Define the following "commutators" recursively:
- $[g,h] = g^{-1}h^{-1}gh$
- $[g_1, \dots, g_m] = [g_1, [g_2, \dots, g_m]]$ for all $m \geq 3$
Let $G_i$ be the lower central series of a group, $G$. Given a $n\in \mathbb N$, let $G^n = \langle g^n \mid g \in G \rangle$.
I am asked to show that if $G$ is $s$-step nilpotent (i.e. $G_{s+1} = \{e\}$ and $G_s \neq \{e\}$), then:
$$(G_s)^{n^s} \subset (G^n)_s \subset (G_s)^n$$
Here is what I have managed to show so far:
$G_s \subset Z(G)$, so $G_s$ is abelian, and thus: $(G_s)^n = \langle [g_1, \dots, g_s]^n \mid g_i \in G\rangle$
Hence also $(G_s)^{n^s} = \langle[g_1, \dots, g_s]^{n^s} \mid g_i \in G \rangle$
$(G^n)_s = \langle [g_1^n, \dots, g_s^n \mid g_i \in G]\rangle$
$G_k, G^l$ are characteristic subgroups of $G$ for all $k,l \in \mathbb N$
However, now I am unsure what I could do to make progress on this question.
It is my immediate thought to show that the generators of each of these three subgroups are contained in the group claimed to be above it. However, I am struggling to work with these extended commutators and I don't actually see how I might be able to show something like that.
For example, suppose we wanted to show the second inclusion. Then it suffices to show that for every choice of $g_1, \dots, g_s \in G$, there exists $h_1, \dots, h_s \in G$ such that:
$[g_1^n, \dots, g_s^n] = [h_1, \dots, h_s]^n$
To this end:
$[g_1^n, \dots, g_s^n]$ may be thought of as a word of length $2^{s+1} - 2^{s-1} -2$ for $s \geq 2$ in which, for each $i \leq s-1$, $g_i^{-n}, g_i^{n}$ appear $2^i$ times between them, and $g_s^{-n}, g_s^n$ appear $2^{s-1}$ times between them.
Similarly, $[g_1, \dots, g_s]^n$ is a word of length $n(2^{s+1} - 2^{s-1} - 2)$ for $s \geq 2$ and for each $i \leq s-1,$ we have $g_i^{-1},g_i$ appearing $2^in$ times between them, and $g_s^{-1}, g_s$ appear $2^{s-1}n$ times between them.
Viewing these both as words in $g_i^{\pm1}$, we see they have the same length, so the thought has occurred to me that perhaps we can shuffle the terms, gaining lower degree commutators along the way, until eventually we have a commutator that is the identity by nilpotency of $G$.
Using a small example of $s=2, n= 2$:
$[g^2,h^2] = g^{-1}g^{-1}h^{-1}h^{-1}gghh = g^{-1}h^{-1}g^{-1}h^{-1}gghh[[[[[[g^{-1},h^{-1}],h^{-1}],g],g],h],h] = g^{-1}h^{-1}g^{-1}h^{-1}gghh = g^{-1}h^{-1}g^{-1}h^{-1}ghgh[[g,h],h] = g^{-1}h^{-1}[g,h]gh = g^{-1}h^{-1}gh[g,h][[g,h],gh] = [g,h]^2$
We see that this works. However, because it is quite messy, I can't quite demonstrate that this generalises well to the cases where $s$ and $n$ are larger.
I wanted to ask:
- Firstly if there is a better way to achieve what it is that I wanted to show
- Is there some form of standard notation that might make what I'm doing here neater, as part of the problem I'm experiencing it that it is hard to keep track of things when the numbers get large.
- Does this then show equality as opposed to containment? If so, have I made a mistake, or is that simply the case?
You should use the elementary fact that if $[a, [a, b]] = 1$, then $[a^{t}, b] = [a, b]^{t}$. Now for $g_{i} \in G$ we have $$ [g_{1}^{u}, [g_{2}, \dots , g_{s}]^{v}] = [g_{1}, [g_{2}, \dots , g_{s}]]^{u v} $$ as $$ [g_{1}, [g_{2}, \dots , g_{s}]] \in G_{s} \le Z(G) $$ Now proceed by induction on $s$. Looking at the quotient group $G/G_{s}$, we will have $$ [g_{2}^{n}, \dots , g_{s}^{n}] = [g_{2}, \dots , g_{s}]^{n^{s-1}} z $$ for some $z \in G_{s} \le Z(G)$, so that $$ [g_{1}^{n}, [g_{2}^{n}, \dots , g_{s}^{n}]] = [g_{1}^{n}, [g_{2}, \dots , g_{s}]^{n^{s-1}} z] = [g_{1}^{n}, [g_{2}, \dots , g_{s}]^{n^{s-1}}] = [g_{1}, [g_{2}, \dots , g_{s}]]^{n^{s}}. $$