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    http://www2.imperial.ac.uk/~mwl/m1p2/M1P2SH2.PDF

    How do I do 2(e) and 2(f)?
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    e is easy - if we're working with permutations of 7 things, restrict to thinking about permutations on 4 of them. yeah?

    for f, note 7 = 3+4.
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    (e) Any element of S_6 can have order n=1,2,3,4,5,6 and any cycle of length n will have order n.

    If you have two or more disjoint cycles in a subgroup of S_n then the order of the subgroup is the least common multiple of the lengths of the cycles. You can use this to answer part (f).
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    So

    for (e) I just need a 4-cycle (and the other two elements go to themselves)

    for (f) I need a 3-cycle and a 4-cycle.
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    (Original post by gangsta316)
    So

    for (e) I just need a 4-cycle (and the other two elements go to themselves)

    for (f) I need a 3-cycle and a 4-cycle.
    Yes.
 
 
 
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