CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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

Statement

A planar graph (Plane embeddings of finite simple graphs, their faces, facial boundary walks and lengths (counting a bridge twice), and planar graphs) contains no subgraph that is a subdivision of $K_{5}$ or $K_{3, 3}$ in the sense of Vertex and edge deletion, edge contraction, graph minors, subdivisions and topological minors.

Facts & Assumptions

Given: A planar graph $G$ .

[L1]

$K_{5}$ and $K_{3, 3}$ are nonplanar ( $K_{5}$ and $K_{3, 3}$ are nonplanar).

[F1]

A subdivision repeats edge subdivision zero or more times (Vertex and edge deletion, edge contraction, graph minors, subdivisions and topological minors).

Proof

technique · contradiction

1.1

In a plane drawing of a subdivision, suppressing a degree-two subdivision vertex replaces its two incident polygonal edge arcs by their concatenation and preserves a plane drawing. Repeating this suppression shows that planarity of a subdivision implies planarity of the original graph.

2.1

Suppose $G$ contained a subdivision of $K_{5}$ or $K_{3, 3}$ . A subgraph of a planar graph inherits a plane drawing, and step 1.1 would turn that drawing into a plane drawing of the corresponding original graph, contradicting [L1].

assume-contrastep 1.1L1F1discharge-contradiction∎

Depends on

Used by

Dependency tree · two levels

12 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., Corollary 4.2.11 (standard reference, not scraped)

A planar graph contains no subdivision of K5K5​ or K3,3K3,3​