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

We can understand the structure of $G$ using the following normal series: $1 \le \langle z \rangle \le \langle x^2, z \rangle \le \langle x^2, xy, z \rangle \le \langle x,y,z \rangle$

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.