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Homework Help: Determining the number of subgroups

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data

    How many subgroups does //567,000 have? (don't know the smart-text for a group)

    2. Relevant equations

    3. The attempt at a solution

    From the notes I'm looking at, he did something like
    [tex]2^3 * 3^4 * 5^3 * 7 = 567,000[/tex]
    and somehow this equates to 128 subgroups.

    I have no clue how he came to this though... could someone give me a hand on finding the number of subroups in a group?
     
  2. jcsd
  3. Mar 5, 2009 #2

    Dick

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    You mean the cyclic group Z_567000, yes? It's just the number of divisors of 567000. I don't think it's 128. How many are there?
     
  4. Mar 5, 2009 #3
    Thanks for the reply!!

    I'm not exactly sure how to find the number of divisors... I think there's a way to do something with the primes, but I never really understood it.
     
  5. Mar 5, 2009 #4

    Dick

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    The number of divisors is the number of integers of the form 2^i*3^j*5^k*7^l. i is in {0,1,2,3}, j is in {0,1,2,3,4} etc. Do you get my drift? Each choice is a different number because of unique prime factorization. How many are there?
     
  6. Mar 5, 2009 #5
    I still don't understand =(
    Maybe if we used a smaller number for me to get the hang of it? sorry
     
  7. Mar 5, 2009 #6

    Dick

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    Ok. How many divisors of 12? 12=2^2*3^1. So the divisors are 2^i*3^j where i is in {0,1,2} and j is {0,1}. That's 3 choices (for i) times 2 choices (for j)=6. And that is the number of divisors of 12, right? Same idea for 567000.
     
  8. Mar 5, 2009 #7
    So for 567,000 I'm getting:

    [tex]2^3 * 3^4 * 5^3 * 7 = 210 subgroups?[/tex]
     
  9. Mar 6, 2009 #8

    Dick

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    No. How did you get 210?
     
  10. Mar 6, 2009 #9
    2 * 3 * 5 * 7 ? =/
     
  11. Mar 6, 2009 #10

    Dick

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