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

Statement
It is possible to have a 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 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.
 * $$H'$$ is not a powering-invariant subgroup of $$G'$$: $$H'$$ inside $$G'$$ 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.