Using the classification of finite simple groups, the author proves that if a finite group \(A\) acts fixed-point-freely on a finite group \(G\) and that either \(A\) is cyclic or \((|G|,|A|)=1\), then \(G\) is solvable. The proof is dependent on the following lemmas, each of interest in itself. (1) Let \(H\) be an \(A\)-invariant subgroup of \(G\). Then \(A\) acts fixed-point-freely on \(H\) and if \(H\trianglelefteq G\), fixed- point-freely on \(G/H\). (2) Suppose that \(A\) and \(G\) both act upon a finite set \(\Omega\). If \(G\) is transitive on \(\Omega\) and \((\omega g)^\alpha=\omega^\alpha g^\alpha\) for all \(\omega\in\Omega\), \(g\in G\), and \(\alpha \in A\), there is a unique element of \(\Omega\) fixed by \(A\). (3) If \(G\) has a cyclic Sylow \(p\)-subgroup, then \(G\) has a normal \(p\)-complement [observed first by F. Gross: Can. J. Math. 20, 1300-1307 (1968; Zbl 0186.31803)].