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Sylow Subgroups

  1. Aug 26, 2010 #1
    Hi Guy's,
    I know this is not for home work questions however I have had no luck in that section.

    Have I done enough to show that 10 cannot be a sub-group of order 324


    1. The problem statement, all variables and given/known data

    Let G be a group of order 324. Show that G has subgroups of order 2,
    3, 4, 9, 27 and 81, but no subgroups of order 10.


    2. Relevant equations

    Sylow showed that if a prime power divides the order of a finite group G, then G has a subgroup of order .

    3. The attempt at a solution


    I can see that G can have the subgroup 2 because [itex]2^n n=1 = 2[\latex]
    subgroup 3 because [itex]3^n n=1 = 3[\latex] divides 324
    subgroup 4 because [itex]2^n n=2 = 4[\latex] divides 324
    subgroup 9 because [itex]3^n n=2 = 9[\latex] divides 324
    subgroup 27 because [itex]3^n n=3 = 27[\latex] divides 324
    subgroup 81 because [itex]3^n n=4 = 81[\latex] divides 324

    I know that 10 does not divide 324 in Z

    Is that enough to show that the sub group can't be order 10 ?
     
  2. jcsd
  3. Aug 26, 2010 #2

    Office_Shredder

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    Yes, see Lagrange's theorem
     
  4. Aug 26, 2010 #3
    thanks alot
     
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