2026-07-072026-07-07http://salesiana.dossiersoluciones.com/handle/123456789/166776Let X be a finitistic space with non-trivial cohomology groups H^in(X;Z)=Z with generators v_i, where i = 0, 1, 2, 3. We say that X has cohomology type (a, b) if v_1^2 = av_2 and v_1v_2 = bv_3 . In this note, we determine the mod 2 cohomology ring of the orbit space X/G of a free action of G = Z_2 on X, where both a and b are even. In this case, we observed that there is no equivariant map S^m --> X for m > 3n, where S^m has the antipodal action. Moreover, it is shown that G can not act freely on space X which is of cohomology type (a, b) where a is odd and b is even. We also obtain the mod 2 cohomology ring of the orbit space X/G of free action of G = S^1 on the space X of type (0, b).11 pagesAlgebraic Topology57S17 (Primary) 57S25 (Secondary)text