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Group theory: True or false

  1. Jan 26, 2009 #1

    tgt

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    1. The problem statement, all variables and given/known data
    Every nontrivial subgroup H of the symmetric group with 9 elements containing some odd permutation contains a transposition.



    It does seem the case that if a subgroup of H of the symmetric group with 9 elements contain an odd permutation then certainly a transposition must be apparent (there might be more but surely one is apparent).

    Have I misread the question?
     
  2. jcsd
  3. Jan 26, 2009 #2

    Dick

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    There are odd elements that don't consist of a single transposition. Some are the product of three transpositions. Or more. And there is no symmetric group with 9 elements. So you must mean H has 9 elements. And if H contains a transposition then it has a element of order 2. Or do you mean S_9? Still not true. The more I think about this the less sense it makes. Are you sure that's the real question?
     
    Last edited: Jan 26, 2009
  4. Jan 27, 2009 #3

    tgt

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    Maybe I have. It's asking whether every nontrivial subgroup of S9 containing an odd permutation must contain a single transposition. The answer is no if we consider the group {I, (12)(34)(56)}.
     
  5. Jan 27, 2009 #4

    Dick

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    Ok, right. That's a subgroup of S9 and contains no transposition. I'm still fixated on H being having 9 elements. Sorry.
     
    Last edited: Jan 27, 2009
  6. Jan 27, 2009 #5

    tgt

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    symmetric group with 9 element is S_9 which certainly exits. Why do you say it doesn't?

    http://en.wikipedia.org/wiki/Symmetric_group

    The answer to the OP is false as shown by a counter example above.
     
  7. Jan 27, 2009 #6

    tgt

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    Yes.

    You've probably mistaken the wording in the OP. The symmetric group with 9 elements is obviously not good use of words. I really mean S_9 which has 9! elements.
     
  8. Jan 27, 2009 #7

    Dick

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    I agree. There is a subgroup of order 2 with no transposition. As you're example points out. I edited the previous reply.
     
  9. Apr 5, 2011 #8
    This is an interesting problem:
    Its Question 13d of "A First Course in Abstract Algebra by John B Fraleigh".
    Of course the solution in the back of the book is wrong because it says that the statement is false; when in fact it is true. He also says that A_3 is a commutative group WHEN ITS CLEARLY NOT! thats question 13g. This is an excellent book for finding mistakes ..... if you can find them.
     
  10. Apr 6, 2011 #9

    Deveno

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    A3 IS commutative, as is any group of order 3.

    A3 = {I, (1 2 3), (1 3 2)}

    it is cyclic, since 3 is prime.
     
  11. Apr 6, 2011 #10
    Yeah you're right .... thanks for pointing that out. My bad
     
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