A polycyclic presentation for the \(q\)-tensor square of a polycyclic group (Q2285741)

In their fundamental paper [Topology 26, 311--335 (1987; Zbl 0622.55009)], \textit{R. Brown} and \textit{J.-L. Loday} defined the non-abelian tensor product \(G \otimes H\) of groups \(G\) and \(H\) acting compatibly on each other. The paper under review deals with a generalisation of this classical construction. \par Given groups \(G, H\) acting compatibly on each other and a non-negative integer \(q\), \textit{D. Conduché} and \textit{C. Rodríguez-Fernández} [J. Pure Appl. Algebra 78, No. 2, 139--160 (1992; Zbl 0795.20035)] defined their \(q\)-tensor product \(G\otimes^q H\) in the context of \(q\)-crossed modules. The construction yields the non-abelian tensor product \(G \otimes H\) of Brown and Loday [loc. cit.] for \(q=0\). If \(G=H\), then \(G\otimes^q G\) is referred as the \(q\)-tensor square. \par Let \(G^\phi\) be an isomorphic copy of \(G\) via an isomorphism \(\phi\). For each \(q \ge1\), let \(\widehat{\mathcal{G}}=\{\hat{k}\mid k \in G\}\) be a set of symbols one for each element of \(G\) and set \(\widehat{\mathcal{G}}=\emptyset\) for \(q=0\). Define an extension \(\nu^q(G)\) of the \(q\)-tensor square \(G\otimes^q G\) via the presentation \[\nu^q(G)= \langle G, G^\phi,\widehat{\mathcal{G}} \mid R, [g, h^\phi]^k[g^k, (h^k)^\phi]^{-1}, [g, h^\phi]^{k^\phi}[g^k, (h^k)^\phi]^{-1}~\text{for all}~g, h,k \in G \rangle,\] where \(R\) is a certain set of relations involving elements of \(\widehat{\mathcal{G}}\) and that of \(G\). The authors give an algorithm for deriving a polycyclic presentation for \(G\otimes^q G\) when \(G\) is polycyclic, via its embedding into the extension \(\nu^q(G)\). They also establish a criterion for computing the \(q\)-exterior center \(Z^\wedge_q(G)\) of a polycyclic group \(G\), which is helpful for deciding whether or not \(G\) is capable modulo \(q\). These results extend similar results of \textit{B. Eick} and \textit{W. Nickel} [J. Algebra 320, No. 2, 927--944 (2008; Zbl 1163.20022)] for the case \(q= 0\).