Coprime commutators in profinite groups
Pith reviewed 2026-05-18 10:57 UTC · model grok-4.3
The pith
If the coprime commutators of a profinite group are covered by countably many procyclic subgroups, then its pronilpotent residual is finite-by-procyclic.
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
Core claim
If the set of coprime commutators of a profinite group G is covered by countably many procyclic subgroups, then γ_∞(G) is finite-by-procyclic. In particular, it follows that G is finite-by-pronilpotent-by-abelian. The proof uses the identification that the subgroup generated by all coprime commutators equals the pronilpotent residual γ_∞(G).
What carries the argument
The covering of the set of coprime commutators by countably many procyclic subgroups, which is used to deduce the finite-by-procyclic structure of the pronilpotent residual γ_∞(G).
If this is right
- γ_∞(G) is finite-by-procyclic.
- G is finite-by-pronilpotent-by-abelian.
- Finiteness-type conditions on coprime commutators continue to impose strong global restrictions on the structure of profinite groups.
Where Pith is reading between the lines
- The same covering hypothesis might be tested on other residual subgroups or on pro-p groups specifically.
- One could ask whether a finite covering (instead of countable) would force an even stronger conclusion such as finiteness of γ_∞(G).
- The result suggests looking for analogous statements when commutators are replaced by other words or when the group is required to be finitely generated.
Load-bearing premise
That the subgroup generated by all coprime commutators equals the pronilpotent residual γ_∞(G).
What would settle it
A profinite group in which the coprime commutators lie in a countable union of procyclic subgroups yet the pronilpotent residual fails to be finite-by-procyclic.
read the original abstract
By a coprime commutator in a profinite group $G$ we mean any element of the form $[x, y]$, where $x,y\in G$ and $(|x|,|y|)=1$. It is well-known that the subgroup generated by the coprime commutators of $G$ is precisely the pronilpotent residual $\gamma_\infty(G)$. There are several recent works showing that finiteness conditions on the set of coprime commutators have strong impact on the properties of $\gamma_\infty(G)$ and, more generally, on the structure of $G$. In this paper we show that if the set of coprime commutators of a profinite group $G$ is covered by countably many procyclic subgroups, then $\gamma_\infty(G)$ is finite-by-procyclic. In particular, it follows that $G$ is finite-by-pronilpotent-by-abelian.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper proves that if the set of coprime commutators in a profinite group G is covered by countably many procyclic subgroups, then the pronilpotent residual γ_∞(G) is finite-by-procyclic. It further concludes that G itself is finite-by-pronilpotent-by-abelian, building on the well-known fact that the subgroup generated by coprime commutators equals γ_∞(G).
Significance. If the result holds, it adds a new countable covering condition to the recent literature on how restrictions on coprime commutators control the structure of profinite groups and their residuals. The theorem supplies a concrete hypothesis that forces γ_∞(G) to have a finite normal subgroup with procyclic quotient, yielding a structural decomposition for G.
major comments (1)
- [Abstract] Abstract and the final paragraph of the introduction: the deduction that γ_∞(G) finite-by-procyclic implies G is finite-by-pronilpotent-by-abelian requires an explicit argument that a finite normal subgroup N of γ_∞(G) can be taken G-normal (or that the image of γ_∞(G) in G/N remains the pronilpotent residual of G/N). Without this, the 'in particular' claim does not follow directly from the main theorem on γ_∞(G).
Simulated Author's Rebuttal
We thank the referee for the careful reading of the manuscript and for identifying this point concerning the deduction in the abstract and introduction. We agree that an explicit argument is required to justify passing from the structure of γ_∞(G) to the claimed structure of G, and we will revise the paper to supply it.
read point-by-point responses
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Referee: [Abstract] Abstract and the final paragraph of the introduction: the deduction that γ_∞(G) finite-by-procyclic implies G is finite-by-pronilpotent-by-abelian requires an explicit argument that a finite normal subgroup N of γ_∞(G) can be taken G-normal (or that the image of γ_∞(G) in G/N remains the pronilpotent residual of G/N). Without this, the 'in particular' claim does not follow directly from the main theorem on γ_∞(G).
Authors: We accept the referee's observation. In the revised version we will insert a short paragraph (or a brief lemma) immediately after the statement of the main theorem. Let N be a finite subgroup of γ_∞(G) that is normal in γ_∞(G) with γ_∞(G)/N procyclic. Because G is profinite and N is finite, the conjugation action yields a continuous homomorphism G → Aut(N). The group Aut(N) is finite, so the kernel C_G(N) is open of finite index in G. Consequently N possesses only finitely many distinct G-conjugates. The normal closure N^G is therefore generated by finitely many finite subgroups and is itself finite. Moreover N^G ⊴ G and N^G ≤ γ_∞(G). The quotient γ_∞(G)/N^G is a quotient of the procyclic group γ_∞(G)/N and hence procyclic. We then verify that the image of γ_∞(G) in G/N^G coincides with the pronilpotent residual of G/N^G: the quotient (G/N^G)/(γ_∞(G)/N^G) ≅ G/γ_∞(G) is pronilpotent by definition of the residual, and any smaller normal subgroup of G/N^G whose quotient is pronilpotent would lift to a normal subgroup of G properly contained in γ_∞(G) whose quotient is pronilpotent, contradicting the minimality of γ_∞(G). With this justification in place the 'in particular' statement follows at once. revision: yes
Circularity Check
No significant circularity; derivation relies on standard facts
full rationale
The paper states as well-known that the subgroup generated by coprime commutators equals γ_∞(G), then proves the covering hypothesis implies γ_∞(G) is finite-by-procyclic. No quoted step reduces a prediction or central claim to a fitted input, self-definition, or self-citation chain by construction. The argument uses external properties of profinite groups and is self-contained against benchmarks; the 'in particular' structural claim for G is a separate deduction whose validity is a correctness issue, not circularity.
Axiom & Free-Parameter Ledger
axioms (1)
- domain assumption The subgroup generated by the coprime commutators of G is precisely the pronilpotent residual γ_∞(G)
Reference graph
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