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Abstract Algebra Proof Using the First Isomorphism Theory

  1. Dec 15, 2012 #1
    1. The problem statement, all variables and given/known data
    See attatchment. I couldn't upload the picture.

    2. The attempt at a solution
    I have the following:

    Define mapping f: ℝ2 -> ℝ as follows:
    f(x,y) = 3x - 4y

    Claim: f is a homomorphism

    Pick any (x,y) in ℝ2. Then f(x,y) = f(x)*f(y) = 3x - 4y = (x+x+x)-(y+y+y+y) = x+x+x-y-y-y-y = f(x*y). Hence, it perserves the operation.

    Claim: f is onto.
    Pick any (x,y) in ℝ2 such that (x,y) = (1,0). Then...

    I don't think I'm going the right way on showing f is onto.
     

    Attached Files:

  2. jcsd
  3. Dec 15, 2012 #2

    Zondrina

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    So for your map [itex]f[/itex] :

    By the first isomorphism theorem, if you can show f is a homomorphism that is either surjective or injective. Then the map f is an isomorphism such that [itex]f: G/Ker(f) → f(G)[/itex].

    So if f is already a homomorphism say, and you've shown its either 1-1 or onto, then it follows its an isomorphism right? Now could you tell me why its an isomorphism?

    Also, welcome to PF.

    Hint : What is the structure of N? Is it finite or infinite?
     
    Last edited: Dec 15, 2012
  4. Dec 15, 2012 #3
    Thanks. Glad to be part of PF.
    This is my final answer.
    Define mapping f: ℝ2 -> ℝ as follows: f(x,y) = 3x - 4y

    Claim: f is a homomorphism

    Pick any X, X1 ∈ ℝ2 in which X = (x1, y1) and x1 = (x2, y2). Then:

    f(x*x1 ) = f(x1+y1, x2+y2) = f(x1+y1) +f(x2 +y2) = f(x+x1 ). Hence, f perserves the operation.

    Claim: f is onto

    Pick any elemnet of b ∈ ℝ. Let x = (x0, y0) ∈ ℝ2 such that f(x) = 3x0 - 4y0 = b.

    Can be rewritten as x = (b+4y0)/3 or ((b+4)/3 , y0) which ∈ ℝ2 . this implies that f is onto. Since ker(f) = N, the first isomorphism theory implies that ℝ2 / N is isomorphic to ℝ.

    How does that sound!?
     
  5. Dec 15, 2012 #4

    Zondrina

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    Change the x's in this quote into big X and big X1 :

    As for you other question :

    Well, to show f is onto, you have to show that for any [itex]b \in f(N)[/itex], there exists an [itex]a \in N/Ker(f)[/itex] such that [itex]f(a) = b[/itex] right?
     
  6. Dec 15, 2012 #5
    I don't understand. Did I not show it was onto?
     
  7. Dec 15, 2012 #6

    micromass

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    Your proof wasn't correct. You said immediately to take x such that f(x)=b. But you have to show that such an x actually exists, you can't assert it as true.

    You need to start by taking an element b and then actually finding an x such that f(x)=b.
     
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