Sylow’s theorems in S_4

Finds the Sylow 2- and 3-subgroups of the symmetric group S_4 (order 24 = 2^3 * 3) and checks the counting constraints n_p = 1 (mod p) and \(n_p \mid |G|/p^k\).

from mathematicskit.abstract_algebra import PermutationGroup, sylow_subgroups

Sylow subgroups of S_4

s4 = PermutationGroup(4)
for p in (2, 3):
    result = sylow_subgroups(s4, p)
    m = s4.order // result.sylow_order
    print(f"p = {p}: Sylow order {result.sylow_order}, n_{p} = {result.count}")
    print(f"  n_p mod p = {result.count % p} (should be 1), n_p divides {m}: {m % result.count == 0}")
    for h in result.subgroups:
        print(f"    {h}")

p = 2: Sylow order 8, n_2 = 3
  n_p mod p = 1 (should be 1), n_p divides 3: True
    [(0, 1, 2, 3), (0, 1, 3, 2), (1, 0, 2, 3), (1, 0, 3, 2), (2, 3, 0, 1), (2, 3, 1, 0), (3, 2, 0, 1), (3, 2, 1, 0)]
    [(0, 1, 2, 3), (0, 3, 2, 1), (1, 0, 3, 2), (1, 2, 3, 0), (2, 1, 0, 3), (2, 3, 0, 1), (3, 0, 1, 2), (3, 2, 1, 0)]
    [(0, 1, 2, 3), (0, 2, 1, 3), (1, 0, 3, 2), (1, 3, 0, 2), (2, 0, 3, 1), (2, 3, 0, 1), (3, 1, 2, 0), (3, 2, 1, 0)]
p = 3: Sylow order 3, n_3 = 4
  n_p mod p = 1 (should be 1), n_p divides 8: True
    [(0, 1, 2, 3), (0, 2, 3, 1), (0, 3, 1, 2)]
    [(0, 1, 2, 3), (2, 1, 3, 0), (3, 1, 0, 2)]
    [(0, 1, 2, 3), (1, 2, 0, 3), (2, 0, 1, 3)]
    [(0, 1, 2, 3), (1, 3, 2, 0), (3, 0, 2, 1)]