Center not is quotient-local powering-invariant in solvable group

This article gives the statement, and possibly proof, of the fact that in a group satisfying the property solvable group, the subgroup obtained by applying a given subgroup-defining function (i.e., center) need not satisfy a particular subgroup property (i.e., quotient-local powering-invariant subgroup)
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Statement

It is possible to have a solvable group $G$ with center $Z(G)$ such that $Z(G)$ is not a quotient-local powering-invariant subgroup of $G$. Explicitly, let $\varphi :G\to G/Z(G)$ be the quotient map. Then, there exists a prime number $p$ and an element $g\in G$ such that $g$ has a unique $p^{th}$ root in $G$, but $\varphi (g)$ has multiple $p^{th}$ roots in $G/Z(G)$.

Proof

Define:

$G:=\langle x,y,z\mid y^{2}=x^{2}z,[x,z]=[y,z]=1\rangle$

We can understand the structure of $G$ using the following normal series:

$1\leq \langle z\rangle \leq \langle x^{2},z\rangle \leq \langle x^{2},xy,z\rangle \leq \langle x,y,z\rangle$

The successive quotients are $\mathbb {Z} ,\mathbb {Z} ,\mathbb {Z} ,\mathbb {Z} /2\mathbb {Z}$. More details below:

• $\langle z\rangle$ is the center and the quotient group $G/\langle z\rangle$ is isomorphic to the amalgamated free product $\mathbb {Z} *_{2\mathbb {Z} }\mathbb {Z}$, with the two pieces generated by the images of $x$ and $y$ and the amalgamated part being the image of $x^{2}$, which coincides with the image of $y^{2}$.
• $\langle x^{2},z\rangle$ is the second center and the quotient group $G/\langle x^{2},z\rangle$ is isomorphic to the free product $\mathbb {Z} /2\mathbb {Z} *\mathbb {Z} /2\mathbb {Z}$, which in turn is isomorphic to the infinite dihedral group (where the images of $x$ and $y$ are both reflections whose product gives a generator for the cyclic maximal subgroup).
• $\langle x^{2},xy,z\rangle /\langle x^{2},z\rangle$ is the cyclic maximal subgroup inside $G/\langle x^{2},z\rangle$.

Consider now the element $g=x^{2}$. We have the following:

• $G$ is solvable: This is obvious from the normal series for $G$ where all the quotients are abelian.
• $g$ has a unique square root, namely $x$, in $G$: This requires some work to show rigorously, and can be demonstrated using a polycyclic presentation with the elements $x,u,z$ where $u=xy$. The idea is to compute the general expression for the square of an arbitrary element that is of the form $x^{m_{1}}u^{m_{2}}z^{m_{3}}$ and deduce that, for the square to equal $g$, we must have $m_{1}=1,m_{2}=m_{3}=0$.
• Under the quotient map by the center, the image of $g$ does not have a unique square root: The center is $\langle z\rangle$, and the quotient group is $\mathbb {Z} *_{2\mathbb {Z} }\mathbb {Z}$. As noted above, the image of $g=x^{2}$ coincides with the image of $y^{2}$ in the quotient, but the images of $x$ and $y$ do not coincide. Thus, the image of $g$ modulo the center does not have a unique square root in the quotient group by the center.