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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!

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  • $\begingroup$ I am afraid that I don't know what $O_{r,s}(1)$ means! $\endgroup$
    – Derek Holt
    Apr 16, 2019 at 23:44
  • $\begingroup$ My guess is that it means: a "constant" which depends only on $r$ and $s$. $\endgroup$
    – verret
    Apr 17, 2019 at 3:39
  • $\begingroup$ @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. $\endgroup$
    – user366818
    Apr 17, 2019 at 9:40

1 Answer 1

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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 $[G:G^n] \le n^{f(r,s)}$.

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