LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every finite-group element has unique commuting p- and p-prime parts

Statement

Let $G$ be finite, let $p$ be prime, and let $g \in G$ have order $p^{a} m$ , where $a \geq 0$ and $\gcd (p, m) = 1$ . There are unique commuting powers $g_{p}, g_{p^{'}}$ of $g$ such that

$g = g_{p} g_{p^{'}},$

$g_{p}$ has $p$ -power order, and $g_{p^{'}}$ is $p$ -regular. Moreover, formation of the two parts commutes with conjugation.

Facts & Assumptions

Given: The finite group, prime, element, and factorization of its order in the Statement.

[F1]

A $p$ -regular element has order prime to $p$ (p-regular and p-singular elements).

[F2]

Coprime integers satisfy Bezout's identity (Bézout's identity: for integers $a, b$ not both zero, $\gcd (a, b)$ is the least positive element of ${a x + b y : x, y \in Z}$ ; in particular $a x + b y = \gcd (a, b)$ has an integer solution).

Proof

1.1

By F2 choose $r, s \in Z$ with $r p^{a} + s m = 1$ , and set $g_{p} = g^{s m}, g_{p^{'}} = g^{r p^{a}} .$ These elements commute because they are powers of $g$ , and their product is $g^{s m + r p^{a}} = g$ . Also $g_{p}^{p^{a}} = 1$ and $g_{p^{'}}^{m} = 1$ , so their orders divide $p^{a}$ and $m$ , respectively. Thus they have the required order types.

F1F2algebra

1.2

Suppose $g = u v$ is another commuting factorization, with $∣ u ∣ = p^{b}$ and $∣ v ∣ = d$ prime to $p$ . Bezout applied to $p^{b}$ and $d$ gives an exponent $e$ satisfying $e \equiv 1 (m o d p^{b})$ and $e \equiv 0 (m o d d)$ ; hence $g^{e} = (u v)^{e} = u$ . Interchanging the two congruences similarly expresses $v$ as a power of $g$ . Consequently $∣ u ∣$ and $∣ v ∣$ divide $∣ g ∣$ , so $∣ u ∣ ∣ p^{a}$ and $∣ v ∣ ∣ m$ .

F2algebra

2.1

Using the exponent from step 1.1 in the factorization $g = u v$ gives $g^{s m} = u^{s m} v^{s m} = u,$ because $v^{m} = 1$ and $s m \equiv 1 (m o d p^{a})$ , hence modulo $∣ u ∣$ . Likewise $g^{r p^{a}} = v$ . Thus $u = g_{p}$ and $v = g_{p^{'}}$ , proving uniqueness.

step 1.1step 1.2algebra

3.1

For $h \in G$ , the pair $(h g_{p} h^{- 1}, h g_{p^{'}} h^{- 1})$ is a commuting $p$ -by- $p^{'}$ factorization of $h g h^{- 1}$ . Uniqueness therefore identifies it with $((h g h^{- 1})_{p}, (h g h^{- 1})_{p^{'}})$ . If $a = 0$ , the formulas give $g_{p} = 1, g_{p^{'}} = g$ ; if $m = 1$ , they give $g_{p} = g, g_{p^{'}} = 1$ . Thus the endpoint cases are included.

step 2.1algebra∎

Depends on

Used by

The p-section of a p-element Definition
Nagao error terms have zero trace on the relevant p-section Lemma

Dependency tree · two levels

18 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

Aschbacher–Kessar–Oliver, Fusion Systems in Algebra and Topology, Part IV section 4.4, p. 272 (standard reference, not scraped)
Craven, The Brauer Correspondence, Chapter 1 section 1.5, pp. 13–14 (standard reference, not scraped)