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Subgroups of Z30xZ5

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data
    3. How many subgroups of order 10 are there in Z30 x Z5?


    2. Relevant equations
    the order of an element must divides the order of the group


    3. The attempt at a solution

    3. I have no idea where to start with this, especially since Z30 x Z5 is not cyclic, since 30 and 5 are not relatively prime.
     
    Last edited: Nov 3, 2011
  2. jcsd
  3. Nov 3, 2011 #2

    mathwonk

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    find at least one subgroup of order 10. or two. do you know what the order of an element is? (in this group?)

    oh you do. you showed this. well how many elements of order 10 are there in a given subgroup of order 10?
     
    Last edited: Nov 3, 2011
  4. Nov 3, 2011 #3
    There are 4 elements of order 10 in a given subgroup of order 10. I did this by brute force though; is there a more elegant way to show this? I concluded that there are 6 subgroups of order 10 that contain the 24 elements of order 10. Now I'm looking at elements of order 2 and 5 to see if I can generate any others.

    I was wondering if there is a better way to find out how many rather than just listing them all.
     
  5. Nov 3, 2011 #4

    mathwonk

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    "I concluded that there are 6 subgroups of order 10 that contain the 24 elements of order 10. Now I'm looking at elements of order 2 and 5 to see if I can generate any others."

    huhh????
     
  6. Nov 3, 2011 #5

    Deveno

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    suppose we had a subgroup H of order 10 in Z30xZ5. 2 and 5 are prime divisors of 10.

    so we have a subgroup A of H of order 2, and a subgroup B of order 5 (cauchy's theorem for abelian groups).

    now H is obviously abelian, so H = AB (chinese remainder theorem, or even the more elementary: |AB| = |A||B|/|A∩B|).

    so H is cyclic, since if a generates A, and b generates B, ab generates H

    ((ab)m = ambm, right? (H is abelian). so if (ab)m = e, 5|m and 2|m, since H∩K = {e}).

    so every subgroup of order 10 of Z30xZ5 is generated by an element of order 10. you already know how many elements of order 10 you have, now just don't count elements that generate the same subgroup of Z30xZ5 as some earlier one on your list of 24.

    for example, any element of order 10 of the form (x,0) gives the same subgroup of order 10, isomorphic to Z10 x {0}.

    (this group is generated by (3,0)).

    use mathwonk's suggestion to speed up the counting....
     
  7. Nov 3, 2011 #6
    ok, so as I said, I have 6 unique subgroups of order ten generated by the 24 elements of order 10 (ie, every 4 elements generate the same group). So, according to the logic from Deveno, this is it, right?
     
  8. Nov 3, 2011 #7
    there are no elements of order 2 in Z5 because 2 does not divide 5. elements in Z5 have order 5 or 1. If the element from Z5 has order 1, the element from Z30 must have order 10 so the lcm=10. If the element from Z5 has order 5, the element from Z30 could have order 2 or 10 for the lcm to be 10. I'm pretty sure that 10&1, 10&5, and 2&5 are the only possibilities.
     
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