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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.

pre-kidney
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    It's not true in general. Have you made any attempt to look for a counterexample? Try the smallest nonabelian group with $H$ the trivial subgroup. But I notice you have the tag nilpotent groups - did you mean to assume that $G$ is nilpotent? It's still not true in general. – Derek Holt Jun 21 '19 at 09:57
  • I don't think that this claim is proven here, as you suggest. – Dietrich Burde Jun 21 '19 at 11:17
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    @DietrichBurde Well no, it can't be proven if it is not true! I think that the relevant commutators are central in the context of that proof. – Derek Holt Jun 21 '19 at 14:14
  • That statement is wrong, but then, that is not what the Proposition uses or claims. You are misunderstanding the argument. – Arturo Magidin Jun 21 '19 at 18:32
  • @DerekHolt thank you for pointing out the simple counterexample. Sometimes the simplest examples are in front of our noses. Despite such a simple counterexample, the question remains: where have I gone wrong in abstracting the argument from Proposition 3? Since there was insufficient space to explore this further in a comment, I have posted an answer below where (guided by your comments) I believe the sticking point in the argument has been crystallized, and my question has slightly morphed. – pre-kidney Jun 22 '19 at 06:04
  • In short, why are the relevant commutators central in the context of that proof? – pre-kidney Jun 22 '19 at 06:05
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    In that proof we are considering $G_c/G_{c+1}$ as a subgroup of $G/G_{c+1}$. It is central by definition of the lower central series. – Derek Holt Jun 22 '19 at 06:49

2 Answers2

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

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?

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 $$ [x,g]^n=\prod_{i=1}^{n-1}\bigl[[x^i,g],x\bigr]\cdot [x^n,g]. $$

pre-kidney
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  • It’s not; but that is not the claim. The claim is that $[x,G][H,G]$ is central in $G/[H,G]$. We are working modulo $G_{c+2}$ (or in your notation, modulo $[H,G]$). – Arturo Magidin Jun 22 '19 at 06:26
  • The actual deduction being made is not that $[x,g]^n\in [H,G]$, but rather that $[x^n,g]\equiv [x,g]^n\pmod{[H,G]}$. Then the conclusion that $[x,g]^n\in [H,G]$ follows because this is true for $[x^n,g]$. – Arturo Magidin Jun 22 '19 at 06:30
  • More precisely, isn't the condition to be verified that $[x[H,G],G/[H,G]]\in Z(G/[H,G])$? Written out explicitly, this condition means that all cosets $g[H,G]$ commute (in $G/[H,G]$) with all cosets $g'[H,G]\in [x[H,G],G/[H,G]]$? And how is this justified exactly? – pre-kidney Jun 22 '19 at 06:32
  • Note that in the Lemma, $x\in G_c$, so that $[x,g]\in H$ (since $G_{c+1}$). You need to show that $[x,g][H,G]$ is central in $G/[H,G]$. But $[x,g]\in H$, so that for all $y\in G$, $[[x,g],y] \in [H,G]$. That means that in $G/[H,G]$, the commutator of $[x,g][H,G]$ and $y[H,G]$ is trivial, so $[x,g][H,G]$ commutes with $y[H,G]$. As $y[H,G]$ is an arbitrary element of $G/[H,G]$, it follows that $[x,g][H,G]$ is central in $G/[H,G]$; or, more simply, that $[x,g]$ is central modulo $[H,G]$. – Arturo Magidin Jun 22 '19 at 06:38
  • That is the computation being made. You dropped a couple of assumptions in your ”crystalization”, like the fact that $[x,g]\in H$ for all $g\in G$... so that you missed the easy fact that the obvious commutator in question is trivial in the quotient, which means the elements commute in the quotient. – Arturo Magidin Jun 22 '19 at 06:40
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    (Your “relevant quote”, for example, is missing the fact that $x\in G_c$) – Arturo Magidin Jun 22 '19 at 06:41
  • That is a very fair point. Does my crystallization become true if one appends the assumption that $[x,G]\subseteq H$? – pre-kidney Jun 22 '19 at 06:43
  • Yes; because then what we are doing is showing that if $N\triangleleft G$ and $x\in N$, then $x[N,G]$ is central in $G/[N,G]$. But this follows because for any $g\in G$, $[x,g]$ is trivial in $G/[N,G]$, so $x[N,G]$ commutes with all $g[N,G]$. – Arturo Magidin Jun 22 '19 at 06:51
  • By the way: the better way to add this information would have been in the question itself, no? – Arturo Magidin Jun 22 '19 at 07:14
  • I did not see a sensible way to do this in a manner that would keep the comments on the question comprehensible. – pre-kidney Jun 22 '19 at 07:16
  • You can put a horizontal line at the bottom, with <hr>; then a note stating that this is being added at a particular time, so that the original question remains and the comments are seen to refer to it. – Arturo Magidin Jun 22 '19 at 07:17
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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.

Arturo Magidin
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  • If I am understanding correctly, this new lemma is the key to providing a simple and self-contained proof that my crystallization + missing condition is true. If so, how does $N$ relate to my $H$? It would greatly clarify the situation for me if you could show how the lemma is applied to simplify the proof of my crystallization + missing condition (which is apparently what the first part of your answer is doing, if I understand it correctly?). By the way, I am grateful you are taking the time to explain this and I don't mean to be obstinate, just hoping to "get" it. – pre-kidney Jun 22 '19 at 06:57
  • @pre-kidney: I was rewriting while you were commenting. – Arturo Magidin Jun 22 '19 at 06:59
  • @pre-kidney (The reason that was not explicit is that as you can see, it is a pretty simple lemma: if you are moding out by all commutators of the form $[H,G]$ then everything in $H$ becomes central in the quotient, because all the relevant commutators have been moded out) – Arturo Magidin Jun 22 '19 at 07:02
  • I am still trying to grok all the symbol-pushing of your argument, even if the intuition makes more sense now. Is the upgraded crystallization true: Let $N\trianglelefteq G$ and let $x,g\in G$ such that $[x,g]\in N$. Then $[x,g]^n\in[x^n,g][N,G]$? – pre-kidney Jun 22 '19 at 07:28
  • Yes, I think that is true, since $[x,g]$ will be central in $G/[N,G]$. – Arturo Magidin Jun 22 '19 at 16:53
  • @pre-kidney: The observation underlying the computations is really that in any group $G$, the bracket $[-,g]$ induces a $\mathbb{Z}$-linear map from $G_c/G_{c+1}$ to $G_{c+1}/G_{c+2}$. – Arturo Magidin Jun 22 '19 at 19:08