Torsion-free not implies powering-injective

Non-solvable version
It is possible to construct a group $$G$$ with the following properties:


 * 1) $$G$$ is a torsion-free group, i.e., it has no non-identity element of finite order. In particular, it is also a $$\pi$$-torsion-free group for every prime set $$\pi$$.
 * 2) $$G$$ is not a powering-injective group for any prime. In other words, for any prime number $$p$$, the map $$x \mapsto x^p$$ is not an injective set map from $$G$$ to itself.

Solvable version
It is possible to construct a solvable group $$G$$ with the following properties:


 * 1) $$G$$ is a torsion-free group, i.e., it has no non-identity element of finite order. In particular, it is also a $$\pi$$-torsion-free group for every prime set $$\pi$$.
 * 2) $$G$$ is not a powering-injective group for the prime 2.

Proof
Let $$G$$ be the amalgamated free product of two copies of the group of rational numbers amalgmated over a shared copy of the group of integers. Explicitly, $$G = \mathbb{Q} *_{\mathbb{Z}} \mathbb{Q}$$.

Then the following are true:


 * $$G$$ is a torsion-free group.
 * The map $$x \mapsto x^p$$ is not injective in $$G$$ for any prime number $$p$$. This is because the generator of the shared $$\mathbb{Z}$$ is a $$p^{th}$$ power of a suitable element in the first free factor, and also of a suitable element in the second free factor.

Solvable example for the case p = 2
Let $$G$$ be the amalgamated free product of two copies of $$\mathbb{Z}$$ with a common $$2\mathbb{Z}$$ identified. Explicitly:

$$G = \langle a,b \mid a^2 = b^2 \rangle$$

$$G$$ is solvable because it has this normal series:

$$1 \le \langle a^2 \rangle \le \langle a^2,ab \rangle \le \langle a,b \rangle$$

with successive quotients $$\mathbb{Z}, \mathbb{Z}, \mathbb{Z}/2\mathbb{Z}$$. The quotient of the whole group by $$\langle a^2 \rangle$$ is a free product of two copies of cyclic group:Z2, which is isomorphic to the infinite dihedral group.

We note that:


 * $$G$$ is a torsion-free group, and in particular, a 2-torsion-free group.
 * The square map in $$G$$ is not injective, because $$a^2 = b^2$$.