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Isomorphic Quotient Groups in Z4 x Z4

  1. Apr 6, 2009 #1
    1. The problem statement, all variables and given/known data

    In Z4 x Z4, find two subgroups H and K of order 4 such that H is not isomorphic to K, but (Z4 x Z4)/H isomorphic (Z4 x Z4)/K

    2. Relevant equations

    3. The attempt at a solution

    I know (Z4 x Z4) has twelve elements (0,0), (1,0), (2,0), (3,0), etc. I can generate subgroups of order 4 by picking an (a,b) in which one of two elements has order 4. For example <(2,1)> = {(2,1), (0,2), (2,3),(0,0)}, call this H. The problem is that anytime I do this, I end up with subgroups which are isomorphic b/c they each have two elements of order 4, one elt of order 2 and the identity.

    So my thought was to use one of the subgroups above and then find a different subgroup of order 4 which is not isomorphic. The group I found is {(0,0), (2,0), (0,2), (2,2)}, call this K.

    Now I have two subgroups which are not isomorphic, but when I figure out (Z4 x Z4)/H and (Z4 x Z4)/K I end up with quotient groups which I do not think are isomorphic.

    (Z4 x Z4)/H = {H, (1,0) + H, (2,0) + H, (3, 0) + H}

    (Z4 x Z4)/K = {K, (0,1) + K, (1,0) + K, (1,1) + K}

    (Z4 x Z4)/H has 3 elements of order 3 and identity

    (Z4 x Z4)/K has 3 elements of order 2 and identity

    Am I calculating the order of these elements incorrectly?
    Any suggestions or insight would be greatly appreciated.
     
    Last edited: Apr 6, 2009
  2. jcsd
  3. Apr 7, 2009 #2

    matt grime

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    First, Z4 x Z4 does not have 12 elements. Are you sure you have the written the question correctly?

    Second, a group is not determined by a list of group element's orders. That is there are non-isomorphic groups where the multi-sets of orders are the same.

    Third, you're not calculating orders correctly. It is impossible to have any elements of order 3 appearing.
     
  4. Apr 7, 2009 #3
    Thank you for the reply. I really appreciate the response and have tried to correct my mistakes below.

    1) The order of Z4xZ4 is 16. I have 16 elements written out, but I failed to double check my post.

    2) I am not sure I am understanding your hint. Here is how I determined my groups.

    For the subgroups of Z4xZ4 I was trying to use the subgroup test. Since I know, Z4xZ4 is a group, isn't it enough to show that any subgroup is closed and inverses exists? I can manually show {(2,1), (0,2), (2,3),(0,0)} and {(0,0), (2,0), (0,2), (2,2)} are both closed under the group operation and contain inverses.

    For the quotient groups I reasoned that since Z4xZ4 was abelian, any subgroup is normal. Therefore I could create a quotient group using any subgroup.

    I understand that having elements of the same order does not make two groups isomorphic (I have to be able to build a map), but isn't looking at the order of elements a way to narrow down the candidates since if two groups are isomorphic they must have the same structure?

    3) I thought I might be doing something wrong with the orders. I did some rereading and here is what I think I need to do. In order to find the order of an element of the quotient group (Z4 x Z4)/K = {K, (0,1) + K, (1,0) + K, (1,1) + K} , say (0,1), I need to find the number of times I have to add (0,1) to get an element of (Z4 x Z4)/K. (0,1) + (0,1) = (0,2), so I believe the order of (0,1) + K is 2.
     
  5. Apr 7, 2009 #4

    matt grime

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    Yes, but I wanted to make sure that you weren't going to use it for when you wanted to show isomorphism (and plenty of people do).

    Now, I'd really like you to verify that you have got the statement of the question correct, since it there are no non-isomorphic H and K with the property you desire.

    There would be if the group were Z_2 X Z_4, try it in this case (and you're thinking and methods are all good).
     
  6. Apr 7, 2009 #5

    I doubled checked and I think I have the problem written correctly. I scanned it just to make sure, here is a link:

    http://yfrog.com/e5problem9j

    Thanks again!

    Don
     
  7. Apr 7, 2009 #6
    You copied it correctly, but that problem is impossible in Z4 x Z4, which you can verify by exhaustion (there are only seven subgroups of order 4).
     
  8. Apr 8, 2009 #7
    Thank you all for the help. After checking with the instructor the book indeed does have an error. The problem should be:

    In Z4 x Z6, find two subgroups H and K of order 4 such that H is not isomorphic to K, but (Z4 x Z6)/H isomorphic (Z4 x Z6)/K
     
  9. Apr 8, 2009 #8
    hey sorry to hijack this but is it true that we can't have an element of order 3 because if the order of [itex]g \in \mathbb{Z}_4 \times \mathbb{Z}_4[/itex] was 3 then we would be able to generate a cyclic subgroup [itex]<g>[/itex] of [itex]\mathbb{Z}_4 \times \mathbb{Z}_4[/itex] which isn't allowed by Lagrange's Theroem as 3 doesn't divide 16, yes?
     
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