Second center not is 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., second center) need not satisfy a particular subgroup property (i.e., local powering-invariant subgroup)
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It is possible to have a solvable group G such that the second center of G is not a local powering-invariant subgroup of G. In other words, there exists an element g in the second center and a natural number n such that the element g has a unique n^{th} root in G but this n^{th} root is not inside the second center.

Related facts


Define (here 1 denotes the identity element):

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 is in the second center, based on the description of the second center as \langle x^2,z \rangle.
  • 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.
  • The unique square root of g is not in the second center: This follows from the explicit description of the second center and the information above.