Index of $G^n$ in $G$, an $s$-step nilpotent group of rank $\leq r$

For a group $$G$$ and $$n \in \mathbb N$$, let $$G^n = \langle g^n \mid g \in G \rangle$$

I am asked to show that if $$G$$ is $$s$$-step nilpotent and of rank at most $$r$$, then $$[G:G^n] \leq n^{O_{r,s}(1)}$$

I believe that the right way to approach this is to find a subset $$X \subset G$$ that definitely contains a complete set of coset representatives of $$G^n$$ in $$G$$, such that $$|X| \leq n^{ O_{r,s}(1)}$$.

Additionally, given that what we want to show is a power of $$n$$, I assume that we need to consider $$O_{r,s}(1)$$ things, each of which have $$n$$ possibilities, in some sense.

However, I am not entirely sure what these $$O_{r,s}(1)$$ things should be.

I have also considered that perhaps this question should be approached from in a more inductive way, since $$G^n \leq G^m, \; \forall m \mid n$$,

Thus if we can show the result for prime powers, $$p$$, then we would be better off since if $$K \leq H \leq G$$, then:

$$[G:K] = [G:H][H:K]$$

Thus: $$[G:G^{pq}] = [G:G^p][G^p : G^{pq}]$$

If we can then show that $$(G^p)^q = G^{pq}$$, then I think we would be done.

However, I can't seem to show the result for primes, nor can I show that $$G^{pq} = (G^p)^q$$.

I wanted to ask if either of these approaches are actually correct, and if so how I might actually go about showing this. Any help you may be able to offer would be very much appreciated, thank you!

• I am afraid that I don't know what $O_{r,s}(1)$ means! Apr 16, 2019 at 23:44
• My guess is that it means: a "constant" which depends only on $r$ and $s$. Apr 17, 2019 at 3:39
• @DerekHolt I'm sorry I should have been more clear, it is what verret has stated. A constant that may depend only on $r$ and $s$. In some sense it is a generalisation of the Big $O$ notation. Apr 17, 2019 at 9:40

Let $$G$$ be any group, and let $$G = \gamma_1(G) \ge \gamma_2(G) \ge \cdots$$ be the lower central series of $$G$$, where $$\gamma_{i+1}(G) := [G,\gamma_i(G)]$$ for $$i \ge 1$$.

Let $$G = \langle x_1,x_2,\ldots,x_r \rangle$$. It can be proved using the commutator laws that, for $$w,x \in G$$ and $$y,z \in \gamma_{i}(G)$$ for some $$i \ge 1$$, we have $$[wx,y] = [w,y][x,y] \bmod \gamma_{i+1}(G)$$ and $$[w,yz] = [w,y][w,z] \bmod \gamma_{i+1}(G)$$.

It follows that, if $$\gamma_i(G)/\gamma_{i+1}(G)$$ is generated by the images in $$\gamma_i(G)$$ of the elements $$y_{i1},y_{i2},\ldots,y_{ik_i}$$ of $$\gamma_i(G)$$, then $$\gamma_{i+1}(G)/\gamma_{i+2}(G)$$ is generated by the images of the elements $$rk_i$$ elements $$[x_j,y_{il}]$$ ($$1 \le j \le r$$, $$1 \le l \le k_i$$) of $$\gamma_{i+1}(G)$$.

So we can take $$k_i=r^i$$ for all $$i$$. (In fact we could take $$k_2 = r(r-1)/2$$.)

Furthermore, since the groups $$\gamma_i(G)/\gamma_{i+1}(G)$$ are all abelian, every element of $$G/\gamma_{i+1}(G)$$ can be written mod $$\gamma_{i+1}(G)$$ as a product of integral powers of the $$r+r^2+r^3+ \cdots +r^i$$ elements $$x_1,x_2,\ldots,x_r,y_{21},\ldots,y_{2k_2},\ldots, y_{i1},\ldots,y_{ik_i}$$ of $$G$$.

All of that is true for any group $$G$$, but if $$G$$ is nilpotent of class $$s$$, then $$\gamma_{s+1}(G)=1$$, so every element of $$G$$ can be written as a product of powers of a list of $$f(r,s) := r+r^2+\cdots+r^s$$ elements.

Then every element can be written mod $$G^n$$ as a product of such powers with exponents in the range $$0 \ldots n-1$$, and hence $$[G:G^n] \le n^{f(r,s)}$$.