# $G$ is 2-vertex-connected graph if and only if every 2 vertices from $G$ lie on a simple cycle.

$G$ is $2$-vertex-connected graph if and only if every $2$ vertices from $G$ lie on a simple cycle.

$\implies$

If $G$ is $2$-vertex connected graph it means for every $v \in V(G), \deg(v) \ge 2$ because if exist any $v_1$ and $\deg v_1 = 1$ then we can erase neighboring vertex and graph will not connected so won't be 2-vertex connected. So if $v \in V(G), \deg(v) \le 2$ then it exist cycle. $\deg_v \ge 2$ so if we start in $v$ we have to back to this vertex.

$\impliedby$ I don't know.

• What definition of two-connected are you using? $G$ is connected and after removing any single vertex from $G$ it is still connected? Specifically, is $K_2$ 2-connected? Jun 30, 2013 at 11:33

Take two vertices $u$ and $v$. We want to find a simple cycle containing them. As you've noted, each vertex must have at least two neighbours. So let $u_1$ and $u_2$ be two neighbours of $u$, and consider the graph we get from deleting $u$. We know by assumption that it is still connected, so we can find paths from $u_1$ to $v$ and $u_2$ to $v$. Joining these together will give us a simple cycle containing $u$ and $v$, so we're done — provided the two paths don't intersect. Can every such path have an intersection? Explain why not.
For the reverse implication, take a vertex $x$ which we want to delete, and two vertices $a$ and $b$ which we want to connect in the resulting graph. Using the fact that we can conjure up a simple cycle containing $a$ and $b$, show that we can get a path in $G \setminus \{ x \}$ containing $a$ and $b$.