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Normal Subgroups of Powerful $p$ -groups

T0 review · 1 major / 5 minor · reviewed 2026-08-14 · deepseek-v4-flash

Pith's one-line read Normal subgroups inside the p-th power subgroup of a powerful p-group are themselves powerfully nilpotent.

desk verdict Terse but correct note closing the expected gap in powerful p-groups; the skeleton is sound and the only real risk is the external order-bound theorem it quotes. read the letter →

arxiv 1908.07030 v1 pith:V44D2ET5 submitted 2019-08-19 math.GR

classification math.GR MSC 20D15
keywords powerfulp-groupspowerfullynilpotentgroupsnormalsubgroupspowercommutatororderboundsfinitecoclass
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

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

The reading

The paper proves that for odd primes, every normal subgroup N of a powerful p-group G with $N \leq G^p$ is powerfully nilpotent; for $p=2$ the same conclusion holds when $N \leq G^4$. Powerfully nilpotent groups are powerful groups with a central series whose new steps satisfy $[H_i,G] \leq H_{i-1}^p$, and their rank and exponent are bounded in terms of a single invariant, the powerful coclass. The theorem upgrades the known fact that such $N$ is powerful to the full layered structure, and its corollary places powerfully nilpotent subgroups at every level of the lower $p$-power series of $G$. This also gives a partial answer to the question of which $p$-groups can appear as subgroups of powerful $p$-groups.

What carries the argument

The load-bearing mechanism is the commutator order bound stated as Theorem 3: in a powerful $p$-group, if $o(x) \leq p^{i+1}$ and $o(y) \leq p^i$, then $o([x^{p^j},y^{p^k}]) \leq p^{i-j-k}$. This lets the proof convert normality and containment in $G^p$ into precise order control on commutators, forcing elements of order $p^2$ to appear centrally in $N$ when $N$ has exponent $p^2$ (Lemmas 9--11). The induction closes with two quotient criteria: a finite $p$-group is powerfully nilpotent if and only if $G/G^{p^2}$ is, and if and only if $G/Z(G)^p$ is.

What would settle it

A direct counterexample would be a powerful $p$-group with odd $p$ and a normal subgroup $N \leq G^p$ such that $N/N^{p^2}$ is not powerfully nilpotent, since Proposition 6 then rules out powerful nilpotence of $N$. Concretely, an exhaustive check of powerful groups of order $3^k$ for $k \leq 8$, testing every normal subgroup contained in $G^3$, would either confirm the theorem or exhibit such $N$.

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Extended reading notes

Core claim

Theorem 12 states that if $p$ is odd, $G$ is powerful, and $N$ is normal in $G$ with $N \leq G^p$, then $N$ is powerfully nilpotent. Theorem 8 covers $p=2$ with $N \leq G^4$. The proof inducts on the order of $G$: first quotient by $N^{p^2}$, which reduces to the case where $N$ has exponent exactly $p^2$; a sequence of lemmas then finds an element of order $p^2$ in the centre of $N$, and quotienting by $Z(N)^p$ gives a smaller powerful quotient to which the inductive hypothesis applies. Proposition 6 and Proposition 7 translate powerful nilpotence of these quotients back to $N$, and the corollary extends the same conclusion to normal subgroups of $G^{p^i}$ contained in $G^{p^{i+1}}$ (with the case $p=2$ shifted by one).

Load-bearing premise

The argument stands on the external order bound stated as Theorem 3, which limits the order of every commutator of powers in a powerful $p$-group; if that bound fails in a single powerful $p$-group, the central-element lemmas that drive the induction collapse.

