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It is trivial that a group $G$ is abelian if and only if every subgroup of $G$ with two generators is abelian (i.e., any two elements commute).

If $G$ is a nilpotent group, every subgroup with two generators must be nilpotent. Is the reciprocal true? More precisely:

Let $G$ be a group and suppose that every subgroup of $G$ generated by two elements is nilpotent (with uniformly bounded class if needed). Is $G$ necessarily nilpotent?

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When $G$ is finite the answer is "yes", since any two elements of coprime order commute, so a Sylow $p$-subgroup is normal for each prime divisor $p$ of the group order. When $G$ is infinite, I don't know, but others might. Each element of $G$ is certainly an Engel element, but there are non-nilpotent groups in which every element is an Engel element, (constructed, for example, by P.M. Cohn).

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If we do not require that the class of a subgroup generated by two elements is bounded by some fixed constant, here is one example.

Consider the infinite direct sum $G = G_1 \oplus G_2 \oplus G_3 \oplus \cdots$ where $G_i$ is nilpotent of class $i$.

Then $G$ is not nilpotent, since it has nilpotent subgroups of arbitrarily large class. But any finitely generated subgroup of $G$ is contained in $G_1 \oplus G_2 \oplus \cdots \oplus G_n$ for some $n$, and this subgroup is nilpotent of class $n$. In particular any subgroup generated by two elements is nilpotent.

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I suspected something like this would kill my hopes unless one bounds the class, thank you for a quick example. – Joel Moreira Jul 29 '14 at 21:30
I wonder whether the statement is true for finitely generated groups. – Derek Holt Jul 30 '14 at 8:18
@DerekHolt: The survey here mentions at page 4 that Golod has constructed a group that is a finitely generated, non-nilpotent Engel group such that every subgroup generated by two elements is finite. Since finite Engel groups are nilpotent, that should give you an example. – Mikko Korhonen Jul 30 '14 at 20:45
@JoelMoreira: It might be a good idea to look up the results about Engel groups, perhaps there is something there that would give you a proof or a counterexample for the bounded class question – Mikko Korhonen Jul 30 '14 at 20:48

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