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Abstract Algebra - Define and Prove

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This looks like an application of the First Isomorphism theorem. I asked one of my classmates how she would go on on doing this and she said that Iwould want to define a homomorphism f: ℝ^{2} -> ℝ where f(ℝ^{2}) = ℝ and ker(f)=N.She also said that given my definition for N, she would suggest using f(x,y)=3x-4y as my homomorphism.

I am having trouble following her advice. Can anyone further explain what I should do to tackle this problem? Thanks!

Re: Abstract Algebra - Define and Prove

What's the problem ? Using the first isomorphism theorem you have following her advice that Rē/N =~ R. It is clear that ker(f) is N.

Re: Abstract Algebra - Define and Prove

Remember isomorphism theorem says, if G, H are groups, and $\displaystyle \phi:G \to H $ is a homomorphism, if $\displaystyle \phi(G) = H $ (every element in H gets hit under the homomorphism), then $\displaystyle \frac{G}{Ker(\phi)} = H $. Clearly, you got your homomorphism from $\displaystyle \mathbb{R}^2 \to \mathbb{R} $, $\displaystyle \phi(x,y) = 3x-4y $. You also got your kernel, which is your set N, all you need to show is that the homomorphism is surjective (onto) $\displaystyle \mathbb{R} $

Re: Abstract Algebra - Define and Prove

I have the following:

Define mapping f: ℝ^{2} -> ℝ as follows:

f(x,y) = 3x - 4y

__Claim:__ f is a homomorphism

Pick any (x,y) in ℝ^{2}. 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 ℝ^{2} such that (x,y) = (1,0). Then...

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

Re: Abstract Algebra - Define and Prove

Your claim of showing f is a homomorphism is a bit off.

An element of your domain $\displaystyle x \in \mathbb{R}^2 $ is off the form $\displaystyle x = (x_0, y_0) $. So to show this group homomorphism, take, $\displaystyle x, \bar{x} \in \mathbb{R}^2 $, so $\displaystyle x = (x_1, y_1), \bar{x} = (x_2, y_2) $. So $\displaystyle \phi(x * \bar{x}) = \phi( (x_1 + x_2, y_1 + y_2) ) = \phi((x_1, y_1)) + \phi((x_2, y_2)) $

As for onto:

Say i pick an element in $\displaystyle p \in \mathbb{R} $, i just want an $\displaystyle x = (x_0, y_0) \in \mathbb{R}^2 $ such that $\displaystyle \phi(x) = 3x_0 - 4y_0 = p $. Doing some manipulation we get, $\displaystyle x = \frac{p + 4y}{3} $ and clearly $\displaystyle (\frac{p + 4y}{3}, y) \in \mathbb{R}^2 $. So $\displaystyle \phi(x) = \phi((\frac{p + 4y}{3}, y) = p $ for any p they picked, so we can indeed hit any element in $\displaystyle \mathbb{R} $, so it is onto.

Re: Abstract Algebra - Define and Prove

Thank you, thank you, thank you, jackncoke. You have a way of explaining things. Thanks!

Re: Abstract Algebra - Define and Prove

E-mailed my professor and she said the following:

Second part is not correct. You said immediately to take x such that f(x)=p. 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.

Re: Abstract Algebra - Define and Prove

R^{2} is a vector space. the map f(x,y) = 3x - 4y is a linear map. linear maps are "vector space homomorphisms" meaning, among other things, that they are group homomorphisms of the underlying abelian group of the vector space of their domain.

N is the null space of a linear map, meaning it is also the kernel of a group homomorphism. by the rank-nullity theorem (this is the "vector space analogue" of the first isomorphism theorem for groups), 2 = dim(N) + dim(im(f)). if we can find ONE non-zero pair (x,y) in N and f(x,y) in im(f), this means this sum is 2 = 1 + 1.

clearly, (4,3) is in N, and is non-zero. also, f(1,1) = 3 - 4 = -1. thus dim(N) = 1, and dim(im(f)) = 1. since im(f) is a 1-dimensional subspace of R, it is all of R: that is, f is onto. hence AS VECTOR SPACES:

R^{2}/N and R are isomorphic, meaning they are isomorphic as (abelian) groups.

now one can do this directly (as jakncoke did) solely within "the group arena":

f((x,y) + (x',y')) = f((x+x',y+y')) = 3(x+x') -4(y+y') = 3x + 3x' - 4y - 4y' = 3x - 4y + 3x' - 4y' = f((x,y)) + f((x',y')) <--f is a homomorphism

f((4-a,3-a)) = 3(4-a) - 4(3-a)) = 12 - 3a - 12 + 4a = a <--f is onto

so why did i use vector spaces? to show that group theory's usefulness actually extends BEYOND groups, and that sometimes "extra structure" actually makes things EASIER.

Re: Abstract Algebra - Define and Prove

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Re: Abstract Algebra - Define and Prove

Quote:

Originally Posted by

**bondvalencebond** E-mailed my professor and she said the following:

Second part is not correct. You said immediately to take x such that f(x)=p. 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.

That is what i did! Look, I took a $\displaystyle p \in \mathbb{R} $ (Any random number p), and i said, the element $\displaystyle (\frac{p+4y_0}{3}, y_0) $ for some arbitrary number $\displaystyle y_0$, lets just say $\displaystyle y_0 = 1 $. So $\displaystyle (\frac{p+4}{3}, 1) \in \mathbb{R}^2$ , The **said** element, $\displaystyle \phi( (\frac{p+4}{3}, 1) ) = 3*\frac{p+4}{3} - 4 = p $. I just took an element from my domain group, and i showed you that that element maps to the arbitary element you gave me from the codomain group.

Re: Abstract Algebra - Define and Prove

Okay, maybe I didn't make myself clear with my professor. Thank you for explaing once again.

Re: Abstract Algebra - Define and Prove

some other values you might have used:

(x,y) = (-p,-p)

(x,y) = (p/3,0)

(x,y) = (0,-p/4)

in general: any vector (x,y) of the form:

a(4,3) - (p,p) will do.

jakcoke's solution has a = (p+1)/3 (or more generally: (p+y_{0})/3), my solution has a = 1. i just don't like fractions.