## Commutators in metabelian groups. (Commutateurs dans les groupes métabéliens.)(French)Zbl 0769.20015

Let $$G$$ be a group and let $$G'$$ be the commutator subgroup of $$G$$. Then $$c(G)$$ is defined as the least positive integer $$k$$ such that every element of $$G'$$ is the product of $$k$$ commutators, if such an integer exists, otherwise $$c(G) = \infty$$. The authors prove that if $$G_ n$$ is the free metabelian group on $$n$$ generators $$(n \geq 2)$$, then $$c(G_ 2) = 2$$ and $$\left[{n\over 2}\right] \leq c(G_ n) \leq n$$, where $$\left[{n\over 2}\right]$$ is the integral part of $${n\over 2}$$. Moreover, if $$G$$ is the free metabelian group on a countable set of generators, then $$c(G) = \infty$$. As a consequence $$c(G)$$ is finite for every polycyclic group $$G$$.

### MSC:

 20F12 Commutator calculus 20F05 Generators, relations, and presentations of groups 20F16 Solvable groups, supersolvable groups 20E05 Free nonabelian groups
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### References:

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