Prove that a k-regular connected graph with edge chromatic number = k is 2-connected. Prove that a $k$-regular connected graph with edge chromatic number = $k$ is $2$-connected.
I think from handshaking lemma that k must be odd but i don't see how this helps.
any help much appreciated
 A: Hint:


*

*If you would pick one particular color, edges of that color would form a perfect matching.

*Union of any two disjoint perfect matchings forms a set of disjoint simple cycles spanning the whole graph.

*Assume there is a cut vertex $u$, and its removal would form at least $2$ connected components, pick some two colors $\alpha$ and $\beta$ with edges leading to different components.

*Union of matchings implied by $\alpha$ and $\beta$ forms cycles, one of which passes through $u$, hence, $u$ couldn't have been a cut vertex.


I hope this helps $\ddot\smile$
A: We only have to prove that a graph $G$ that is $k$-regular and $k$-edge-colorable is not $1$-(vertex)-connected, since we know that $G$ is connected and have to prove that it is (at least) $2$-connected. 
First some definitions and observations:
Suppose to the contrary that $G$ is $1$-connected (and not $2$-connected). Then, we can find a vertex $A$ such that $G-\{A\}$ (The graph $G$ without the vertex $A$ and the edges joining $A$) is disconnected. Now look at a component of $G-\{A\}$ and add $A$ to it. Cal that subgraph of $G$ $S$. We know that every vertex in $S$, apart from $A$, has exactly $k$ neighbours. Say that $l=|V(S)|$. We know that $A$ is connected to $S$ and has at least one neighbour not in $S$. Thus, WLOG, we can assume that $A$ has an edge not in $S$ that has color $1$. Also, we can assume that $A$ has no edge not in $S$ that has color $k$.
Now the proof:
Every single vertex of the $l$ vertices in $S$ must be connected to an edge of color $i$ for every $1\leq i\leq k$. We know that $A$ already has an edge of color $1$ in $G-S$. Thus, the $l-1$ remaining vertices all need to have an edge with color $1$, and since an edge connects two vertices, $l-1$ has to be even. But we know that $A$ still needs an edge of color $k$. Thus, we need to add some edges of color $k$ to exactly $l$ points. Thus, $l$ must be even. Since $l\neq l-1\mod 2$, we have a contradiction, and thus, the assumption that $G$ is $1$- (and not $2$)-connected must have been false. We conclude that every $k$-regular graph that is $k$-edge-colorable is at least $2$-vertex-connected.
