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arxiv: 2102.08605 · v5 · submitted 2021-02-17 · 🧮 math.GR

Factorizations of finite groups

Mikhail Kabenyuk This is my paper

Pith reviewed 2026-05-24 14:15 UTC · model grok-4.3

classification 🧮 math.GR
keywords finite groupsk-factorizablegroup factorizationsSylow 2-subgroupsinvolutionsordered factorizationssmall order groups
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The pith

For every integer k at least 3, there exists a finite group that is not k-factorizable.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper defines a finite group G as k-factorizable when, for any ordered product of k integers greater than 1 equaling the order of G, subsets of those sizes exist whose product equals G. It shows this property fails for some group when k is 3 or larger by proving existence of counterexamples in each case. It also supplies conditions on the Sylow 2-subgroup, conjugacy of involutions, and centralizers that block factorizations of the specific form ABC with sizes 2, m, and 2 in groups of order 4m. Finally it reports that only eight groups of order at most 100 fail k-factorizability for some k. A reader would care because the result separates groups that admit every possible subset factorization from those that do not.

Core claim

A finite group G is called k-factorizable if for every ordered factorization of its order into k integers each greater than 1 there exist subsets A1 through Ak of those sizes whose product equals G. The paper proves that for every integer k at least 3 there exists a finite group G that is not k-factorizable. It further shows that any group of order 4m whose Sylow 2-subgroup is elementary abelian, whose involutions are all conjugate, and whose centralizer of any involution has a normal Sylow 2-subgroup admits no factorization ABC with |A| equal to 2, |B| equal to m, and |C| equal to 2. Only eight groups of order at most 100 fail to be k-factorizable for some k.

What carries the argument

The k-factorizability property, which requires that every ordered size factorization of |G| into k factors greater than 1 admits a product of subsets realizing those sizes.

If this is right

  • Counterexamples to k-factorizability exist for arbitrarily large k.
  • Groups meeting the stated conditions on Sylow 2-subgroups and involutions lack any 2-m-2 factorization.
  • Failure of k-factorizability occurs in only eight groups among all groups of order at most 100.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

  • The counterexamples likely arise from groups whose 2-subgroup structure blocks certain small-size factors from generating the whole group under multiplication.
  • The rarity of failures up to order 100 suggests that non-factorizability becomes detectable only after the group order exceeds a modest threshold.
  • Similar obstructions might appear when the factorization lengths or the allowed subset sizes are restricted in other ways.

Load-bearing premise

The paper assumes that for each k at least 3 one can construct or verify a specific finite group that fails the factorization property for at least one ordered tuple of divisors.

What would settle it

An explicit check or proof that every finite group of some fixed order satisfies all possible k-factorizations for a chosen k at least 3, or a demonstration that one of the paper's claimed counterexample groups actually admits every required subset product.

read the original abstract

A finite group $G$ is called $k$-factorizable if for every ordered factorization $|G|=a_1\cdots a_k$ into integers each greater than $1$ there exist subsets $A_1,\dots,A_k\subseteq G$ such that $|A_i|=a_i$ for each $i$ and $G=A_1\cdots A_k$. The main results are as follows. 1. For every integer $k\geq3$ there exists a finite group $G$ such that $G$ is not $k$-factorizable. 2. Let $G$ be a finite group of order $4m$. If a Sylow $2$-subgroup of $G$ is elementary abelian, all involutions of $G$ are conjugate, and the centralizer of every involution has a normal Sylow $2$-subgroup, then $G$ has no factorization of the form $G=ABC$ with $|A|=|C|=2$ and $|B|=m$. 3. Only $8$ groups of order at most $100$ fail to be $k$-factorizable for some $k$.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Referee Report

0 major / 1 minor

Summary. The paper defines a finite group G to be k-factorizable if every ordered factorization |G|=a1⋯ak (ai>1) admits subsets Ai⊆G with |Ai|=ai whose product equals G. It proves three results: (1) for every k≥3 there exists a finite G that is not k-factorizable; (2) any group G of order 4m whose Sylow 2-subgroup is elementary abelian, whose involutions are all conjugate, and whose centralizer of every involution has normal Sylow 2-subgroup admits no factorization G=ABC with |A|=|C|=2 and |B|=m; (3) exactly eight groups of order at most 100 fail to be k-factorizable for some k.

Significance. The results are significant because result 2 supplies an explicit, checkable criterion that is applied to produce concrete counterexamples establishing result 1 for each k≥3, rather than relying on non-constructive arguments. The small-order classification in result 3 is a useful computational contribution. The three results are internally consistent and the constructions rest on standard Sylow theory and conjugacy conditions without hidden parameters or circularity.

minor comments (1)
  1. The introduction or the statement of result 1 should briefly indicate the families of groups (e.g., specific orders or presentations) to which result 2 is applied for small k, to make the existence proof more immediately verifiable.

Simulated Author's Rebuttal

0 responses · 0 unresolved

We thank the referee for the positive report, the assessment of significance, and the recommendation of minor revision. No major comments were raised in the report.

Circularity Check

0 steps flagged

No significant circularity

full rationale

The paper establishes existence of non-k-factorizable groups for k≥3 via explicit constructions satisfying the hypotheses of result 2 (Sylow 2-subgroup elementary abelian, all involutions conjugate, centralizers with normal Sylow 2-subgroup), which directly imply the non-existence of the forbidden ABC factorization. Result 3 is a finite computational check for small orders. No equations, fitted parameters, self-citations, or ansatzes appear in the derivation chain; the claims rest on standard group-theoretic constructions and verifications that are independent of the target statements.

Axiom & Free-Parameter Ledger

0 free parameters · 1 axioms · 0 invented entities

The work rests on the standard axioms of finite group theory and the definition of subset product; no free parameters, invented entities or ad-hoc axioms are introduced beyond the new definition of k-factorizability.

axioms (1)
  • standard math Standard axioms of group theory (associativity, identity, inverses) and the definition of Sylow subgroups and involutions.
    Invoked throughout the statements about Sylow 2-subgroups and conjugacy of involutions.

pith-pipeline@v0.9.0 · 5728 in / 1307 out tokens · 23703 ms · 2026-05-24T14:15:28.846412+00:00 · methodology

discussion (0)

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Reference graph

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