Test elements in direct products of groups (Q2709979)

An element \(g\) of a group \(G\) is called a test element if for any endomorphism \(\phi\colon G\to G\), \(\phi(g)=g\) implies that \(\phi\) is an automorphism. It is easy to check that if \(g=(g_1,\dots,g_n)\) is a test element for a direct product \(G=G_1\times\cdots\times G_n\), then (a)~each \(g_i\) is a test element for \(G_i\), and (b)~\(\{g_1,\dots,g_n\}\) is an independent set, which means that for no \(i\neq j\) is there a homomorphism \(\psi\colon G_i\to G_j\) with \(\psi(g_i)=g_j\). The first main result of this paper is that these two conditions are sufficient if in addition each \(G_i\) has cyclic centralizers and \(g\) is in the commutator subgroup of \(G\).NEWLINENEWLINENEWLINETo provide examples, the authors prove that in the free group \(F_r\) on the set \(\{x_1,\dots,x_r\}\) with \(r>1\), the elements \(w_{k,r}=x_1^k\cdots x_r^kx_1^{-k}\cdots x_r^{-k}\) are test elements when \(k>1\), and that \(\{w_{p_1,r_1},\dots,w_{p_n,r_n}\}\) is an independent set of elements of \(F_r\) if the \(p_i\) are distinct odd primes. A similar result holds for surface groups. The authors note that hyperbolic groups containing torsion need not have test elements, but conjecture that all torsionfree hyperbolic groups do have them.