theorem 1 is sharp for the complete bipartite graph
4 ⌋ edges. Clearly, every bipartite graph is a triangle-free graph. On the other hand, in 1974, Andrásfai, Erdős and Sós [1] found the minimum degree condition which forces a triangle-free graph to be bipartite.
Theorem 2 If G is a simple triangle-free graph with n vertices and δ(G) > 2 5 n, then G is bipartite. Also, a similar result for triangle-free graphs was obtained by Erdős, Fajtlowits and Staton [2] in 1991.
Theorem 3 If G is a simple triangle-free graph with no three vertices having equal degree, then G is bipartite.
In 1989, Erdős, Pach, Pollack and Tuza [3] investigated the connection between the diameter and minimum degree of connected triangle-free graphs. In particular, they proved the following
.
In this short note we prove that if G is a simple triangle-free graph with n ≥ 3 vertices, without a perfect matching, and having a minimum degree at least n-1 2 , then G is isomorphic either to C 5 or to
.
Theorem 5 Let G be a graph on n ≥ 3 vertices satisfying the conditions:
u 2 , . . . , u r-1 . Since r -1k < r -2, there is an edge u 1 v j ∈ E(G), 1 ≤ j ≤ k. Now, note that w, u 1 and v j form a triangle, which contradicts (c).
Thus, r = 2. Since n = 2r + 1, we imply that G is isomorphic to C 5 . The proof of the theorem is completed.
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