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Permutation/Alternating Group

  1. Apr 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Let Qn = {x in Sn | x=y2 for some y in Sn}.

    Is it true or false that Qn = An?


    2. Relevant equations



    3. The attempt at a solution
    I'm inclined to say it's false because of the strangeness of S2. Shouldn't A2 contain both the identity and (12)? But (12)(12) is the identity so it seems there is no way to get (12) by the definition of Q. Is this correct or am I missing something?
     
  2. jcsd
  3. Apr 16, 2009 #2
    no S2 has 2!=2 elements, A2 has 2!/2=1 elements (the identity)
     
  4. Apr 16, 2009 #3
    Oh yes, I can't believe I overlooked that. But I'm still not sure whether Q is equal to the alternating group...
     
  5. Apr 17, 2009 #4

    matt grime

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    Do you know of any set of elements that generates A_n?

    Forget that - that's wrong. I thought you had to look at the group generated by Q_n since it is not even clear that Q_n is a subgroup of S_n, nevermind being equal to A_n.

    Instead, have you tried to see if it's true for n=3,4 or 5?
     
    Last edited: Apr 17, 2009
  6. Apr 17, 2009 #5

    matt grime

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    Ok, let me go further: have you checked if it is true for n=6? (perhaps via a computer algebra package).

    If you don't have one (a computer algebra package I mean), let's try thinking about why it might be false (rather than finding a counter example).

    If g has order k, what is the order of g^2? (There are two cases: k even and k odd.)
     
  7. Apr 17, 2009 #6
    It seems like it is true from the examples I worked out. For instance, the square of a 5-cycle is a 5-cycle, and in general the square of a cycle of odd length n will produce another cycle of odd length n.
    For even n, I am getting that for length 2n you get two n-cycles.
    So I am thinking it is probably true. I can't seem to find any reasoning why it wouldn't be at this point.
     
  8. Apr 17, 2009 #7

    matt grime

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    Well, I'm afraid you're wrong: Q_n isn't even a group in general, nevermind being A_n.

    I think you have the hint the wrong way round. Suppose that I have element of order 4 in A_n. What order must its 'square root' have if it were to exist.
     
  9. Apr 17, 2009 #8
    Okay, here is what I think may be a counterexample:
    the permutation (1234)(56) is in A6, but it cannot be produced by squaring any element in S6. So yes, you're right Qn is not equivalent to An.
     
  10. Apr 17, 2009 #9

    matt grime

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    How do you know it is not the square of any element in S_6? (It isn't, but you need a proof of this fact.)
     
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