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Finding which direct sum of cyclic groups Z*n is isomorphic to

  1. Feb 14, 2013 #1
    I always see problems like "how many structurally distinct abelian groups of order (some large number) are there? I understand how we apply the theorem which tells us that every finite abelian group of order n is isomorphic to the direct sum of cyclic groups. We find this by looking at the prime factorization of n, etc. (like for 600, 600 = 2^3 * 3 * 5^2, so we know there are 6 combinations possible and so 6 structurally distinct abelian groups of order 600)

    however, what if I'm going the other way around? What if I have a group of order say something simple like 4, takes Z*(8) for example. I want to put this in terms of a direct sum of cyclic groups, and I know there are 4 options. How do I find out which one it is isomorphic to? I know I can do brute force things like take the cyclic subgroups generated by each of the elements and see what the order of each element is. But is there any hard and fast way which will tell me, like maybe based on the prime factorization of n, so that when I'm dealing with a large number (like Z*(700)) I'd be able to tell which direct sum of cyclic groups its isomorphic to?
  2. jcsd
  3. Feb 14, 2013 #2
    I will note, I am fairly certain that if I know the generating set of a group, then I can find an answer to this question. For example, Z*(8) is not cyclic, but I know its generating set is (3)(5). From this I can figure out that Z*(8) is isomorphic to C_2 * C_2. But how do I do this for arbitrarily large values of n? How do I find the generating set? Does it have to do with the cosets of a certain cyclic group?
  4. Feb 14, 2013 #3
    I found a paper which addresses my questions of how to find the group which is the isomorphic direct sum and wanted to share it for future reference


    The only thing I can't figure out still is how to find the group presentation, or the generating set. My guess is that if you know which direct sum of cyclic groups its isomorphic to, then if we enumerate all the cyclic subgroups of Z*n, we can select then based on which elements have the right orders.
    Last edited: Feb 15, 2013
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