Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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.

Commutator identities in a group whose derived subgroup is central

Statement

Let G be a group (Group and abelian group) whose derived subgroup [G,G] is contained in the centre Z(G) (Commutators [g,h]=ghg1h1 and the commutator subgroup [G,G], The center Z(G) of a group). If [G,G]Z(G) then [xy,w]=[x,w][y,w], [x,yw]=[x,y][x,w], and [xn,y]=[x,y]n=[x,yn] for every integer n and all x,y,wG, integer powers being those of Powers gn: natural exponents in a monoid and integer exponents in a group, with g0=e.

Facts & Assumptions

Given: A group G with [G,G]Z(G), and elements x,y,wG.

[F1]

For g,hG the commutator is [g,h]:=ghg1h1, and [G,G]:={[g,h]:g,hG} is the subgroup generated by all commutators (Commutators [g,h]=ghg1h1 and the commutator subgroup [G,G]).

[F2]

Z(G):={zG:zg=gz for every gG} (The center Z(G) of a group).

Proof

technique · direct
1.1

Every commutator [g,h] lies in [G,G], hence in Z(G), so it commutes with every element of G and may be moved to any position in a product without changing that product.

F1F2given
2.1

Expanding, [xy,w]=(xy)w(xy)1w1=xywy1x1w1=x(ywy1w1)wx1w1=x[y,w]wx1w1=[y,w]xwx1w1=[y,w][x,w]=[x,w][y,w], the last two equalities moving the central factor [y,w] past x and then past [x,w].

F1L2step 1.1algebra
2.2

Expanding in the second variable, [x,yw]=x(yw)x1(yw)1=xywx1w1y1=(xyx1)(xwx1w1)y1=(xyx1)[x,w]y1=[x,w]xyx1y1=[x,y][x,w], where the central factor [x,w] is moved past y1 and then past [x,y].

F1L2step 1.1algebra
3.1

For nN the identity [xn,y]=[x,y]n follows by induction: at n=0 both sides are the identity, since x0 is the identity and [e,y]=eye1y1=e, and [xn+1,y]=[xnx,y]=[xn,y][x,y]=[x,y]n[x,y]=[x,y]n+1.

F1L1step 2.1algebra
4.1

For a negative integer n put k=n; then e=[x0,y]=[xkxk,y]=[xk,y][xk,y] by step 2.1, so [xn,y]=[xk,y]=([x,y]k)1=[x,y]k=[x,y]n, and the identity [x,yn]=[x,y]n is obtained the same way from step 2.2.

L1step 2.1step 2.2step 3.1algebra

Remarks

The hypothesis [G,G]Z(G) is exactly the vanishing of the third term of the lower central series: γ3(G)=[G,[G,G]] is trivial precisely when every commutator commutes with every element of G (Subgroup commutators and the lower central series).

The first two identities are additivity in each variable separately, and they fail without the hypothesis: in general [xy,w]=x[y,w]x1[x,w], and the conjugating factor is what the hypothesis removes.

Depends on

Used by

Dependency tree · two levels

26 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