# Center not is intermediately 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., intermediately local powering-invariant subgroup)
View all such subgroup property satisfactions OR View more information on subgroup-defining functions in solvable groups

## 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 $H$.

## 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 $Z(G)$.