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EXAMPLE
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a(4) = 4 since the subgroups of S_4 up to conjugation as computed by GAP are:
H(1) = { ()}
H(2) = { (), (1,3)(2,4)}
H(3) = { (), (3,4)}
H(4) = { (), (2,3,4), (2,4,3)}
H(5) = { (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}
H(6) = { (), (3,4), (1,2), (1,2)(3,4)}
H(7) = { (), (1,2)(3,4), (1,3,2,4), (1,4,2,3)}
H(8) = { (), (3,4),(2,3), (2,3,4), (2,4,3), (2,4)}
H(9) = { (), (3,4), (1,2), (1,2)(3,4), (1,3)(2,4), (1,3,2,4), (1,4,2,3), (1,4)(2,3)}
H(10) = { (), (2,3,4), (2,4,3), (1,2)(3,4), (1,2,3), (1,2,4), (1,3,2), (1,3,4), (1,3)(2,4), (1,4,2), (1,4,3), (1,4)(2,3)}
H(11) = { (), (3,4), (2,3), (2,3,4), (2,4,3), (2,4), (1,2), (1,2)(3,4), (1,2,3), (1,2,3,4), (1,2,4,3), (1,2,4), (1,3,2), (1,3,4,2), (1,3), (1,3,4), (1,3)(2,4), (1,3,2,4), (1,4,3,2), (1,4,2), (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3)}
Only H(1), H(2), H(5) and H(7) contain neither 2-cycle nor 3-cycle and the largest of these groups has order 4.
I use here the GAP convention of writing cycles with commas.
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