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

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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.