Use Sylow Solution By Sylow Theorems, n2 | 3, and n2 = 1 (mod 2) ==> n2 = 1 or 3

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By Sylow Theorems, n2 | 3, and n2 = 1 (mod 2) ==> n2 = 1 or 3 n3 | 8, and n3 = 1 (mod 3) ==> n3 = 1 or 4. ---------------- Sylow 2-subgroups of S4 are necessarily of order 2^3 = 8; and are isomorphic to D4 (check!): One of them (call it H) is generated by (1234) and (13)(24)................................................ All other such subgroups must be conjugates. There are two others (as predicted): We find that they are (after some hunting): (12) H (12)^(-1) and (14) H (14)^(-1)..................... (You can find the generators to these by conjugating the ones for H with either (12) or (14).) ................................. Sylow 3-subgroups of S4 are necessarily of order 3; hence any such subgroup is generated by a 3-cycle..................................... They are {(1), (123), (132)}, {(1), (124), (142)}, {(1), (134), (143)}, and {(1), (234), (243)}. I hope this helps!

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