Scott complexity and adjoining roots to finitely generated groups. (Q355389)

Let \(G\) be a finitely generated group with Grushko decomposition \(G_1*\cdots*G_p*F_q\). The Scott complexity of \(G\) is the ordered pair \(\chi_{\text{fd}}(G)=(q-1,p)\). An element of a group \(G\) is indivisible if it is not a proper power.NEWLINENEWLINE Several generalizations of the well-known fact that a commutator in a free group is not a proper power are given. In particular, an improved Shenitzer's theorem is proved: Let \(G=A*_{\langle t\rangle}B\) be a free product with amalgamation where \(A\) and \(B\) are free of rank \(n\) and \(m\), respectively. If \(t\) is indivisible in at least one of \(A\) or \(B\) and \(G\) maps onto a free group of rank \(n+m-1\), then \(G\) is free and \(t\) is primitive in at least one of \(A\) or \(B\). If \(t\) is not indivisible in either factor then \(G\) is a free product of a free group and an amalgamated product of two infinite cyclic groups.