Adam> Bill It's impossible if they have to be distinct up to isomorphism, but otherwise this is a solution: 1: {e} 2: C_2 2: S_2 3: C_3 3: A_3 4: C_4 4: C_2 x C_2 5: C_5 6: C_6 6: S_3 6: D_6 7: C_7 8: Q_8 Total number of elements = 57. (where C_n = cyclic group of order n; D_2n = dihedral group of order 2n; S_n = symmetric group of order n!; A_n = alternating group of order n!/2; Q_8 = quaternions {±1, ±i, ±j, ±k} under multiplication; x denotes (exterior) direct product.) Sincerely, Adam P. Goucher ------- Brilliant! --rwg