# Prove that a simple critically 3-chromatic graph without isolated vertices has $\Delta(G)=2$

Can any one help me prove - A simple critically 3-chromatic graph without isolated vertices has $\Delta(G)=2$.

I tried to do it by contradiction and show that if a vertex $v$ has degree 3 or more then the graph $G$ is not critically 3-colourable.

Take a vertex $u$ and its neighbours $v_1,v_2,v_3,...,v_n$. Then $G-u$ or $G-v_i$ is not 2 colourable because... I'm not sure to finish it off? Or I could be heading in the wrong direction?

Thanks.

A critically $3$-chromatic graph is in particular $3$-chromatic, so it is not bipartite. This means it has an odd cycle.
If there was an edge that does not belong to that cycle we could remove it and the graph would still be $3$-chromatic. The same happens if there where a vertex not in the cycle.
Therefore the graph is the cycle, and of course cycles satisfy $\Delta(G)=2$