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Orders of elements in a quotient group.

  1. May 25, 2013 #1
    1. The problem statement, all variables and given/known data

    I want to find the orders of the elements in [itex]Z_8/(Z_4 \times Z_4)[/itex], [itex](Z_4 \times Z_2)/(Z_2 \times Z_2)[/itex], and [itex]D_8/(Z_2 \times Z_2)[/itex].

    2. Relevant equations



    3. The attempt at a solution
    The elements of [itex]Z_2 \times Z_2[/itex] are (0,0), (1,0), (0,1), (1,1), and the elements of [itex]Z_8[/itex] are of course 1, 2, 3, 4, 5, 6, and 7.

    So elements of [itex]Z_8/(Z_2 \times Z_2)[/itex] must be of the form…

    [tex] (0,0), (1,0), (0,1), (1,1) [/tex]
    [tex]1+(0,0), 1+ (1,0), 1+ (0,1), 1+ (1,1) [/tex]
    [tex]2+(0,0), 2+(1,0),2+ (0,1), 2+(1,1) [/tex]
    [tex]3+(0,0), 3+(1,0), 3+(0,1), 3+(1,1) [/tex]

    [tex]7+(0,0), 7+(1,0), 7+(0,1), 7+(1,1) [/tex]

    But since the number we are adding to [itex]Z_2 \times Z_2[/itex] is mod 8, none of these are equal. So I don’t know how we have 2 elements [itex]Z_2 \times Z_2[/itex].

    I’m having the same problem with [itex]D_8/(Z_2 \times Z_2)[itex].
    We know that [itex]D_8 = \{e, b, b^2, b^3, a, ba, b^2a, b^3a \}[/itex]

    [tex] (0,0), (1,0), (0,1), (1,1) [/tex]
    [tex]b+(0,0), b+(1,0), b+ (0,1), b+(1,1)[/tex]

    [tex]b^3a+(0,0), b^3a+(1,0), b^3a+(0,1), b^3a+(1,1)[/tex]

    But, for example, we know that [itex]b[/itex] has order 4, right?
     
    Last edited: May 25, 2013
  2. jcsd
  3. May 26, 2013 #2

    Dick

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    Homework Helper

    Your first question doesn't make sense to me whatsoever. The quotient G/H is defined when H is a subgroup of G. [itex]Z_2 \times Z_2[/itex] is not a subgroup of ##Z_8##. It doesn't even have a subgroup isomorphic to [itex]Z_2 \times Z_2[/itex]. ##D_8## does so I think you can make some sense of that.
     
  4. May 27, 2013 #3
    Thanks a lot.
     
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