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Math Help - Disjoint Cycles

  1. #1
    Junior Member
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    Disjoint Cycles

    P=
    (1234567
    3125746)

    q=
    (1234567
    5641723)

    r=
    (1234567
    7631425)

    s=
    (1234567
    5713264)

    Find pq(^-1)sr(^2) write the answer as a product of disjoint cylcles. Find its order Parity and Inverse.

    I understand parts of the question but for the majority I am unable to do it.
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  2. #2
    Junior Member beebe's Avatar
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    Re: Disjoint Cycles

    Which parts, specifically, are you unable to do?
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  3. #3
    Junior Member
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    Re: Disjoint Cycles

    (132)(4576)*(15734)(26)
    start with 1
    (1
    q sends 1 to 5 and p sends 5 to 7
    (17- because p.q(1) = p(5) = 7
    so the composite permutation begins (17...)
    then see what p.q(7) is and so on
    q sends 7 to 3 and p sends 3 to 2
    (172
    q sends 2 to 6 and p sends 6 to 4
    (1724
    q sends 4 to 1 and p sends 1 to 3
    (17243
    q sends 3 to 4 and p sends 4 to 5
    (172435
    q sends 5 to 7 and p sends 7 to 6
    (1724356
    just to check, q sends 6 to 2 and p sends 2 to 1
    (1724356)-7 digits so completed as pq

    To find sr we do the same
    Start with 1
    Sr
    1
    R takes 1 to 7 and s takes 7 to 4
    14
    R takes 4 to 1 and s takes 1 to 5
    145
    R takes 5 to 4 and s takes 4 to 3
    1453
    R takes 3 to 3 and s takes 3 to 1
    14531


    this is as far as ive got
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  4. #4
    MHF Contributor

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    Re: Disjoint Cycles

    ok, well, i do not know if for you:

    pq means first do q, and then do p, or vice-versa (different authors use different conventions). i will assume pq(x) = p(q(x)) (do q first).

    so first we have to calculate pq-1sr2. this is tedious.

    first we do r twice:

    1→7→5
    2→6→2
    3→3→3
    4→1→7
    5→4→1
    6→2→6
    7→5→4

    then we apply s:

    1→5→2
    2→2→7
    3→3→1
    4→7→4
    5→1→5
    6→6→6
    7→4→3

    then we apply q-1, which is:

    1→4
    2→6
    3→7
    4→3
    5→1
    6→2
    7→5, to get:

    1→2→6
    2→7→5
    3→1→4
    4→4→3
    5→5→1
    6→6→2
    7→3→7

    finally, we apply p to get:

    1→6→4
    2→5→7
    3→4→5
    4→3→2
    5→1→3
    6→2→1
    7→7→6, so that pq-1sr2 = (1 4 2 7 6)(3 5)

    this is the product of a disjoint 5-cycle and 2-cycle, so has order lcm(2,5) = 10.

    alternatively, we have pq-1sr2 = (1 3 2)(4 5 7 6)(1 4 3 7 5)(2 6)(1 5 2 7 4 3)(1 7 5 4)(2 6)(1 7 5 4)(2 6).

    thus pq-1sr2(1) = p(q-1(s(r(r(1)))))

    = p(q-1(s(r(7)))) = p(q-1(s(5))) = p(q-1(2)) = p(6) = 4 so we start:

    (1 4......

    then we calculate pq-1sr2(4) = p(q-1(s(r(r(4)))))

    = p(q-1(s(r(1)))) = p(q-1(s(7))) = p(q-1(4)) = p(3) = 2, so we continue:

    (1 4 2.....

    next, we want to know pq-1sr2(2) = p(q-1(s(r(r(2)))))

    = p(q-1(s(r(6)))) = p(q-1(s(2))) = p(q-1(7)) = p(5) = 7, so we are up to:

    (1 4 2 7....

    AND.....pq-1sr2(7) = p(q-1(s(r(r(7)))))

    = p(q-1(s(r(5)))) = p(q-1(s(4))) = p(q-1(3)) = p(7) = 6, so we have:

    (1 4 2 7 6.....

    and quickly, now! pq-1sr2(6) = p(q-1(s(r(r(6)))))

    = p(q-1(s(r(2)))) = p(q-1(s(6))) = p(q-1(6)) = p(2) = 1, so our first cycle closes:

    (1 4 2 7 6)(....

    the next number not in the first cycle is 3, so we calculate:

    pq-1sr2(3) = p(q-1(s(r(r(3)))))

    = p(q-1(s(r(3)))) = p(q-1(s(3))) = p(q-1(1)) = p(4) = 5, and by bijectivity,

    we must have pq-1sr2(5) = 3, so we get (1 4 2 7 6)(3 5), as we did above.

    5-cycles are even, and (3 5) is a transposition, therefore pq-1sr2 is odd.
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