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For a finite $p$-group $G$, let $\mho_i(G)$ denote the subgroup generated by $p^i$-powers of elements of $G$.

It is well-known that $\mho_i(\mho_j(G))$ can differ from $\mho_j(\mho_i(G))$ and from $\mho_{i+j}(G)$; for example, in the groups of order $2^8$ and ID between $515$ and $524$ have $\mho_1(\mho_2(G))=\mho_3(G)=1$ but $\mho_2(\mho_1(G))=C_2$.

However, it seems that $\mho_i(\mho_j(G))\subseteq\mho_j(\mho_i(G))$ whenever $i\le j$.

I have not been able to find this in the literature, nor do I have a proof.

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    $\begingroup$ (I'm sure this is standard terminology/notation in $p$-group theory, but I'll just leave this Wikipedia link here for anyone else who was confused by the unusual word "agemo": en.wikipedia.org/wiki/Omega_and_agemo_subgroup ) $\endgroup$
    – Sam Hopkins
    Commented Aug 20, 2023 at 13:13
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    $\begingroup$ @SamHopkins, re, although the cuteness of the name is undercut slightly by the fact that the letter is upside down, not backwards (which would be no change). But I guess mho indicates how these matters are to be handled, so probably I should take it up with Lord Kelvin instead of the $p$-group theorists 😄. $\endgroup$
    – LSpice
    Commented Aug 20, 2023 at 21:28

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Well, unfortunately it's not true :(

I just ran the calculation on the free $2$-nilpotent group on two generators and class $8$, for which the groups $\mho_1(\mho_2(G))$ and $\mho_2(\mho_1(G))$ are incomparable (and both strictly containing $\mho_3(G)$).

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  • $\begingroup$ Class 8: you mean exponent 8? Class usually refers to nilpotency class (which is 2 here). $\endgroup$
    – YCor
    Commented Aug 24, 2023 at 8:48
  • $\begingroup$ However I did the calculation on the free 2-step-nilpotent exponent-8 group on 2 generators (which has order $2^8$) and it seems to me that both $\mho_1(\mho_2(G))$ and $\mho_2(\mho_1(G))$ are trivial therein. I might have misunderstood something. $\endgroup$
    – YCor
    Commented Aug 24, 2023 at 13:32
  • $\begingroup$ No, you understood correctly: the $2$-nilpotent refers to the prime $2$, while $8$ is the class. Start with $G_1=F$ the free group on $2$ generators, and define $G_{i+1}=[G_i,F]\mho_1(G_i)$. I did the calculations in $F/G_8$. $\endgroup$
    – grok
    Commented Aug 25, 2023 at 10:29
  • $\begingroup$ I indeed understood incorrectly as I often used $k$-nilpotent to mean $k$-step-nilpotent. OK, I understand now (that's "8-step-(2-nilpotent)" in some sense). This is a quite big group then, what's its order? $\endgroup$
    – YCor
    Commented Aug 25, 2023 at 12:13
  • $\begingroup$ It has order $2^{167}$, and can be produced in GAP with the commands LoadPackage("anupq"); and Pq(FreeGroup(2):Prime:=2,ClassBound:=8); $\endgroup$
    – grok
    Commented Aug 28, 2023 at 8:17
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These are quite complete notes on finite $p$-groups: Fernández-Alcober - An introduction to finite $p$-groups: regular $p$-groups and groups of maximal class. You should be able to find the claim in Section 2.1.

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  • $\begingroup$ I do not find the result in a quick scan of Section 2.1—but I am not a $p$-groups expert, so probably I just skimmed over it. Do you have a more specific result number? $\endgroup$
    – LSpice
    Commented Aug 20, 2023 at 21:26
  • $\begingroup$ @LSpice Well, it can be deduced from Hall combination theorem and divisibility properties of binomial coefficients. I could try to write out this mess, but I hope that somebody comes up with an already written reference. $\endgroup$
    – Denis T
    Commented Aug 20, 2023 at 22:13
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    $\begingroup$ @DenisT, re, sure, I was just asking @‍MatteoVannacci if they could point to a specific result number in this already written reference. $\endgroup$
    – LSpice
    Commented Aug 20, 2023 at 22:22
  • $\begingroup$ @DenisT, I'd love to see how the mess turns out --- there could be something to salvage in it. What I really want is $\mho_i(\mho_j(G))\equiv\mho_{i+j}(G)$ modulo some higher commutators, and indeed this does follow from messy binomial coefficients and Hall's theorem. However, see my answer, the claim itself is too much to be true. $\endgroup$
    – grok
    Commented Aug 24, 2023 at 8:48

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