LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Distinct minimal normal subgroups centralize one another

Statement

Let $G$ be a finite group, and let $M, N ⊴ G$ be distinct minimal normal subgroups. Then every element of $M$ commutes with every element of $N$ . Equivalently, $[M, N] = 1$ .

Facts & Assumptions

Given: A finite group $G$ and distinct minimal normal subgroups $M, N ⊴ G$ .

[L1]

Using the convention of Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ , put $[M, N] : = ⟨ [m, n] = m n m^{- 1} n^{- 1} : m \in M, n \in N ⟩ .$

[A1]

Because $M$ and $N$ are normal in $G$ , the subgroup $[M, N]$ is normal in $G$ and is contained in both $M$ and $N$ .

Proof

technique · direct

1.1

By [A1], the subgroup $[M, N]$ is a normal subgroup of $G$ contained in $M$ . Since $M$ is minimal normal, either $[M, N] = 1$ or $[M, N] = M$ .

givenA1

2.1

The same argument with $N$ shows that either $[M, N] = 1$ or $[M, N] = N$ . Because $M \neq N$ , the subgroup $[M, N]$ cannot equal both $M$ and $N$ .

givenA1step 1.1

3.1

Therefore $[M, N] = 1$ . By [L1], every commutator $[m, n]$ is trivial, so $m n = n m$ for all $m \in M$ and $n \in N$ .

L1step 2.1∎

Depends on

Used by

Dependency tree · two levels

7 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

James E. Humphreys, A Course in Group Theory, Chapter 16 (standard reference, not scraped)