Derived subgroup of free nilpotent group need not be free nilpotent

Statement

It is possible to have a free nilpotent group (in fact, we can choose it to be a finitely generated free nilpotent group) whose derived subgroup is not a free nilpotent group.

In fact, the free group of nilpotency class four on any generating set of size more than one has the property that the derived subgroup is not a free nilpotent group.

Why smaller nilpotency class doesn't work

• For class zero or one, the derived subgroup is trivial, which is free of nilpotency class zero, or equivalently, free of any nilpotency class on an empty generating set.
• For class two or three, the derived subgroup is abelian (Note that this follows from the fact that derived length is logarithmically bounded by nilpotency class, which says that if the class is $c$, the derived length is $\leq \log _{2}c+1$). In the finitely generated case, it's easy to see that it must therefore be a free abelian group. The result is also true in the case of infinite generating sets, though not as obvious.

Proof

We will provide the proof here for finite size of generating set, though the idea also works with infinite size generating set.

Suppose $G$ is the free group of class four on $n$ generators, with $n>1$. Consider the subgroup $H=\gamma _{2}(G)=[G,G]$. Then, we have:

• $\gamma _{2}(H)=[H,H]=[[G,G],[G,G]]\leq \gamma _{4}(G)$ (using lower central series is strongly central), and $\gamma _{3}(H)=1$. $H$ has class at most two. Note that $H$ is not abelian, this can be worked out by observing that there exist actual 2-generated nilpotent groups whose derived subgroup is non-abelian.
• We also know that $[\gamma _{2}(G),\gamma _{3}(G)]\leq \gamma _{5}(G)=1$ (we use that lower central series is strongly central). In particular, this means that $\gamma _{3}(G)\leq Z(\gamma _{2}(G))=Z(H)$.
• In particular, this means that $Z(H)$ is strictly bigger than $\gamma _{2}(H)$.
• If $H$ were a free nilpotent group it would be a free class two group and that would force $Z(H)=\gamma _{2}(H)$, so therefore $H$ is not nipotent.

We can also verify numerically that the ranks of the abelianization and derived subgroup of $H$ don't fit the formula for the ranks in a free class two group. If $H$ were free of class two and its abelianization had rank $m$, its derived subgroup would have rank $(m^{2}-m)/2$.

However, we know that since $\gamma _{2}(H)\leq \gamma _{4}(G)$, the rank of the abelianization of $H$ is at least the rank of the abelian group $H/\gamma _{4}(G)=\gamma _{2}(G)/\gamma _{4}(G)$, which is the sum of the ranks of $\gamma _{2}(G)/\gamma _{3}(G)$ and $\gamma _{3}(G)/\gamma _{4}(G)$, so we get:

• ${\frac {n^{2}-n}{2}}+{\frac {n^{3}-n}{3}}$ is less than or equal to the rank of the abelianization of $H$.
• The rank of the derived subgroup of $H$ is at most ${\frac {n^{4}-n^{2}}{4}}$.

These clearly don't work out for $n>2$:

$n$	${\frac {n^{2}-n}{2}}+{\frac {n^{3}-n}{3}}$ (lower bound on $m$, rank of abelianization)	Corresponding lower bound on ${\frac {m^{2}-m}{2}}$	${\frac {n^{4}-n^{2}}{4}}$ (upper bound on rank of derived subgroup)	Shows it isn't free?
2	3	3	3	Doesn't show it, but it's not free (the bounds aren't tight)
3	11	55	18	Yes, shows it isn't free
4	26	325	60	Yes, shows it isn't free