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Homework Help: Show ZXZ/<1,1> is an infinite cyclic group.

  1. Jan 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Show ZXZ/<1,1> is an infinite cyclic group.

    2. Relevant equations

    3. The attempt at a solution
    <1,1> = {...(-1,-1), (0,0), (1,1),...}

    implies ZXZ/<1,1> = {<1,0>+<1,1>, <0,1>+<1,1>} which is isomorphic to ZXZ.

    But ZXZ is not cyclic, is my description of the factor group wrong , is it not isomorphic to ZXZ, or am I missing other points?
  2. jcsd
  3. Jan 11, 2009 #2
    what characterize an infinite cyclic group?

    1. generated by a single element
    2. infinite order

    can you put an element canonically into a certain form?
    so say you have (m, n), what can I write it as?
  4. Jan 11, 2009 #3
    if (m,n) is an element of one of the left cosets of the factor group then,
    if m=n (m,n) is in <1,1>. if m/=n then (m,n) = (m,n) + <1,1>. (because (0,0) is
    generated by <1,1>). I'm not sure what this implies though.
  5. Jan 11, 2009 #4
    so actually <1,0> + <1,1> generated all elements of ZXZ. Is it acceptable to say ZXZ/<1,1> is isomorphic to Z+0Z which is isomorphic to Z? (Which is generated by <1> or <-1>)
  6. Jan 11, 2009 #5
    <1,1> is just the identity. So, yeah.
  7. Jan 11, 2009 #6
    ... this is the additive group, (0,0) is the identity, and I'm using the notation <1,1> to imply that this is the cyclic generator. I hope that clears up any confusion.

    I'm not sure if the conclusion I've come to is correct.

    ZXZ/<1,1> = {Z+<1,1>}
    Clearly Z is infinite and cyclic generated by <1> or <-1>.
    So, ZXZ/<1,1> is a cyclic infinite group.
  8. Jan 11, 2009 #7
    well, (0,0)=(1,1) since you are modding out the diagonal set.
    (1,0) is the generator (or using coset notations (1,0)+<(1,1)>). Your conclusion is definitely correct (as in post 4)
  9. Jan 11, 2009 #8
    awesome thanks.
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