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Quotient groups of permutations

  1. May 4, 2012 #1
    hey guys,

    I just want grasp the whole concept of quotient groups,

    I understand say, D8/K where K={1,a2}

    I can see the quotient group pretty clearly without much trouble however I start to get stuck when working with larger groups, say S4

    For instance S4/L where L is the set of (xx)(xx) elements of S4 including e,

    Without going through and doing all the calculations I know that the order of the quotient group will be 6, however I can't see what the group will be,

    L is a normal subgroup of S4 since each element in L contains its conjugate so left and right cosets equal.

    I first thought the quotient group would be something like

    {1,(xx)L,(xx)(xx)L,(xxx)L,(xxxx)L,?} (different length cycles x L) but there are only 5 different length cycles.

    Is there a way to find these quotient groups without having to multiply L but every element of S4?
     
  2. jcsd
  3. May 4, 2012 #2


    We can try the following approach: in the quotient group [itex]\,\,S_4/L\,\,[/itex] , with [itex]L:=\{(1),\, (12)(34),\, (13)(24),\, (14)(23)\}\,\,[/itex] , we have

    that [itex]\,\,(12)L\neq (123)L\,\,[/itex] , since [itex]\,\,(12)(123)^{-1}=(12)(132)=(13)\notin L[/itex] .

    Thus, by Lagrange's theorem the group [itex]\,\,S_4/L\,\,[/itex] has order 6 and by the above it isn't abelian, thus it is isomorphic with...

    DonAntonio
     
  4. May 4, 2012 #3
    Sorry I don't follow,

    Doesn't having (12)L ≠ (123)L just mean there two elements of S4/L and the rest shows (13)L≠L, but (13)L could equal (some other permuation in S4)L,

    Is there no way to find all the elements of S4/L without doing gL for all gεS4 and finding which ones are equal, that would give 24 cosets and 18 of the cosets would have to be duplicates of the remaining 6 to give an order 6 group right?

    I could give a guess, is it isomorphic to D6?

    I was just kind of hoping there was a fast way to find quotient groups, without, in this case, having to multiply L by 24 permutations and find which of these cosets are equal

    edit

    oh i think i worked it out,

    start with (12)L

    then you get {(12),(34),( 1 4 2 3 ),( 1 3 2 4 )}


    (12)(13)(24) = (1423), so (13)(24)=(12)(1423) then (13)(24)(1423)=(12)

    so (13)(24)(1324) = (12)

    and its a normal subgroup so left and right cosets are equal so (12)L = (1324)L

    similar argument you can get (12)L=(1423)L=(34)L

    so you find 4 cosets in one, so you'll only have to do 6 calculations to find the quotient group
     
    Last edited: May 4, 2012
  5. May 5, 2012 #4


    Ok, perhaps I wasn't clear enough. What I thought is that you'd check the elements [itex]\,\,(12)(123)L\,\,,\,\,(123)(12)L\,\,[/itex] and reached the

    (correct, even) conclusion that they're different, thus making the quotient a non-abelian group without checking all the different 24 products, or even close.

    And yes, the quotient is isomorphic with [itex]\,\,D_6\cong S_3\,\,[/itex] , though (the most common notation is the second one), and the

    more usual name for the dihedral group of order 2n is [itex]\,\,D_n\,\,,\,\,not\,\,D_{2n}[/itex]

    DonAntonio
     
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