Editorial extensions

If this is right

  • Every normal subgroup of a powerful $p$-group that lies inside $G^p$ (or $G^4$ when $p=2$) is not only powerful but powerfully nilpotent, so it carries the full layered central-series structure.
  • By Corollary 13, normal subgroups of $G^{p^i}$ contained in $G^{p^{i+1}}$ (with the analogous shift for $p=2$) are powerfully nilpotent, making powerful nilpotence abundant in the lower $p$-power series.
  • Because rank and exponent of a powerfully nilpotent group are bounded by functions of its powerful coclass, the theorem brings quantitative control to deep normal subgroups of powerful $p$-groups.
  • The result partially answers the question of which $p$-groups embed in powerful $p$-groups: those that arise as normal subgroups inside $G^p$ have the strongly restricted structure of powerful nilpotence.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • The proof suggests that the only obstruction to powerful nilpotence in a powerful $p$-group lives outside the first power subgroup; normal subgroups that reach into $G^p$ lose the freedom to be wild, so the boundary case $N \not\leq G^p$ is where any counterexamples would have to hide.
  • One testable extension is whether the induction still works after replacing $G^p$ by lower terms $G^{p^k}$, which would broaden the corollary into a fuller characterization of which normal subgroups of a powerful $p$-group are powerfully nilpotent.
  • For $p=2$, the shift from $G^2$ to $G^4$ points to the quotient $N/N^4$ as the complete obstruction; computing the upper powerfully central series of that quotient may yield a criterion for powerful nilpotence of arbitrary normal subgroups in powerful $2$-groups.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Referee Report

1 major / 5 minor

Summary. The paper proves that if G is a finite powerful p-group and N is a normal subgroup of G contained in G^p (for odd p) or in G^4 (for p=2), then N is powerfully nilpotent. The proof reduces the problem to the case where N has exponent p^2, uses commutator-order estimates from Fernández-Alcober's Theorem 3 to show that N then contains a nontrivial central element of order p^2 (Lemma 11), and finishes by induction on |G| together with two reduction criteria (Propositions 6 and 7). The p=2 case is handled by proving that N/N^4 is abelian, so that N is a powerful 2-group and hence powerfully nilpotent by a result from [8].

Significance. The main theorem is a natural strengthening of the earlier result of González-Sánchez and Jaikin-Zapirain, which showed only that such a subgroup N is powerful. It also gives a partial answer to Mann's question about which p-groups occur as subgroups of powerful p-groups. The proof is concise, and the commutator estimates are applied carefully; the role of the external Theorem 3 is explicitly identified. If the cited results are correct, the central claim follows from the presented argument. The paper is a well-focused research note rather than a survey or a computational study.

major comments (1)
  1. [Section 2, Proposition 7] The proof of Proposition 7 is a single sentence, but this reduction is load-bearing: it is used in the induction step of Theorem 12 to conclude that N is powerfully nilpotent from the powerful nilpotence of N/Z(N)^p. Please provide a full proof, or a precise citation to [8]. The nontrivial point is that when one lifts the upper powerfully central series from G/Z(G)^p to G, one must verify that the lifted subgroups H_i satisfy [H_i,G] ≤ H_{i-1}^p at every step; this requires an inductive argument using Z(G)^p ⊆ H_{i-1}^p, and it is not immediate from the definition alone.
minor comments (5)
  1. [Theorem 3] Please give the exact lemma or proposition number in [2] where Theorem 3 is proved, since Lemmas 9, 10, and 11 and Theorem 8 all rely on this bound.
  2. [Lemma 9] In the proof of Lemma 9, the application of Theorem 3 uses j=k=1; please state this explicitly so the reader can verify the exponent subtraction i-j-k = 2-1-1 = 0.
  3. [Theorem 12] After the sentence 'for otherwise N would be abelian by Lemma 9', please add explicitly that an abelian p-group is powerfully nilpotent, so the argument may indeed assume that the exponent of N is exactly p^2.
  4. [Corollary 13] The corollary is labelled 'immediate' but would be clearer if it noted that G^{p^i} is itself a powerful p-group and that Theorem 12 (or Theorem 8 for p=2) is applied with H = G^{p^i}.
  5. [Throughout] The text contains several spacing and typographical artifacts such as 'o f', 'powerf ul', and 'subg roup'; these should be corrected in the final version.

