Are all even order graphs with maximum degree $\ge\frac{|V(G)|}{2}$ is Class 1(edge-colorable(chromatic index) with $\Delta(G)$ colors, where $\Delta(G)$ is maximun degree)? Here, $|V(G)|$ denotes the number of vertices in the graph.
I think yes, because, by Erdos-Posa theorem on the number of maximal disjoint circuits in a graph, we have that any graph has a matching of size at least $min(\Delta(G), \frac{|V(G)|}{2})$. Now, consider a regular graph of degree $\Delta(G)$. Then, by the Erdós-Pósa theorem, it will have a perfect matching(as $\Delta(G)\ge\frac{|V(G)|}{2})$, in fact, $\Delta$ perfect matchings(I think), and thus, should be 1-factorizable, that is must be Class 1. Thus, any graph with maximum degree at least $\Delta(G)$ should be Class 1. Any hints. Thanks beforehand.
This is not true. Start with a complete graph on $n-1$ vertices ($n$ even), and then subdivide one edge by adding a new vertex $v$ to make a graph on $n$ vertices. This graph has $(n-1)(n-2)/2+1=\frac{n^2-3n+4}{2}$ edges, and maximum degree $n-2$. Now colour as many edges as you can with $n-2$ colours. Since only two edges meet $v$, only two colour classes can contain $n/2$ edges, and the other classes contain at most $(n-2)/2$ edges each. So there are at most $n+(n-4)(n-2)/2=\frac{n^2-4n+8}{2}$ edges coloured, which is not enough if $n\geq 6$.
For regular graphs this is essentially the $1$-factorisation conjecture, which is known to be true for sufficiently large $n$ by a result of Csaba, Kühn, Lo, Osthus and Treglown. I don't know how large $n$ has to be for the proof to work, but probably very large; I believe it's still unknown whether it's true for all $n$.