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Homework Help: Finding isomorphism

  1. Nov 17, 2004 #1
    find an isomorphism from from the group of integers under addition to the group of even integers under addition.

    I know, very simple question, but I dont know what Im doing here......

    the hint in the book says to try n to 2n. I thought of that too, since it specificaly says integers to even integers.

    the books says to prove injective, surjective, and phi(x,y) = phi(x) phi(y).

    so what do I do? start x = 2y and prove x = y?

    I think i'm wronng...
     
  2. jcsd
  3. Nov 17, 2004 #2
    You have n to 2n so try defining a function that creates the isomorphism:

    f(x) = 2x

    Once you have that, the rest follows:

    injective: f(x) = y & f(x') = y now show that x = x'
    surjective: you know that if y is an even integer then it is equal to 2x for some x, where x is an integer...

    the last part is showing that f(x+y) = f(x) + f(y)...
     
  4. Nov 17, 2004 #3

    I understand by reading the book what all the steps ask me to do, but I dont know what they mean by "define a map or function." Like, what do I map from what to what?

    do I do 2x = 2y, and then go through all the steps? what if they ask to find an isomorphism from integers to odd integers, or something? do I do 3x = 3y?

    Basically, I dont know what the hint "try n to 2n" means. How am I supposed to use that....

    sorry, really newbie at this.
     
  5. Nov 18, 2004 #4
    The map or function is f(x) = 2x...

    To map to odd integers use f(x) = 2x+1, this is not a group though because it is not closed: 3+3=6...
     
  6. Nov 18, 2004 #5

    Galileo

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    The question asks you to find (define) an isomorphism from [itex]\mathbb{Z}[/itex] to
    [itex]2\mathbb{Z}[/itex]:
    [tex]f:\mathbb{Z} \rightarrow 2\mathbb{Z}[/tex].
    The 'hint' (which basically gives the answer) is: try f(x)=2x.

    What you have to check now is:
    Injectivity: [itex]f(x)=f(y) \Rightarrow x=y[/itex]
    Surjectivity: for every even number y there exist an integer x, such that f(x)=y.
    Homomorphic property: f(x+y)=f(x)+f(y).
     
  7. Nov 18, 2004 #6
    I think that whats confusing you is that both have to have the same number of elements. Since both groups are infinite, it doesn't matter.

    Try thinking of it this way: the integers under addition represent how many $2 bills you have and the even integers represent how many $1 bills you have...
     
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