On projective planes of order 15 admitting a collineation of order 7 (Q1580265)

Let \(\Pi\) be a projective plane of order 15 with a collineation group \(G.\) The author shows that \(G\) cannot have 21 elements. As a corollary he obtains that if 7 divides \(|G|,\) then \(|G|\) divides \(2^6\cdot 7.\) \textit{C. Y. Ho} [Geom. Dedicata 27, No. 1, 49-64 (1988; Zbl 0646.51014)] claims that \(|G|\) cannot be 49, but his proof is incorrect. The author shows the result to be true. In the proofs a computer is used.