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Symmetric Group

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex]G=S_6[/itex] acting in the natural way on the set [itex]X = \{1,2,3,4,5,6\}[/itex].

    (a)(i) By fixing 2 points in [itex]X[/itex], or otherwise, identify a copy of [itex]S_4[/itex] inside [itex]G[/itex].

    (ii) Using the fact that [itex]S_4[/itex] contains a subgroup of order 8, find a subgroup of order 16 in [itex]G[/itex].

    (b) Find a copy of [itex]S_3 \times S_3[/itex] inside [itex]S_6[/itex]. Hence, or otherwise, show that [itex]G[/itex] contains a subgroup of order 9.

    (c) Find a subgroup of order 5 in [itex]G[/itex].

    3. The attempt at a solution

    How do I find a subgroup of S6 which is isomorphic to S4 by fixing 2 elements of X?

    For (a)(i) I know:

    [itex]S_4 = \{ e, (12), (13), (14), (23), (24), (34), (123), (132), (142),[/itex]
    [itex]\;\;\;\;\;\;\;\;\;\;\;\;(124), (134), (143), (234), (243), (1234), (1243), (1324),[/itex]
    [itex]\;\;\;\;\;\;\;\;\;\;\;\;(1342), (1423), (1432) , (12)(34) , (13)(24) , (14)(23) \}[/itex]

    As written, this can be regarded as a subgroup of S6.

    Now, if I want to fix 5 and 6, what permutations in S6 does that leave me with? And conversely, if I write S4 as I've done above, what is the action of each element of S4 on 5 and 6?

    For (a)(ii) I know that the following is a subgroup of [itex]S_4[/itex] of order 8: [tex] \{ e, (24), (13), (12)(34), (14)(23), (1432), (1234), (1324) \}[/tex]

    How do I find the subgroup of order 16 by building it up from this subgroup of order 8?
     
    Last edited: Feb 15, 2012
  2. jcsd
  3. Feb 15, 2012 #2

    Dick

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    If you regard the elements of your S4 as acting on the elements {1,2,3,4} of S, then the don't act on {5,6} at all. 5 and 6 are fixed. You could take your subgroup of order 8 and add a permutation to it that doesn't fix {5,6} to make a larger group.
     
  4. Feb 15, 2012 #3
    Yeah I see how to do it.

    For (b) what are the elements of the direct product [itex]S_3 \times S_3[/itex] ?
     
  5. Feb 15, 2012 #4

    Dick

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    You found a copy of S4 inside of S6. Try and find a copy of S3 inside of S6 in the same way. Now try and find two copies of S3 in S6 that act on disjoint subsets of {1,2,3,4,5,6}.
     
    Last edited: Feb 15, 2012
  6. Feb 15, 2012 #5
    Well a copy of [itex]S_3[/itex] in [itex]S_6[/itex] is [tex] \{ e, (12), (23), (13), (123), (132) \}[/tex]

    [itex]S_3 \times S_3 = \{ (a,b) : a,b \in S_3 \}[/itex] with operation [itex](a,b)(c,d)=(ac,bd)[/itex]

    I'm not getting how to express cyclic permutations of [itex]S_3 \times S_3[/itex]
     
    Last edited: Feb 15, 2012
  7. Feb 15, 2012 #6

    Dick

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    That's one S3. Try using {4,5,6} as elements to permute to get the second S3 in the cross product.
     
  8. Feb 15, 2012 #7
    OK so [tex]S_3 \times S_3 = \{ e, (12), (23), (13), (123), (132) \} \times \{ e, (45), (56), (46), (456), (465) \}[/tex]
     
  9. Feb 15, 2012 #8

    Dick

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    Yes, both of your S3 groups act on {1,2,3,4,5,6} and it's a direct product because they don't interfere with each other. They commute.
     
  10. Feb 15, 2012 #9
    So is this a copy of [itex]S_3 \times S_3[/itex] in [itex]S_6[/itex]?

    How do I find a subgroup of order 9 from this?
     
  11. Feb 15, 2012 #10

    Dick

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    Yes, it is. Use the same sort of idea you used to get a subgroup of order 16 from a subgroup of order 8. Think about it.
     
  12. Feb 15, 2012 #11
    Well what is the order of [itex]S_3 \times S_3[/itex]?
     
  13. Feb 15, 2012 #12

    Deveno

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    the order of S3 x S3 is 36. perhaps if you picked a subgroup of order 3 from each "factor", you might have a subgroup of order ___?
     
  14. Feb 15, 2012 #13
    So can I just say that [itex]A=\{ e, (12), (23) \}[/itex] is a subgroup of [itex]S_3[/itex] and so is [itex]B=\{ e, (45), (56)\}[/itex].

    Then [itex]A \times B[/itex] is a subgroup of [itex]S_3 \times S_3[/itex] which in turn is a subgroup of [itex]S_6[/itex] of order 9?
     
    Last edited: Feb 15, 2012
  15. Feb 15, 2012 #14

    Dick

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    [itex]A=\{ e, (12), (23) \}[/itex] is not a subgroup.
     
  16. Feb 15, 2012 #15

    Deveno

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    A is not a subgroup of S3, neither is B.

    neither one of these 3-element sets is closed under multiplication. for example:

    (1 2)(2 3) = (1 2 3) (applying (2 3) first), which is not in A.
     
  17. Feb 15, 2012 #16
    The alternating group [itex]A_3[/itex] is a subgroup:

    [itex]A=\{ e, (123), (132) \}[/itex]

    [itex]B=\{ e, (456), (465) \}[/itex]

    So [itex]A \times B < S_3 \times S_3 < S_6[/itex]

    So [itex]A \times B < S_6[/itex]

    How do I go about finding a subgroup of order 5?
     
  18. Feb 15, 2012 #17

    Dick

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    Are you thinking about these before you ask the question? Give me your thoughts on that.
     
  19. Feb 15, 2012 #18
    Well can I do something similar to the previous 2 parts? i.e. identify a copy of a certain group inside S6 and deduce a subgroup of order 5? I can't think of any groups to use though!
     
  20. Feb 15, 2012 #19

    Dick

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    It's not very similar to the other parts. Does S6 have an element of order 5?
     
  21. Feb 15, 2012 #20
    (12345)(6) is an element of order 5
     
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