User:Stixme/Sandbox

Lemma
For any 3-connected graph $G$ with $|G| > 4$, exists an edge $e$ such that $G \thinspace / \thinspace e$ is again 3-connected.

Proof
Proof by contradiction.

Suppose such edge doesn't exist. Then for every edge $xy \in G\:$, $G \thinspace / \thinspace xy$ contains a separating set $S$ such that $|S| \leq 2$.

Since $\kappa \left( G \right) \geq 3$, the contracted $v_{xy} \in G \thinspace / \thinspace xy$ lies in $S$ and $|S| = 2$

Thus, $\exists z, \: z \in G, \: z \notin \{x,y\}$, such that $\{v_{xy}, z\}$ separates $G \thinspace / \thinspace xy$

Then any two vertices separated by $\{v_{xy}, z\}$ in $G \thinspace / \thinspace xy$ are also separated in $G$ by $T:=\{x,y,z\}$.

Since no proper subset $T$ separates $G$, every vertex in $T$ has a neighbour in every component $C$ of $G-T$.

Let us choose the edge $xy$, the vertex $z$, and the component $C$ so that $|C|$ is as small as possible, and pick a neighbour $u$ of $z$ in $C$. By assumption, $G \thinspace / \thinspace zu$ is again not 3-connected, so again there is a vertex $w$ such that $\{ z, u, w \}$ separates $G$, and as before every vertex in $\{ z, u, w \}$ has a neighbour in every component of $G − \{ z, u, w \}$

As $x$ and $y$ are adjacent, $G − \{ z, u, w \}$ has a component $D$ such that $D \cap \{ x, y \} = \emptyset$. Then every neighbour of $u$ in $D$ lies in $C$ (since $u ∈ C$), so $D \cap C \neq \emptyset$ and hence $D \nsupseteqq C$ by the choice of $D$.

This contradicts the choice of $xy, z$ and $C$.