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Algebra, find orbits and stabilizers

  1. Dec 14, 2015 #1
    Hello I have the solution of a problem and I don't understand it

    1. The problem statement, all variables and given/known data

    We know that every subgroup L<S10 acts on [1, 10] := {1, 2,..., 10} by the formula π • i = π(i). Consider L the subgroup of S10 generated by the permutation p = (1, 2, 3, 4)(4, 5)(8, 9, 10).
    Find the orbit and stabilizer of 2

    2. Relevant equations
    Orb(x)={l•x ,l ∈ L }
    Stab(2)={ l ∈ L , l•x=x}

    3. The attempt at a solution
    Here's the solution
    L=< { p=(1, 2, 3, 4)(4, 5)(8, 9, 10) = (1, 2, 3, 4,5)(8, 9, 10) } > ⊂ S10 is generated by a 5-cycle and a 3-cycle so |L|=15 because(1, 2, 3, 4,5) &(8, 9, 10) don't commute.
    Orb(2) = { 1, 2, 3, 4 , 5 }
    Stab(2)={ Identity, p5, p10 }


    I don't understand how they found the orbit. By definition Orb(2)= L • 2 := {l • 2; l ∈ L}. l • 2= l(2) and thn I don't know what to do.
    For the Stabilizer why they take p5=(8,10,9) and p10=(8,9,10) ? The order of the permutation is 15 so why Stab(2) isn't { Identity, p15} ?
    Thanks
     
  2. jcsd
  3. Dec 14, 2015 #2

    andrewkirk

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    Let ##a=(1\ 2\ 3\ 4\ 5)## and ##b=(8\ 9\ 10)##. The subgroup L is ##\{a^nb^n\ |\ n\in \mathbb{N}\}##. In calculating the orbit of 2 you can ignore the ##b^n## factor because it doesn't act on 2.
    So the orbit of 2 is ##\{a^n(2)\ |\ n\in\mathbb{N}\}## which is ##\{a^n(2)\ |\ n\in\{0,1,2,3,4\}\}## since ##a^n=##Identity.

    ##p^{15}## is equal to the identity, so will not be listed in any stabilizer group.

    From the above we have that ##p^5=b^5, p^{10}=b^{10}## because the ##a## part is the identity, having a cycle of 5. To see why ##b^5=##(8 10 9) apply the permutation ##b## five times and see what happens.
     
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