Finitely generated, nilpotent, torsion-free group that is also radicable

Question

I am currently working with Mal'cev completions, using the following definition: Let $N$ be group that is

1. Nilpotent
2. Torsion-free
3. Finitely generated

Then the Mal'cev completion or radicable hull is the unique group $N^\mathbb{Q}$ (up to isomorphism) that satisfies the four conditions below

1. $N^\mathbb{Q}$ contains $N$ as a subgroup
2. $N^\mathbb{Q}$ is nilpotent and torsion-free
3. $N^\mathbb{Q}$ is radicable, i.e. we can take unique roots in $N^\mathbb{Q}$: $$ \forall x\in N^\mathbb{Q}, m\in \mathbb{Z}_{>0},\ \exists! y \in N^\mathbb{Q}: y^m = x$$
4. For all $x\in N^\mathbb{Q}$, there exists a power $m\in \mathbb{Z}_{>0}$ such that $x^m\in N$

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Now I was wondering if there existed any groups $N$ that were equal to their Mal'cev completion. It's clear that this would be the case if and only if $N$ was finitely generated, nilpotent, torsion-free and radicable. However, I can't seem to find an example of such a group and I don't know if one exists. Does anyone have any ideas?

Thanks in advance

Answer 1

If $N$ is radicable, then the abelianization $N/[N,N]$ is divisible. But a divisible abelian group cannot be finitely generated unless it is trivial, so $N/[N,N]$ would have to be trivial, which means $N$ would have to be trivial if $N$ is nilpotent.