Torsion-free not implies powering-injective

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., torsion-free group) need not satisfy the second group property (i.e., powering-injective group for a set of primes)
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Statement

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:

$G=⟨a,b\mid a^2=b^2⟩$

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

$1\le ⟨a^2⟩\le ⟨a^2,ab⟩\le ⟨a,b⟩$

with successive quotients $\mathbb{Z},\mathbb{Z},\mathbb{Z}/2\mathbb{Z}$. The quotient of the whole group by $⟨a^2⟩$ 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 $a^2=b^2$.