Powering-invariance does not satisfy lower central series condition in nilpotent group

This article gives the statement, and possibly proof, of a subgroup property (i.e., powering-invariant subgroup) satisfying a subgroup metaproperty (i.e., lower central series condition)
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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., powering-invariant subgroup) need not satisfy the second subgroup property (i.e., LCS-powering-invariant subgroup)
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Statement

It is possible to have a nilpotent group $G$, a powering-invariant subgroup $H$, and a positive integer $k$ such that ${\gamma }_k(H)$, the $k^{th}$ member of the lower central series of $H$, is not a powering-invariant subgroup of ${\gamma }_k(G)$, the $k^{th}$ member of the lower central series of $G$.

In fact, for any $k>1$, we can construct an example that works for that value of $k$.

Proof

Proof for the derived subgroup (i.e., $k=2$)

Suppose $G$ is the direct product $UT(3,\mathbb{Q})\times \mathbb{Z}$. The first direct factor here is unitriangular matrix group:UT(3,Q) and the second direct factor is the group of integers. Let $H$ be the subgroup $UT(3,\mathbb{Z})$ inside the first direct factor. Then:

• $H$ is a powering-invariant subgroup of $G$, because $G$ is not powered over any prime.
• $H^{\prime }$ is not a powering-invariant subgroup of $G^{\prime }$: $H^{\prime }$ inside $G^{\prime }$ looks like Z in Q, so it is not powering-invariant.

Proof for arbitrary $k$

Take $G=UT(k+1,\mathbb{Q})\times \mathbb{Z}$ and let $H=UT(k+1,\mathbb{Z})$ inside the first direct factor. Here, $UT$ denotes the unitriangular matrix group.