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Group of order (35)^3

  1. Mar 4, 2008 #1
    [SOLVED] group of order (35)^3

    1. The problem statement, all variables and given/known data
    Show that every group of order (35)^3 has a normal subgroup of order 125.


    2. Relevant equations



    3. The attempt at a solution
    I was trying to use the Third Sylow Theorem to show that there can only be one Sylow 5-subgroup but 21 Sylow 5-subgroups seems to work fine.
     
  2. jcsd
  3. Mar 4, 2008 #2
    35=5x7
    35^3=(5^3=125) x 7^3

    now use fundamental theorem of arithmetic and first and second sylow theorem, and the third too.
     
  4. Mar 4, 2008 #3
    How could the fundamental theorem of arithmetic possibly be useful?
    By the first sylow theorem, there exists a Sylow p-group of order 125. I need to show that there can only be one such subgroup. The third Sylow theorem says that the number of Sylow 5-groups divides 35^3 and is congruent to 1 mod 5. 21 divides 35^3 and is congruent to 1 mod 5. So what am I doing wrong?
     
  5. Mar 4, 2008 #4

    Tom Mattson

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    21 does not divide 35^3. How could it? 35 has no factor of 3. So there, the Fundamental Theorem of Arithmetic is useful here after all!
     
  6. Mar 4, 2008 #5
    Sorry. We need to show that 5n+1 divides 35^2 only if n=0. Assume n is greater than 0. 35^2=(5^2)*(7^2) so when is 5n+1 a factor of (5^2)*(7^2)? Only when it is a factor of 7^2? And it is never a factor of 7^2. Therefore there is only one Sylow 5-group which must then be normal by the Second Sylow Theorem. Is that right?
     
  7. Mar 4, 2008 #6

    Tom Mattson

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    Please don't do that. When you registered you were presented with the PF Global Guidelines right before you clicked the button that says, "I agree."

    Here is a relevant excerpt.

    Color added for emphasis.
     
  8. Mar 5, 2008 #7
    sorry, I knew and read that, but I saw no harm as he already showed his work and I thought he got it, but just needed to get it straight. You're right to delete it though.

    ehrenfest: what you wrote is about right. Can you see why 1 mod 5 never divides 7^3?
     
  9. Mar 5, 2008 #8
    Yes. I actually saw the deleted post in the e-mail notification and it was exactly the same proof I had, except perhaps more clear. :)
     
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