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Every edge-maximal graph of order at least four with no subdivision of $K_{5}$ or $K_{3, 3}$ is three-connected

Statement

Every finite graph of order at least four that is edge-maximal without a subdivision of $K_{5}$ or $K_{3, 3}$ is three-connected (Vertex cuts, edge cuts, vertex connectivity $κ (G)$ and edge connectivity $λ (G)$ , with conventions for complete and one-vertex graphs, The cardinality $∣ A ∣$ of a finite set). The proof uses the separation structure of In an edge-maximal graph with no $K_{5}$ or $K_{3, 3}$ subdivision, a minimum proper separation of order at most two has an adjacent two-vertex separator and edge-maximal sides, the path form of connectivity A finite graph on at least $k + 1$ vertices is $k$ -connected if and only if every two vertices have $k$ internally disjoint paths, the three-connected planar case Every three-connected graph with no $K_{5}$ or $K_{3, 3}$ minor is planar, and the obstruction exclusion A planar graph contains no subdivision of $K_{5}$ or $K_{3, 3}$ , with minor equivalence from A graph has a $K_{5}$ or $K_{3, 3}$ minor exactly when it has a subdivision of $K_{5}$ or $K_{3, 3}$ as a subgraph.

Facts & Assumptions

Given: An edge-maximal obstruction-free graph $G$ of order at least four.

[L1]

A minimum proper separation of order at most two has separator $K_{2}$ , and both induced sides are edge-maximal without either obstruction (In an edge-maximal graph with no $K_{5}$ or $K_{3, 3}$ subdivision, a minimum proper separation of order at most two has an adjacent two-vertex separator and edge-maximal sides).

[L2]

For a finite graph on at least four vertices, three-connectivity is equivalent to the existence of three internally vertex-disjoint paths between every two vertices (A finite graph on at least $k + 1$ vertices is $k$ -connected if and only if every two vertices have $k$ internally disjoint paths).

[L3]

Every three-connected graph without a $K_{5}$ or $K_{3, 3}$ minor is planar (Every three-connected graph with no $K_{5}$ or $K_{3, 3}$ minor is planar).

[L4]

Excluding subdivisions of $K_{5}, K_{3, 3}$ is equivalent to excluding those two minors (A graph has a $K_{5}$ or $K_{3, 3}$ minor exactly when it has a subdivision of $K_{5}$ or $K_{3, 3}$ as a subgraph).

[L5]

A planar graph contains no subdivision of $K_{5}$ or $K_{3, 3}$ (A planar graph contains no subdivision of $K_{5}$ or $K_{3, 3}$ ).

[L6]

Every plane edge on a cycle is incident with two distinct faces (Face frontiers are unions of whole edges; a cycle edge borders two faces and a bridge borders one).

[L7]

Every facial boundary in a two-connected plane graph is a cycle (Every face of a two-connected plane graph is bounded by a cycle).

Proof

technique · induction

1.1

At order four, edge maximality forces $K_{4}$ , which is three-connected. Assume the assertion for smaller orders and suppose $G$ is not three-connected. By [L2] it has a minimum proper separation of order at most two.

baseL2

1.2

By [L1] the separator is an edge $x y$ , and the two induced sides $G_{1}, G_{2}$ are smaller edge-maximal obstruction-free graphs. By the induction hypothesis, each side is a triangle or three-connected. In the latter case [L4] excludes the forbidden minors and [L3] makes the side planar; a triangle is planar as well. In a plane drawing of each side, [L6] puts $x y$ on a face boundary and [L7] makes that boundary a cycle, so it contains another vertex $z_{i}$ .

ihL1L3L4L6L7

2.1

Make the chosen face of each side the outer face, place the two drawings in opposite closed half-planes, and identify their copies of the boundary edge $x y$ . The two outer boundary arcs complementary to $x y$ then lie on one face of the combined drawing and contain $z_{1}$ and $z_{2}$ . Drawing the missing cross-edge $z_{1} z_{2}$ inside that face gives a planar proper supergraph of $G$ . By [L5] it still contains neither forbidden subdivision, contradicting edge maximality.

step 1.2L5construct

3.1

This contradiction rules out the small separator in step 1.1, so $G$ is three-connected. The induction is complete.

step 1.1step 2.1discharge-induction∎

Depends on

Used by

Kuratowski–Wagner theorem: a finite graph is planar exactly when it has neither a K₅ nor a K_3,3 minor, equivalently neither subdivision Theorem

Dependency tree · two levels

30 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

R. Diestel, Graph Theory, 6th ed., Lemma 4.4.5 (standard reference, not scraped)

Every edge-maximal graph of order at least four with no subdivision of K5K5​ or K3,3K3,3​ is three-connected