Abstract Algebra Proof Using the First Isomorphism Theory

[babygotpi]

Homework Statement

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.

 

[STEMucator]

Homework Statement

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.

So for your map $f$ :

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 $f: G/Ker(f) → f(G)$.

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?

 

[babygotpi]

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!?

 

[STEMucator]

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!?

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

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

As for you other question :

Well, to show f is onto, you have to show that for any $b \in f(N)$, there exists an $a \in N/Ker(f)$ such that $f(a) = b$ right?

 

[babygotpi]

I don't understand. Did I not show it was onto?

 

[micromass]

I don't understand. Did I not show it was onto?

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.