All Questions

It is known that for free $k$-step nilpotent group on $r$ generators $N(r, k)$ its integral homology is torsion-free in degrees $\leq 3$ (obvious for 1 and 2, Igusa&Orr computations for 3). ...

Fix some $n \geq 2$. For $k \geq 1$, let $N_k$ be the free $k$-step nilpotent group on $n$ generators, i.e., the quotient of the free group $F_n$ by the $(k+1)^{\text{st}}$ term $\gamma_{k+1}(F_n)$ ...

A group $G$ is said to be nilpotent if $G$ has a central series of finite length, that is, a series of normal subgroups $$ \{1\} = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G $$ ...