p-Divisibility of the Number of Solutions of xp = 1 in a Symmetric Group

For a prime p and for the number a(n) of solutions of xp = 1 in the symmetric group on n letters, ordp\( (a(n)) \geq [n/p] - [n/p^2] \), and especially, ordp\( (a(n)) = [n/p] - [n/p^2] \) provided \( n \equiv 0 \) mod p2. Let r be an integer with \( 1 \leq r \leq p^2 - 1 \). If ordp\( (a(r)) \leq [r/p] + 1 \), then, for each positive integer m, ordp\( (a(mp^2 + r)) = m(p-1)+ ord_p (a(r)) \). Assume that ordp\( (a(r)) = [r/p] + 2 \). If \( a(p^2 + r) \equiv -p^{p-1} a(r) mod \quad p^{p+[r/p]+2} \), then ordp\( (a(mp^2 + r)) = m(p-1) + [r/p] + 2 \); otherwise, there exists a p-adic integer b such tha ordp\( (a(mp^2 + r)) = m(p-1) + [r/p] + 2 + ord_p(m-b) \).