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Let $N$ be a group. Assume that $N$ is torsion-free, finitely generated and nilpotent. I read somewhere that $$ Z(N) \subset [N,N] \iff N \text{ cannot be written as a direct product of groups } N = A \times B \text{ where }A \text{ is non-trivial abelian.}$$ One implication is clear to me: $\implies$. I prove it by contraposition: assume $N$ can be decomposed as $A\times B$ where $A$ is non-trivial abelian. Then it is clear that $$ Z(N) = Z(A) \times Z(B) = A\times Z(B).$$ On the other hand, we have that
$$ [N,N] = [A,A] \times [B,B] = \{1\} \times [B,B].$$ If we had that $Z(N) \subset [N,N]$, we would need that $A\subset \{1\}$, which is clearly not possible as $A$ was non-trivial. So by contraposition, we have proven the first implication.

Now my problem is with the other implication. I don't know how I can prove the converse. I don't even know how to go about doing that. My gut suggests contraposition again, but then I have to use the assumption that $Z(N) \not\subset [N,N]$ to somehow construct a direct product decomposition of $N$ which has a non-trivial abelian factor, and I don't see how I can go about doing something like that.

Does anyone have any suggestions?

Thanks in advance!

Edit: I added the assumptions that $N$ is torsion-free, finitely generated and nilpotent. These are the only groups I am interested in in the context of my research.

noparadise
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    I suggest that you consider the possibility that the result you are trying to prove is not true, and look for a counterexample. – Derek Holt Mar 17 '24 at 12:36
  • Hi @DerekHolt, thank you for your response! I can't seem to find a counterexample. Moreover, I am in fact only interested in groups that are finitely generated, nilpotent and torsion-free. I edited my question to reflect this. Sorry for the confusion! – noparadise Mar 17 '24 at 13:08
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    Keep looking (3-generated 2-step nilpotent groups). – Moishe Kohan Mar 17 '24 at 16:48
  • I suspect the statement is correct if you replace $[G,G]$ with $\sqrt[G]{[G,G]}$, i.e. the set ${ g \in G \mid \exists n \in \mathbb{N}: g^n \in [G,G] }$. That could also be a hint on finding/constructing a counterexample to the original statement... – sTertooy Mar 17 '24 at 17:33
  • It is still not true for torsion-free finitely generated nilpotent groups. – Derek Holt Mar 17 '24 at 19:13

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This claim does not hold. A counterexample (even in the realm of finitely generated, torsion-free and nilpotent groups) is that of the Heisenberg group $H$ of upper triangular $3\times 3$-matrices with entries in $2\mathbb{Z}$. It can easily be seen that in this case $$ Z(H) \not\subset [H,H].$$ Yet, it is easy to check that $H$ does not have any abelian factor.

The claim can however be proven for groups that are nilpotent, torsion-free and radicable. This has to do with Mal'cev completions and the correspondence between Mal'cev completions and rational Lie algebras.

noparadise
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