Commutators and squares in free nilpotent groups. (Q963491)

An element \(x\) of a group is called square whenever \(x=y^2\) for some element \(y\) of the group. It is well-known that every commutator \([a,b]\) is the product of at most three squares and so any element in the commutator subgroup of a group can be written as a product of finitely many squares. The minimum number of squares needed to write an element \(x\) in the commutator subgroup is denoted by \(\text{Sq}(x)\). Now for any group \(G\), \(\text{Sq}(G)\) is defined to be \(\sup\{\text{Sq}(x) \mid x\in[G,G]\}\). Let \(G\) be any \(n\)-generated nilpotent group of class at most \(3\). Then \([G,G]\) is Abelian of finite rank depending only on \(n\). Hence \(\text{Sq}(G)\) is finite and the natural question is to find the exact value of \(\text{Sq}(G)\) or to find a reasonably good upper bound for it. In the paper under review the author computes \(\text{Sq}(F_{2,3})\), where \(F_{2,3}\) is the \(2\)-generated free nilpotent group of class \(3\). It turns out that \(\text{Sq}(F_{2,3})=3\). The main tool in the proofs is this easy fact that: every element of \(F_{2,3}\) appearing in a commutator can be considered as \(x^ay^b[x,y]^c\), where \(x\) and \(y\) are free generators of \(F_{2,3}\) and \(a,b,c\) are integers.