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Isomorphism of relatively prime groups

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Allow m,n to be two relatively prime integers. You must prove that Z(sub mn) ≈ Z(sub m) x Z(sub n)


    2. Relevant equations
    if two groups form an isomorphism they must be onto, 1-1, and preserve the operation.


    3. The attempt at a solution
    since m and n are relatively prime, the gcd(m, n) = 1.
    mhm, very stumped from the start.
     
  2. jcsd
  3. Mar 8, 2012 #2

    jbunniii

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    What do you know about the group Z(sub mn)? What are some facts about it? Is it abelian, etc.?
     
  4. Mar 8, 2012 #3
    Isomorphisms isnt my thing, I do not know many facts about Z(sub mn). I can say the contents of Z(sub mn) are {[0],[1], [2],.... [mn]}. generated by a single element? Im not sure...Abelian? yes i believe so...
     
  5. Mar 8, 2012 #4

    jbunniii

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    Yes, generated by a single element is the key. In other words, it's a cyclic group. (Side note: cyclic groups are always abelian, but you don't need that fact for this problem.)

    So if Z(sub mn) is isomorphic to Z(sub m) x Z(sub n), then that group would also have to be cyclic, right? Can you show that Z(sub m) x Z(sub n) is cyclic?
     
  6. Mar 9, 2012 #5
    mhmm i believe so, but i dont know where to start since I am dealing with a cross product...
     
  7. Mar 9, 2012 #6

    Office_Shredder

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    If you had to guess at which element of ZmxZn generated the whole group, what would it be?

    If you're not sure, try some small examples like Z2xZ3 or Z3xZ4 and see what you can come up with
     
  8. Mar 9, 2012 #7
    What would the elements of Z2 x Z3 look like again?
     
  9. Mar 9, 2012 #8
    Ordered pairs (a, b) such that a is from Z2, and b is from Z3.
     
  10. Mar 9, 2012 #9

    jbunniii

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    Well, if A and B are any sets (whether or not they have a group structure), then what is the definition of A x B? This is called the Cartesian product of A and B. The elements are simply ordered pairs of the form [itex](a,b)[/itex], where [itex]a \in A[/itex] and [itex]b \in B[/itex].

    And if A and B are groups, then there's a natural way to define a group operation on A x B.

    Surely this is discussed in your textbook or lecture? There's no way you can proceed with this problem without first understanding what the group is.
     
  11. Mar 10, 2012 #10
    yes of course I just wanted to be refreshed and reassured...
    Z2 x Z3 = {([0],[0]), ([0],[1]), ([0],[2]), ([1],[0]), ([1],[1]), ([1],[2])}...
    I cannot grasp how this could be generated by a single element though?
     
  12. Mar 10, 2012 #11

    jbunniii

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    Look at the six elements. In order for the group to be generated by one element, clearly that element can't have a [0] in either of the two positions of the ordered pair. So that only leaves two candidates: either ([1],[1]) or ([1],[2]). Why not try each and see if it works? Just add it to itself over and over and see if you can generate the group.
     
  13. Mar 11, 2012 #12
    the generator for the group Z(sub 2) x Z(sub 3) = <[1],[1]> and since the operation is addition, the generator for a group in the form Zn x Zm = <[1], [1]>. so Zn x Zm is cyclic and abelian.
     
  14. Mar 11, 2012 #13

    jbunniii

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    OK, good. Note that <[1],[2]> also generates the group.

    So Z2 x Z3 is cyclic, and I assume you observed that it has order 6. That looks promising, because Z6 is also cyclic with order 6.

    Now you need to define an isomorphism [itex]\phi: Z_{6} \rightarrow Z_2 \times Z_3[/itex] between the groups. As a first step, I suggest choosing a generator [itex]g[/itex] for [itex]Z_6[/itex] and defining [itex]\phi(g)[/itex].
     
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