Theorem: 6 mod 3 - 0 = 0 (Lagrange divisibility for Z_6 / Z_3)

Theorem (Lagrange-Z6/Z3 canonical): the cyclic group Z_6 = {0, 1, 2, 3, 4, 5} contains the subgroup Z_3 = {0, 2, 4} (multiples of 2 mod 6). Lagrange theorem requires |Z_3| = 3 to divide |Z_6| = 6, so 6 mod 3 = 0 identically. Canonical…