Commutators and squares in free groups. (Q1880644)

Let \(\mathbb{F}_2\) be the free group of rank 2 generated by \(x\) and \(y\) and denote by \([a,b]\) the commutator \(aba^{-1}b^{-1}\) of two elements \(a,b\) in a group. The main result of the paper under review is the following Theorem 1.1. The commutator \([x^m,y^n]\) is a product of two squares in \(\mathbb{F}_2=\langle x,y\rangle\) if and only if \(mn\) is even. In the case when \(m=n=1\) this gives a result of \textit{R. C. Lyndon} and \textit{M. Newman} [Proc. Am. Math. Soc. 39, 267-272 (1973; Zbl 0268.20023)] which states that the commutator \([x,y]\) is not a product of two squares. The methods of the paper also give a proof of the following theorem of \textit{M. Akhavan-Malayeri} [Int. J. Math. Math. Sci. 31, No. 10, 635-637 (2002; Zbl 1013.20028)]: \([x,y]^{2n+1}\) is not a product of two squares in \(\mathbb{F}_2=\langle x,y\rangle\). The author constructs a homomorphism \(\varphi\colon[\mathbb{F}_2,\mathbb{F}_2]\to\mathbb{Z}\) which is invariant under the conjugacy action of \(\mathbb{F}_2\). The theorem of Lyndon and Newman follows from the properties of this homomorphism. This is done as follows: Let \(K\) be the wedge of two circles. Then \(\pi_1(K)=\mathbb{F}_2\). Let \(\widetilde K\) be the universal Abelian cover of \(K\). This is the cover corresponding to the commutator subgroup \([\mathbb{F}_2,\mathbb{F}_2]\) of \(\mathbb{F}_2\). \(\widetilde K\) can be identified with a subcomplex of \(\mathbb{R}^2\) whose vertices are \(\mathbb{Z}^2\) and edges join \((i,j)\) either to \((i+1,j)\) or to \((i,j+1)\). Then consider the simplicial homology of \(\widetilde K\). The author first defines a homomorphism \(\theta\colon Z_1(\widetilde K)\to\mathbb{Z}\) from the 1-cycles to the integers which is invariant under deck transformations. The homomorphism \(\varphi\) will be defined in terms of this.