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HELP Find all abelian groups (up to isomorphism)

  1. Nov 24, 2009 #1
    HELP!!! Find all abelian groups (up to isomorphism)!!!

    I am really confused on this topic.
    can you give me an example and explain how you found, pleaseee!
    for example, when i find abelian group of order 20;
    |G|=20
    i will find all factors and write all of them,
    Z_20
    (Z_10) * (Z_2)
    (Z_5)* (Z_2) * (Z_2)

    for higher order such as |G|=200 , i can't do this.(i did and it was wrong)
    can you tell me the difference.. how should i find them???
    thanks by now....
     
  2. jcsd
  3. Nov 24, 2009 #2
    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    Oh, finite ... How boring.
    I read the title "Find all abelian groups" so I came here to tell you to read on Ulm's Theorem. But that's not so interesting for finite groups.
     
  4. Nov 24, 2009 #3
    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    yeah, tell me!
    in the textbooks there is nothing.. i can't find any understandable example. just theorems and proofs! is there anyone can help?
     
  5. Nov 24, 2009 #4
  6. Nov 24, 2009 #5
    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    VeeEight, thanks for this example, but it is not explained how they found..
     
  7. Nov 24, 2009 #6
    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    Here is another link: http://planetmath.org/?op=getobj&from=objects&id=4654

    The important thing to remember here is that Zmn is isomorphic to ZmxZn if and only if gcd (m,n)=1. Therefore, Z6 is isomorphic to Z3xZ2 because gcd(2,3)=1 (and (2)(3)=6). Thus, one way to complete this exercise is to try to decompose your group Z200 into a direct product of groups, ZaxZbxZc... where (a)(b)(c)... = 200 and then apply the above theorem to cancel out the groups that are isomorphic.

    For example, 200 can be decomposed as 200=(25)(8). But gcd (25, 8) = 1 so Z25xZ8 is isomorphic to Z200 by the above theorem. Doing this method of writing out all the decompositions and canceling out the isomorphic groups works but can be frustrating since there is a lot of information to work through. Therefore it is helpful to first find the prime factorization of the order (200 in this case) and work out the problem like they did in the link above.
     
    Last edited: Nov 24, 2009
  8. Nov 24, 2009 #7
    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    VeeEight, this answer and the link were really helpful.. Thanks..
     
  9. Nov 24, 2009 #8
    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    now i get it! by using your information and my homework assignment i got this idea:
    for 200 = (2^3).(5^2)
    so, our factors can be;
    (2^3).(5^2) =Z_8 x Z_25
    (2^1).(2^2).(5^2) = Z_2 x Z_4 x Z_25
    (2^1).(2^1).(2^1).(5^2)= Z_2 x Z_2 x Z_2 x Z_25
    these are for varying exponents of 2.
    then,
    (2^3).(5^1).(5^1) =Z_8 x Z_25
    (2^1).(2^2).(5^1).(5^1) = Z_2 x Z_4 x Z_25
    (2^1).(2^1).(2^1).(5^1).(5^1) = Z_2 x Z_2 x Z_2 x Z_25
    these are for varying exponents of 5

    also, one of the direct product cannot be 200, 100, 50, 40... (Z_200, Z_100, Z_50....), because they are not power of a prime! so, we have only 6 abelian groups up to isomorphism.. Is that true??
     
  10. Dec 8, 2009 #9
    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    All finite abelian groups are of the form (Z_a)x(Z_b)x...x(Z_n) and anything of this form is a finite abelian group.

    So if your group is of order 20, it must be one of (Z_20), (Z_10)x(Z_2), (Z_5)x(Z_4), (Z_5)x(Z_2)x(Z_2). But (Z_mn)=(Z_M)x(Z_n) iff m,n are coprime.

    So (Z_20) is isomorphic to (Z_5)x(Z_4). (Z_4) is not iso to (Z_2)x(Z_2) because 2 and 2 are not coprime. Also, (Z_20) is not iso to (Z_10)x(Z_2).

    So all abelian groups of order 20 are: (Z_20)=iso-to(Z_5)x(Z_4) or (Z_5)x(Z_2)x(Z_2)

    Following in this manner, given any number, n, express it as a product of primes, p_1 .... p_m

    Then an abelian group of order n is of the form (Z_(p_1))x...(Z_(p_m)) except that if we have repeated primes, we may multiply them together, to give a different group which has some factors (Z_(p_i)^j) say. (We can do this for any any j and any of the p_i's)

    So with 20, 20=2x2x5 is a prime factorisation of 20. So we have either (2,2,5) or (4,5) as we had before.

    Hope this makes things simpler.
     
  11. Jul 27, 2010 #10
    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    I too have a confusion regarding this problem:
    Question 1:
    ------------
    "So all abelian groups of order 20 are: (Z_20)=iso-to(Z_5)x(Z_4) or (Z_5)x(Z_2)x(Z_2)"
    Here the order of (Z_5)x(Z_2)x(Z_2) is lcm(5,2,2) = 10. As the isomorphism is order preserving, Z_20 being order 20 and (Z_5)x(Z_2)x(Z_2) being order 10 are not isomorphic,. Please correct me if i am wrong here or missing something!

    Question 2:
    ------------
    Also, in "First Course in Abstract Algebra by Fraleigh" Example:11.13: Find all abelian groups upto isomorphism of order 360?
    Solution for this example is given as:
    1. (Z_2)x(Z_2)x(Z_2)x(Z_3)x(Z_3)x(Z_5) ----> 2^3 * 3^2 *5 ---- order = lcm(2,3,5) = 30
    and then he proceeds with prime Factorization of n = 360.
    As this (Z_2)x(Z_2)x(Z_2)x(Z_3)x(Z_3)x(Z_5) being order 30, how can it be listed under Finite abelian groups of order 360.

    Thanks,
     
  12. Jul 27, 2010 #11

    quasar987

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    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    You are wrong or missing something. Look at Z_2 x Z_3 for example. What are its elements?
    First fix the first factor to 0. The elements of this kind are (0,0), (0,1), (0,2). Now fix the second factor to 1. The elements of this kind are (1,0), (1,1), (1,2). For a total of 3+3 = 2 x 3 = 6. You can generalize this argument to see that Z_n x Z_m x ... x Z_l has n x m x ... x l elements, and not lcm(n, m, ..., l).
     
  13. Jul 27, 2010 #12
    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    Hi,
    In the example above Z2 x Z3 has order 6. Note that order is minimum smallest integer sothat any element in the Group goes to identity element. In Number of elements, m*n*o*p*.... , there may be elements repetitive. For example., Group (Cartesian product) Z4 * Z2 has order 4. Because (1,1) + (1,1) ... (4 times ) .. = (0,0)
    But the product 4*2 = 8elements has repetitions due to modular arithmetic.

    So, the order of Zm * Zn * Zo * ... = lcm(m,n,o,p...)

    THanks,
     
  14. Jul 28, 2010 #13
    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    hmmm.... ok i got it!!!
     
  15. Jul 28, 2010 #14

    quasar987

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    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    The order of a group is just it's cardinality.

    Not to be confused with the order of an element of the group.
     
  16. Jul 28, 2010 #15
    Re: HELP!!! Find all abelian groups (up to isomorphism)!!!

    Yep, you were just confusing your definition of the order of a group. Given that we already have a word for the number of elements in a group, it's cardinality, maybe your definition makes more sense :approve:

    If we were using this definition, then for an integer n I guess that you would just have all groups of the form Z_n x Z_(x_1) x Z_(x_2) x .... x Z_(x_k) where the x_i are all divisors of n.
     
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