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Abelian group as a direct product of cyclic groups

  1. Mar 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider G = {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64} with
    the operation being multiplication mod 65. By the classification of finite abelian groups, this
    is isomorphic to a direct product of cyclic groups. Which direct product?

    2. Relevant equations


    3. The attempt at a solution
    Since this abelian group has 16 elements, up to isomorphism there are three options: ##\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2## or ##\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2##, or ##\mathbb{Z}_8 \times \mathbb{Z}_2##.

    I am not sure how to determine which one...
     
  2. jcsd
  3. Mar 8, 2017 #2

    fresh_42

    Staff: Mentor

    It would be helpful to know the orders of the elements. And what about ##\mathbb{Z}_{16}## and ##\mathbb{Z}_4 \times \mathbb{Z}_4 \,## ?
     
  4. Mar 8, 2017 #3
    Oops, forgot about those.

    Orders:
    |1|=1
    |8|=57
    |12|=38
    |14|=14
    |18|=47
    |21|=31
    |27|=53
    |31|=21
    |34|=44
    |38|=12
    |44|=34
    |47|=18
    |51|=51
    |53|=27
    |57|=8
    |64|=64
     
  5. Mar 8, 2017 #4

    fresh_42

    Staff: Mentor

    I haven't checked whether this is a group at all or whether mod 65 makes sense here, but shouldn't the order of an element divide the order of the group?

    And ##\mathbb{Z}_{65} = \mathbb{Z}_5 \times \mathbb{Z}_{13}##. Do you know which elements in this groups are multiplicative invertible, i.e. what is ##\mathbb{Z}_{65}^*\,##?
     
  6. Mar 8, 2017 #5
    It's given that its an abelian group.

    I made a mistake with the orders. I tried to calculate them again, and got the following:

    |1|=1
    |8|=4
    |12|=4
    |14|=2
    |18|=4
    |21|=4
    |27|=4
    |31|=4
    |34|=4
    |38|=4
    |44|=4
    |47|=4
    |51|=2
    |53|=4
    |57|=4
    |64|=2

    However, this looks like I am doing something wrong. Because if the group is abelian, then there should be an element of order 16, right?
     
  7. Mar 8, 2017 #6

    fresh_42

    Staff: Mentor

    Does the Klein 4-group have an element of order 4? If there is an element of order 16, wouldn't it generate the group? And what about your own list of possible groups? Do they contain elements of order 16?

    Which of the groups on your list plus the two missing ##\mathbb{Z}_{16}## and ##\mathbb{Z}_4 \times \mathbb{Z}_4## can you rule out now?
     
  8. Mar 8, 2017 #7
    Oops, I was confusing cyclic and abelian groups. Cyclic groups are abelian, but abelian groups are not necessarily cylic.

    Knowing the orders, what should I do next? Should I look at the candidate isomorphic groups to see if they have orders of just 4, 2, and 1 for the identity?
     
  9. Mar 8, 2017 #8

    fresh_42

    Staff: Mentor

    See my edit line. You will be left with two possible groups. I would take a subgroup of order 4 and factor it out. Then a group of 4 elements will be left, and there are only two.
     
  10. Mar 8, 2017 #9
    So I see that we are left with ##\mathbb{Z}_4 \times \mathbb{Z}_4## and ##\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2##.

    I determined that the orders of the elements of the former exactly match ##G##. Is this sufficient to say that G is isomorphic to ##\mathbb{Z}_4 \times \mathbb{Z}_4## or do I need to check the orders of the latter too?
     
  11. Mar 8, 2017 #10

    fresh_42

    Staff: Mentor

    I think this is sufficient. You simply don't have enough elements of order 2 for the latter group. In addition if you look at the units in the ring ##\mathbb{Z}_{65}## you get:
    $$G \leq \mathbb{Z}_{65}^* \cong \mathbb{Z}_{5}^* \times \mathbb{Z}_{13}^* \cong \mathbb{Z}_{4} \times \mathbb{Z}_{12} \cong \mathbb{Z}_{4} \times \mathbb{Z}_{4} \times \mathbb{Z}_{3}$$
    and you have to place a group of order 16 in it.
     
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