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: Theorem 12 is proved from the external commutator bound of Fernández-Alcober via induction; self-citations are tool citations only.

full rationale

The derivation of Theorem 12 is not circular. Lemmas 9, 10, and 11 each reduce a commutator of pth powers directly to the external Theorem 3, a commutator-order bound proved by Fernández-Alcober, not by the present author. For example, Lemma 10 applies Theorem 3 with i=3, j=2, k=1 to obtain [g^{p^2},h^p]=1, and Lemma 11 uses the same bound to force [[[b,a^p],b],b]=1. The induction in Theorem 12 then uses two quotient criteria from the author's prior work: Proposition 6 (powerfully nilpotent iff the p^2-power quotient is powerfully nilpotent) and Proposition 7 (powerfully nilpotent iff the quotient by Z(G)^p is powerfully nilpotent). These are general structural criteria for the class of powerfully nilpotent groups, not restatements of the theorem being proved, and Proposition 7 is even given a one-line proof from the definition. The p=2 case invokes the prior result that powerful 2-groups are powerfully nilpotent from [8], but this is a published external classification, and it is not used in the odd-prime main proof. No fitted parameter is renamed as a prediction, no uniqueness theorem is imported to force a choice, and no definition is circularly tied to the target conclusion. The central claim therefore has independent mathematical content derived from an external commutator bound and induction.

Assumptions & free parameters 0 free parameters · 9 assumptions · 0 invented entities

The proof is a standard group-theoretic derivation. It introduces no free parameters and no invented entities. It does rely on a package of established theorems about powerful p-groups, several from the author's earlier work, which are treated as black boxes.

assumptions (9)
  • domain assumption All groups considered are finite p-groups.
    Section 2, first sentence; the theorem is stated only in this category, and the proof uses finiteness for induction on order.
  • standard math For a powerful p-group, G^{p^k} is exactly the set of p^k-th powers, is powerfully embedded, and G^p is generated by p-th powers of generators.
    Theorem 1 from [5, Lubotzky-Mann], used throughout to write elements of G^p as p-th powers.
  • standard math Powerfully embedded subgroups satisfy [M^{p^i}, N^{p^j}] = [M,N]^{p^{i+j}}.
    Lemma 2 from [7, Shalev], used to justify inclusions such as [G, G^p] ≤ G^{p^2}.
  • standard math Fernandez-Alcober's bound: if o(x) ≤ p^{i+1} and o(y) ≤ p^i, then o([x^{p^j}, y^{p^k}]) ≤ p^{i-j-k}.
    Theorem 3 is the engine of Lemmas 9, 10, 11 and the p=2 case in Theorem 8.
  • standard math A normal subgroup N ≤ G^p of a powerful p-group is powerful, with the analogous statement N ≤ G^4 for p=2.
    Theorem 4 from [3, Gonzalez-Sanchez and Jaikin-Zapirain], cited in the introduction as the starting point.
  • standard math G is powerfully nilpotent if and only if G/G^{p^2} is powerfully nilpotent.
    Proposition 6 from [8], used in the induction step of Theorem 12.
  • standard math G is powerfully nilpotent if and only if G/Z(G)^p is powerfully nilpotent.
    Proposition 7 from [8], used at the end of Theorem 12.
  • standard math Every powerful 2-group is powerfully nilpotent.
    Result from [8, page 81], used to finish the p=2 case in Theorem 8.
  • standard math For a class-2 group, [x^n, y] = [x,y]^n [x,y,x]^{n choose 2}.
    Used in Lemma 11 to compare [b^p, a^p] with [b,a^p]^p.

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Cite this review

Pith. "Pith review of Normal Subgroups of Powerful $p$ -groups." pith.science (2026). https://pith.science/paper/V44D2ET5

@misc{pith2026190807030,
  author       = {Pith},
  title        = {Pith review of: Normal Subgroups of Powerful $p$ -groups},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/V44D2ET5}},
  note         = {Machine review of arXiv:1908.07030}
}
abstract

In this note we show that if $p$ is an odd prime and $G$ is a powerful $p$-group with $N\leq G^{p}$ and $N$ normal in $G$, then $N$ is powerfully nilpotent. An analogous result is proved for $p=2$ when $N\leq G^{4}$.

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Reference graph

Works this paper leans on

9 extracted references · 9 canonical work pages

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    Gunnar Traustason and James Williams, Powerfully nilpotent groups , Journal of Algebra 522 (2019), 80 – 100

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    Shalev, On almost fixed point free automorphisms , Journal of Algebra 157 (1993), no

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  1. [9]

    DOI: 10.22108/ijgt.2019.113217.1507

    James Williams, Omegas of agemos in powerful groups , International Journal of Group Theory, –, In press. DOI: 10.22108/ijgt.2019.113217.1507

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