Let F be a finitely generated free group and $\gamma_m$ the lower central series. Why is $\gamma_m(F)/\gamma_{m+1}(F)$ torsionfree? I know it is abelian, but I couldn't find out more about it, as multiple commutators get very messy.
2026-03-29 12:03:44.1774785824
A lemma about free groups
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It's not just torsionfree, it's free abelian. The basic commutators of weight $m$ are a basis. This is proven in Hall's Group Theory book, and is Hall's Basis Theorem (due to Phillip Hall).