Upper central series member operator commutes with root set operator for torsion-free nilpotent group

Statement
Suppose $$G$$ is a nilpotent group and $$\pi$$ is a set of primes such that $$G$$ is $$\pi$$-torsion-free. Suppose $$H$$ is a subgroup of $$G$$. For any subgroup $$K$$ of $$G$$ denote by $$\sqrt[\pi]{K}$$ the set of roots of elements of $$K$$ by $$\pi$$-numbers in $$G$$. Then, for any natural number $$n$$, we have:

$$Z^n(\sqrt[\pi]{H}) = \sqrt[\pi]{Z^n(H)}$$

Related facts

 * Divisible subset generates divisible subgroup in nilpotent group
 * Root set of a subgroup for a set of primes in a nilpotent group is a subgroup

Breakdown without the torsion-free assumption
Consider the group $$G$$ given as quotient of UT(3,Q) by a central Z, i.e., $$G = UT(3,\mathbb{Q})/\mathbb{Z}$$. Explicitly:

$$\left \{ \begin{pmatrix} 1 & a_{12} & \overline{a_{13}} \\ 0 & 1 & a_{23} \\ 0 & 0 & 1 \\\end{pmatrix} \mid a_{12},a_{23} \in \mathbb{Q}, \overline{a_{13}} \in \mathbb{Q}/\mathbb{Z} \right \}$$

with the matrix multiplication defined as:

$$\begin{pmatrix} 1 & a_{12} & \overline{a_{13}} \\ 0 & 1 & a_{23} \\ 0 & 0 & 1 \\\end{pmatrix}\begin{pmatrix} 1 & b_{12} & \overline{b_{13}} \\ 0 & 1 & b_{23} \\ 0 & 0 & 1 \\\end{pmatrix} = \begin{pmatrix} 1 & a_{12} + b_{12} & \overline{a_{12}b_{23}} + \overline{a_{13}} + \overline{b_{13}} \\ 0 & 1 & a_{23} + b_{23} \\ 0 & 0 & 1 \\\end{pmatrix}$$

where $$\overline{a_{12}b_{23}}$$ is understood to be the image of $$a_{12}b_{23}$$ under the quotient map $$\mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$$.

Take $$H$$ to be the subgroup:

$$\left \{ \begin{pmatrix} 1 & a_{12} & \overline{0} \\ 0 & 1 & a_{23} \\ 0 & 0 & 1 \\\end{pmatrix} \mid a_{12},a_{23} \in \mathbb{Q}, \right \}$$

Take $$n = 1$$ and $$\pi$$ to be the set of all primes. We then have:


 * $$\sqrt[\pi]{H} = G$$
 * $$Z(\sqrt[\pi]{H}) = Z(G)$$, which is the subgroup comprising matrices with $$a_{12} = a_{23} = 0$$.
 * $$Z(H) = H$$
 * $$\sqrt[\pi]{Z(H)} = G$$, which is the whole group.

Thus:

$$\sqrt[\pi]{Z(H)} \ne Z(\sqrt[\pi]{H})$$

The obvious direction
The containment $$Z^n(\sqrt[\pi]{H}) \le \sqrt[\pi]{Z^n(H)}$$ is obvious. This direction does not use $$G$$ being nilpotent or $$\pi$$-torsion-free.

Given: A group $$G$$, a prime set $$\pi$$, a natural number $$n$$, an element $$g \in Z^n(\sqrt[\pi]{H})$$

To prove: $$g \in \sqrt[\pi]{Z^n(H)}$$

Proof:

The difficult direction
The proof idea is to induct downward with the upper central series. We use the fact that the iterated commutator map is multilinear, and use torsion-freeness to assert that appropriate roots are also in the upper central series member.

Given: A group $$G$$ of nilpotency class $$c$$, a prime set $$\pi$$ such that $$G$$ is $$\pi$$-torsion-free, a natural number $$n$$, an element $$g \in \sqrt[\pi]{Z^n(H)}$$

To prove: $$g \in Z^n(\sqrt[\pi]{H})$$

Proof: We use downward induction on $$n$$. For simplicity, we will give here only the proof for $$n = c - 1$$, and skip the induction details.

Consider the $$c$$-fold left-normed commutator map of the form:

$$T:(x_1,x_2,\dots,x_c) \mapsto [\dots [[x_1,x_2],x_3],\dots,x_c]$$

This map is a homomorphism in each coordinate holding the other coordinates fixed. Note that this fact is very specific to the class being $$c$$. It fails for higher class. Moreover, the set of values for $$x_1$$ for which the output is always the identity for the other inputs restricted to $$H$$ is precisely the subgroup $$Z^{c-1}(H)$$.

Note that we already have that $$g \in \sqrt[\pi]{H}$$, so the only thing to show is that the following is always the identity element:

$$T(g,x_2,\dots,x_c)$$

where each $$x_i$$ in in $$\sqrt[\pi]{H}$$.

We see that if we replace each input by a suitable power of it, the first input lands inside $$Z^{c-1}(H)$$ and the remaining inputs land inside $$H$$. Thus, a suitable $$\pi$$-multiple of $T(g,x_2,\dots,x_c)$ is the identity. Since $$G$$ is $$\pi$$-torsion-free, this forces $$T(g,x_2,\dots,x_c)$$ to be the identity, so we obtain the result.

For the inductive step, when proving the result for $$n = i - 1$$ based on knowledge for $$n = i$$, consisder the $$i$$-fold left-normed commutator map, but restrict attention to the first input as being in $$Z^i(H)$$ and the remaining inputs as being arbitrary elements of $$H$$.