Center not is intermediately 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., intermediately local powering-invariant subgroup)
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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.

Related facts

Opposite facts

Proof

Further information: amalgamated free product of Z and Z over 2Z

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).