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Math Help - order of group G

  1. #1
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    order of group G

    ord(G) = ( p^2)*(q^2)
    ( p , q = prime number)
    G not has normal subgroup order p^2
    I want to prove :

    ord(G)=36
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  2. #2
    Member Haven's Avatar
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    Do you know sylow's third theorem?

    Also, do you know the theorem that states that every group of order p^2 is Abelian?
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  3. #3
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    sorry, I don't know the answer!!
    can you help me?
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  4. #4
    Member Haven's Avatar
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    If you don't know Sylow's Third Theorem, then this problem is considerably harder.

    However, trying to solve this problem, I can't help but feel that there might be some information missing from the problem.

    Can you explain to me what you have covered in terms of group theory. What concepts/theorems have you learned thus far?
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Haven View Post
    If you don't know Sylow's Third Theorem, then this problem is considerably harder.

    However, trying to solve this problem, I can't help but feel that there might be some information missing from the problem.

    Can you explain to me what you have covered in terms of group theory. What concepts/theorems have you learned thus far?
    The theorem does not hold. Use Sylow's theorem on a group of order 100=2^2 5^2.
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  6. #6
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    Quote Originally Posted by Swlabr View Post
    The theorem does not hold. Use Sylow's theorem on a group of order 100=2^2 5^2.

    I think the OP meant: let G be a group of order p^2q^2\,,\,\,p>q primes, and such that G has no normal subgroup of order p^2 ; then, it must be that |G| = 36.

    We can argue as follows: any sbgp. of order p^2 in G is a Sylow p-sbgp. of G, so it is non-normal iff there's more than one such sbgps. Now, we know that the number n_p of such sbgps. is congruent to 1\!\!\pmod p and that it divides q^2 , so we get that:

    n_p=q^2\Longrightarrow q^2=1\!\!\pmod p\Longrightarrow q=\pm 1\!\!\pmod p\Longrightarrow q=-1\!\!\pmod p (the other option is impossible: why?) , and as p>q we must have that p=q+1\Longrightarrow q=2\,,\,p=3 and we're done.

    Tonio
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Oh - I missed that `not'.
    Last edited by mr fantastic; July 12th 2010 at 04:43 AM.
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