For a prime p and for the number a(n) of solutions of x p = 1 in the symmetric group on n letters, ord p (a(n)) ≥ [n/p] − [n/p 2 ], and especially, ord p (a(n)) = [n/p] − [n/p 2 ] provided n ≡ 0 mod p 2 . Let r be an integer with 1 ≤ r ≤ p 2 − 1. If ord p (a(r)) ≤ [r/p] + 1, then, for each positive integer m, ord p (a(mp 2 + r)) = m(p − 1) + ord p (a(r)). Assume that ord p (a(r)) = [r/p] + 2. If a(p 2 + r) ≡ −p p−1 a(r) mod p p+[r/p]+2 , then ord p (a(mp 2 + r)) = m(p − 1) + [r/p] + 2; otherwise, there exists a p-adic integer b such that ord p (a(mp 2 + r)) = m(p − 1) + [r/p] + 2 + ord p (m − b).