CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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Every graph has the Erdős–Hajnal property if and only if every prime graph does

Statement

The following are equivalent.

Every finite simple graph has the Erdős–Hajnal property (The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class).
Every prime graph (Prime graphs: those whose only modules are the trivial ones) has the Erdős–Hajnal property.

Facts & Assumptions

Given: The two assertions above.

[L1]

A graph on at least two vertices is prime exactly when it is not obtained by substituting one graph on at least two vertices for a vertex of another such graph (A graph with at least two vertices is prime exactly when it is not obtained by substituting one graph on at least two vertices for a vertex of another graph on at least two vertices).

[L2]

If $H_{1}$ and $H_{2}$ have the Erdős–Hajnal property, then the graph obtained by substituting $H_{2}$ for a vertex of $H_{1}$ has the Erdős–Hajnal property (Alon–Pach–Solymosi: if $H_{1}$ and $H_{2}$ have the Erdős–Hajnal property, so does the graph obtained from $H_{1}$ by substituting $H_{2}$ for a vertex).

[L3]

A graph on $0$ or $1$ vertices is prime, since every module is trivial (Prime graphs: those whose only modules are the trivial ones, The cardinality $∣ A ∣$ of a finite set).

Proof

technique · induction

1.1

If every graph has the Erdős–Hajnal property, then in particular every prime graph has it.

given

1.2

For the converse direction, assume every prime graph has the Erdős–Hajnal property, and induct on the number of vertices of a graph $G$ .

base

1.3

If $∣ V (G) ∣ \leq 1$ , then $G$ is prime by [L3], so the assumption covers $G$ .

L3base

1.4

Fix $n \geq 2$ and assume inductively that every graph with fewer than $n$ vertices has the Erdős–Hajnal property whenever every prime graph does. Let $G$ have $n$ vertices. If $G$ is prime, the assumption on prime graphs covers it. Otherwise [L1] gives graphs $G_{1}, G_{2}$ with at least two vertices each such that $G = G_{1} [a \to G_{2}]$ for some vertex $a$ of $G_{1}$ .

L1ih

2.1

In the non-prime case of step 1.4, both $G_{1}$ and $G_{2}$ have fewer than $n$ vertices, so the induction hypothesis gives the Erdős–Hajnal property for both, and then [L2] gives it for $G$ .

step 1.4L2ih

3.1

Steps 1.3, 1.4 and 2.1 prove that every $n$ -vertex graph has the Erdős–Hajnal property, so the induction closes.

step 1.3step 1.4step 2.1discharge-induction∎

Depends on

Used by

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Dependency tree · two levels

39 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

Y. Huang, Q. Ju, and X. Zhou, Erdős-Hajnal beyond the five-vertex path, sec. 1.2 (standard reference, not scraped)