“…The orbit of s 21 inside M ′ for the subgroup generated by 3 and ∶= (1, 2, 3)(4, 5) has at least 7 elements, hence it equals M ′ . Since (3,4) sends s 21 to − 32 , follows that M is a single 5 -orbit. Hence the stabilizer of s 21 has cardinality 5: but we know that it contains (1,2,3,4,5).…”