# cut-vertex and Hamiltonian graphs

Given a graph $G$ with no cut-vertices, does it directly imply that $G$ is Hamiltonian?

It is known that if a graph $G$ is nonseparable (thus, no cut-vertices) then every two distinct vertices in $G$ lies in a common cycle.

Is it ALWAYS possible that the cycle referred to that result is a Hamiltonian cycle?

Thanks

• False. See the Petersen graph. – Parcly Taxel Oct 16 '16 at 9:14
• Oh yes. I forgot that one. Thanks – mathislove Oct 16 '16 at 9:17