Commutator power lying in commutator subgroup

Question

The argument used in the proof of Proposition 3 of this math.SE answer appears to prove the following claim:

Let $G$ be a group and let $H\subseteq G$ be a normal subgroup. Let $n\geq 0$ and let $x,g\in G$ such that $x^n\in H$. Then $[x,g]^n\in [H,G]$.

Here $[H,G]$ denotes the group generated by the commutators $[h,g]$ with $h\in H$ and $g\in G$.

I was unable to establish the claim, even in the concrete case $n=2$, and began to doubt its veracity. I am looking for either a direct proof of this claim or a counterexample.

Answer 1

EDIT: See bottom of this post for a complete resolution of my question.

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I am posting this as an answer purely for reasons of space in responding to the comments above.

First, let me begin by quoting verbatim (as it exists as of 6am UTC June 22 2019) the relevant portion of the proof of Proposition 3 referenced above:

there exists $n\gt 0$ such that $x^n\in G_{c+1}$. Moreover, since $G_{c+1}/G_{c+2}$ is abelian, by Lemma 2 we have that $[x^n,g] \equiv [x,g]^n\pmod{G_{c+2}}$. But since $G_{c+1}/G_{c+2}$ is abelian, $[h,g]\equiv 1$ if $h\in G_{c+1}$. Since $x^n\in G_{c+1}$, then $[x,g]^n \equiv [x^n,g] \equiv 1 \pmod{G_{c+2}}$

Here $G$ is nilpotent, $c$ is an arbitrary natural number, $G_{\bullet}$ denotes the lower central series, and there is an assumption of finite generation by torsion elements. However, these facts are not cited at all in the portion of the argument above. Since I am trying to crystallize my understanding of the argument, I abstracted away the details of the situation as follows.

Let $H=G_{c+1}$. The only two properties of the objects used in the argument are, in order of appearance:

1. $x^n\in H$
2. $H/[H,G]$ is abelian

(Note that these two properties are both reiterated twice in the quote above.)

I observed that if $H$ is an arbitrary normal subgroup of any group $G$, then $[H,G]$ is a normal subgroup of $H$, and moreover $H/[H,G]$ is abelian. Indeed, to see that $[H,G]\subseteq H$ it suffices to observe that $[h,g]=h\cdot h^g\in H$ for all $h\in H$ and $g\in G$, since $H$ is normal. Then, observe that $[h,g]^s=[h^s,g^s]\in H$, again by normality of $H$. Since $[H,G]$ is generated by elements of the form $[h,g]$, it follows that $[H,G]\trianglelefteq H$. Finally, $H/[H,G]$ is abelian since for all $h,h'\in H$ we have that $[h,h']\in [H,G]$ and thus we have the equality of cosets $hh'[H,G]=h'h[H,G]$.

Consequently my model of the situation satisfies all the explicit claims made in the above argument. However (thank you for pointing out an explicit counterexample Derek Holt) the deduction made by the above quote (which in my notation is $[x,g]^n\in [H,G]$) is incorrect. Derek Holt further suggests that the key property left unstated in this argument is that the relevant commutators are central. Indeed, such a hypothesis is necessary in order to apply Lemma 2 in the argument, although it is never justified. Thus, my question apparently reduces to the following:

Why is $[x,G]\subseteq Z(G)$ in the context of Proposition 3?

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EDIT:

The discussion in this thread allowed me to construct a solution of the desired form. I had missed a key condition when formulating the abstract form of the argument, which now reads as follows.

Claim. Let $N\trianglelefteq G$ be a normal subgroup of a group $G$ and let $x,g\in G$ be elements satisfying $[x,g]\in N$. Then for all $n\geq 0$, $$[x,g]^n\in [N,G][x^n,g].$$

Proof. Using $[x^n,g]=x[x^{n-1},g]x^{-1}[x,g]$ and induction, we have that $[x^n,g]\in N$ for all $n\geq 0$. The same identity also yields that $$ [x,g]^n=\prod_{i=1}^{n-1}\bigl[[x^i,g],x\bigr]\cdot [x^n,g]. $$

Answer 2

Your quote (in the answer you post to expound) misses one important assumption, namely that $x\in G_c$.

In your notation, $H=G_{c+1}$. So in your “crystalization” you lack the assumption that $[x,g]\in H$ for all $g\in G$.

Lemma. Let $N\triangleleft G$ and $z\in N$. Then $z[N,G]$ is central in $G/[N,G]$.

Proof. It suffices to show that for all $g\in G$, $[z[N,G],y[N,G]]$ is trivial in $G/[N,G]$. But this is equivalent to showing that $[z,y][N,G]$ is trivial in $G/[N,G]$, which is equivalent to showing that $[z,y]\in [N,G]$; which is immediate since $z\in N$ and $g\in G$. $\Box$

So, what you have is: $x\in G$, $[x,G]\subseteq H$, and we want to show that for all $g\in G$, $[x,g]$ is central in $G/[H,G]$. But this follows from the Lemma above, just taking $N=H$ and $z=[x,g]$.

From this it follows using commutator identities that $$\begin{align*} [x^2,g][H,G] &= [x,g]^x[x,g][H,G]\\ &= [x,g][H,G]^{x[H,G]}[x,g][H,G]\\ &= [x,g][H,G][x,g][H,G]\\ &= [x,g]^2[H,G]. \end{align*}$$

More generally, inductively, that $[x^n,g][H,G] = [x,g]^n[H,G]$ for all positive integers $n$.

Since we also have that $x^m\in H$, then $[x^m,g]\equiv 1\pmod{[H,G]}$. Therefore, $[x,g]^m \equiv [x^m,g]\equiv 1\pmod{[H,G]}$, which is what that Lemma concludes.