Center not is intermediately local powering-invariant in solvable group

Statement
It is possible to have a solvable group $$G$$ such that the center $$Z(G)$$ is not an intermediately local powering-invariant subgroup of $$G$$, i.e., there exists an intermediate subgroup $$H$$ of $$G$$ such that $$Z(G)$$ is not a local powering-invariant subgroup of $$H$$.

Opposite facts

 * Upper central series members are intermediately local powering-invariant in nilpotent group

Proof
Consider the group $$G = \mathbb{Z} *_{2\mathbb{Z}} \mathbb{Z}$$, explicitly given as:

$$G := \langle x,y \mid x^2 = y^2 \rangle$$

with the subgroup:

$$H = \langle x \rangle$$


 * The center of $$G$$ is the subgroup $$Z(G) = \langle x^2 \rangle$$, which is the amalgamated copy of $$2\mathbb{Z}$$.
 * $$G$$ is solvable: In fact, $$G/Z(G)$$ is isomorphic to the infinite dihedral group, which is a metacyclic group.
 * The subgroup $$H = \langle x \rangle$$ is isomorphic to $$\mathbb{Z}$$, with $$Z(G)$$ living as $$2\mathbb{Z}$$ inside it.
 * $$Z(G)$$ is not local powering-invariant in $$H$$: To see this, note that the element $$x^2 \in Z(G)$$ has unique square root $$x \in H$$ but this square root is not in $$Z(G)$$.