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TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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The Schreier refinement theorem

Statement

Any two finite subnormal series of a group have equivalent refinements (Subnormal and normal series, factors, refinements, and equivalence).

Facts & Assumptions

Given: Subnormal series G=G0Gm=1 and G=H0Hn=1.

[F1]

A refinement inserts subgroup terms, and two series are equivalent when their nontrivial factors can be paired up to isomorphism after repetitions are deleted (Subnormal and normal series, factors, refinements, and equivalence).

[L1]

For AA and BB, the butterfly constructions give normal adjacent terms and isomorphic quotient factors (The Zassenhaus butterfly lemma).

Proof

technique · direct
1.1

For 0i<m and 0jn, set Gi,j:=Gi+1(GiHj). Then Gi,0=Gi and Gi,n=Gi+1; [L1] applied to Gi+1Gi and Hj+1Hj gives Gi,j+1Gi,j.

givenL1
1.2

For 0j<n and 0im, set Hj,i:=(GiHj)Hj+1. Concatenating these chains gives a subnormal refinement of the H-series.

givenL1F1
2.1

Concatenating the finite chains Gi,0Gi,n over i=0,,m1 yields a subnormal refinement of the G-series, possibly with repeated adjacent terms.

step 1.1F1
2.2

For every cell (i,j), [L1] identifies Gi,j/Gi,j+1 with Hj,i/Hj,i+1. Thus the two refinements have their mn displayed factors paired by (i,j)(j,i).

step 1.1step 1.2L1
3.1

Deleting repeated adjacent terms deletes exactly the trivial factors on both sides of each paired cell, so the remaining factors are still paired and isomorphic; the refinements are equivalent.

step 2.1step 2.2F1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 14 results over 6 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources