TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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Every prime graph on at least four vertices contains an induced P_4

Statement

If $G$ is a prime graph with at least four vertices, then $G$ contains an induced copy of $P_{4}$ .

Facts & Assumptions

Given: A prime graph $G$ with $∣ V (G) ∣ \geq 4$ .

[L1]

A prime graph has no nontrivial module (Prime graphs: those whose only modules are the trivial ones, Modules of a graph, and the trivial modules).

[L2]

A graph is a cograph if and only if it is $P_{4}$ -free (The cographs are exactly the P_4-free graphs).

[L3]

Every nontrivial cograph is disconnected or has disconnected complement (Every nontrivial cograph is disconnected or has disconnected complement).

[L4]

Every union of connected components is a module, and so is every union of anticomponents (Every union of connected components is a module, and so is every union of anticonnected components).

[F1]

If a partition of a set with at least four elements has at least two nonempty parts, then some proper union of its parts has cardinality between $2$ and $∣ V (G) ∣ - 1$ : either one part already has at least two elements, or else all parts are singletons and the union of two of them does.

Proof

technique · direct

1.1

Suppose for contradiction that $G$ contains no induced $P_{4}$ . Then [L2] shows that $G$ is a cograph. Since $∣ V (G) ∣ \geq 4$ , the graph is nontrivial, so [L3] gives that $G$ is disconnected or $\overline{G}$ is disconnected.

L2L3givenassume-contra

2.1

If $G$ is disconnected, its connected components form a partition of $V (G)$ into at least two nonempty parts. By [F1], choose a proper union $M$ of component vertex sets with $2 \leq ∣ M ∣ \leq ∣ V (G) ∣ - 1$ . Then [L4] makes $M$ a module of $G$ , and the cardinality bounds say that it is nontrivial. This contradicts [L1].

step 1.1L1L4F1discharge-contradiction

2.2

If $\overline{G}$ is disconnected, then the anticomponents of $G$ form a partition of $V (G)$ into at least two nonempty parts. Again [F1] gives a proper union $M$ of anticomponent vertex sets with $2 \leq ∣ M ∣ \leq ∣ V (G) ∣ - 1$ , and [L4] makes $M$ a nontrivial module of $G$ , contradicting [L1].

step 1.1L1L4F1discharge-contradiction

3.1

Both alternatives from step 1.1 are impossible, so the assumption was false. Therefore $G$ contains an induced copy of $P_{4}$ .

step 2.1step 2.2∎

Depends on

Used by

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Sources

Tero Harju, Lecture Notes on Combinatorial Structures in Graph Theory, Theorem 5.1 and Section 5.2 (standard reference, not scraped)