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Let $G$ be a group. We let $G_1:=G$, we let $G_2 := [G,G]$ the commutator subgroup (i.e. the subgroup generated by all the $[a,b]$ such that $a,b\in G$). Inductively one can define $G_k = [G_{k-1},G]$.

We say that $G$ is $k$-step nilpotent if $G_{k+1}$ is trivial.

Let $G$ be a $k$-step nilpotent group. Let $g\in G$ be an element such that $g^n=1$ and let $s\in G$.

Can we conclude that $[s,g]$ is of finite order?

The answer is yes for small $k$'s. If $k=1$, then $G$ is abelian and there's nothing to prove. If $k=2$ this is also true, for simplicity we prove this for $n=2$ (the same proof holds for general $n$)

we have $$[s,g]^2 = sgs^{-1}g^{-1} [s,g]$$

Since $G$ is $2$-step nilpotent, $[s,g]$ commutes with everything, in particular with $g^{-1}$ so we have

$$[s,g]^2 = sgs^{-1} [s,g]g^{-1}=sgs^{-1}sgs^{-1}g^{-1}g^{-1}=sg^{2}s^{-1}g^{-2}=ss^{-1}=1$$

Can we prove this for higher order of nilpotency?

Note that this argument can be generalized, using it one can show that for any $s\in G_{k-1}$, $[s,g]^n=1$.

Yanko
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    Please note that in general, $G^{(n)}$ denotes the terms of the derived series, not the lower central series; that is, $G^{(1)} = G’ = [G,G]$,. and $G^{(n+1)} = [G^{(n)},G^{(n)}]$. It is much more common to use subscripts for the lower central series, $G_1=G$, $G_{n+1}=[G_n,G]$. – Arturo Magidin Sep 14 '19 at 04:20
  • Yes, this is true in a nilpotent group. One can prove it using commutator collection and basic commutators. You can probably get it as a consequence of some product formulas that appear in R.R. Struik’s papers on nilpotent products of cyclic groups. I don’t have them handy right now, but I’ve one something along these lines; the proof usually goes by “backwards induction”, where it is clear that a basic commutator in the last term of the lower central series that involves $s$ has exponent $n$, and then one shows it holds for basic commutators in $G_k$ if it holds in $G_{k+1}$. – Arturo Magidin Sep 14 '19 at 04:31
  • I can give you the appropriate theorems and references on Monday, rather than try to develop the theory from scratch here (which I guess I can do, but would take me a while). – Arturo Magidin Sep 14 '19 at 05:02
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    For the question: if $g,s$ are in a nilpotent group and $g$ is torsion, then $[g,s]=g(sg^{-1}s^{-1})$ is a product of 2 torsion elements, and torsion elements form a subgroup in a nilpotent group. – YCor Sep 14 '19 at 06:26
  • @ArturoMagidin by "$s$-step nilpotent", mean that $G_{s+1}={1}$ and nothing more; then (like everybody) say class is $s$ if $s\ge 0$ is minimal for this property. I encourage not to further restrict the definition of $s$-step nilpotent (so that this class is a group variety). So $0$-step nilpotent means trivial, $1$-step nilpotent means abelian, $2$-step nilpotent means central-by-abelian, $3$-step nilpotent means central-by-(central-by-abelian), etc. Next, to be precise, class $1$ means abelian and not trivial, class 2 means that $[G,G]$ is central and not trivial, etc. – YCor Sep 14 '19 at 06:29
  • @YCor: $s$-step = class [at most] $s$. Fine. In any case, the OPs nomenclature is incorrect (note he says abelian is “2-step nilpotent”). – Arturo Magidin Sep 14 '19 at 06:37
  • @YCor: Is there a proof that product of torsion is torsion in nilpotent groups that does not use the commutator collection formulas? – Arturo Magidin Sep 14 '19 at 06:40
  • @ArturoMagidin You proved it yourself here – Derek Holt Sep 14 '19 at 07:00
  • @DerekHolt: Ha! Yeah; been a while, though.... – Arturo Magidin Sep 14 '19 at 07:03

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Modulo correcting the nomenclature, as noted in comments, $$[x,y] = xyx^{-1}y^{-1} = x(yx^{-1}y^{-1})$$ is a product of two conjugates of $x$; if $x$ is torsion, then this is a product of two torsion elements.

In a nilpotent group, the torsion elements form a subgroup. See here, but note that the answer there uses a different commutator convention, namely that $[x,y]=x^{-1}y^{-1}xy$.

Please note the comments on common notation (which clashes with yours) and nomenclature (your definition of $s$-step nilpotent is off).

Arturo Magidin
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  • Thanks for the proof and for the correction of the notations. I will edit my question to fit the right notations – Yanko Sep 14 '19 at 10:11
  • It seems from the proof that one could even get some stronger result that if $x,y$ are of order $n$ and $G$ is $k$-step nilpotent then $xy$ is of order at most $n^k$. Do you think that could be true? – Yanko Sep 16 '19 at 13:40
  • @Yanko: You can find the exact results that one can get in the following paper: "On the orders of products and commutators in prime-power groups", by T.E. Easterfield, Proc. Cambridge Philos. Soc. 36 (1940), pp. 14-26, MR 1,104b. – Arturo Magidin Sep 16 '19 at 14:04
  • The short answer is that one can certainly bound the order, but your calculation is too naive; for $k$ small compared to the primes that divide $n$, one can do much better; for $k$ large compared to the primes, you have expressions that depend on $k$ but can be made much more precise. – Arturo Magidin Sep 16 '19 at 14:07
  • I'll look up for that, thanks – Yanko Sep 16 '19 at 20:04