# A circuit without K4 minor has a cutset of size at most 2

I want to prove the following claim:

A connected graph, $$G=(V,E)$$ with $$|V|\geq 4$$ and without a $$K_4$$-minor has a cut-set of cardinality at most $$2$$.

For the proof, I divided the problem to multiple cases:

• $$G$$ is a tree
• $$G$$ is a circuit
• G includes a circuit

I could easily show the first case, but stuck at the second case. How can we prove the part when $$G$$ is itself a circuit, with at least $$4$$ notes, and without a $$K_4$$ minor?

Every $$3$$-connected graph contains a $$K_4$$-minor.
Proof: Let $$G$$ be a $$3$$-connected graph with no $$K_4$$ as subgraph. Assume two distinct vertices $$u,v \in V(G)$$. $$G$$ is $$3$$-connected so there exist three disjoint paths $$P$$,$$Q$$ and $$R$$ from $$u$$ to $$v$$, and vertices $$p \in V(P)−\{u, v\}$$ and $$q \in V(Q) − \{u, v\}$$. By connectivity there exists a path $$S$$ from $$p$$ to $$q$$ in $$G-\{u,v\}$$. Now let S be the shortest such path. If $$z$$ in the interior of $$S$$ belongs to $$V(P) \cup V(Q) \cup V(R)$$, then $$S$$ is not a shortest path, and a subpath of S with ends $$p$$ and $$z$$ or $$q$$ and $$z$$ contradicts the choice of $$S$$. therefore $$S$$ is internally disjoint from $$P$$, $$Q$$, and $$R$$. Now $$P \cup Q \cup R \cup S$$ is a $$K_4$$-subdivision in $$G$$, and it's contradiction!
Corollary: If $$G$$ be a graph with no $$K_4$$-minor then $$G$$ contains a vertex of degree at most $$2$$.
Therefore $$G$$ has a cut-set of cardinality at most $$2$$
• A $3$-connected planar graph is a circuit graph, and moreover, a $3$-connected planar graph with one vertex removed is also a circuit graph. If $G$ is $3$-connected graph it contains $k_4$-minor otherwise Menger's theorem says that in a finite graph, the size of a minimum cut set is equal to the maximum number of disjoint paths that can be found between any pair of vertices. Therefore $G$ has a cut-set of cardinality at most $2$. Commented Mar 10, 2019 at 21:55