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.
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: ℝ2 -> ℝ where f(ℝ2) = 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!
Remember isomorphism theorem says, if G, H are groups, and is a homomorphism, if (every element in H gets hit under the homomorphism), then . Clearly, you got your homomorphism from , . You also got your kernel, which is your set N, all you need to show is that the homomorphism is surjective (onto)
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.
Your claim of showing f is a homomorphism is a bit off.
An element of your domain is off the form . So to show this group homomorphism, take, , so . So
As for onto:
Say i pick an element in , i just want an such that . Doing some manipulation we get, and clearly . So for any p they picked, so we can indeed hit any element in , so it is onto.
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.
R2 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:
R2/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.
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+y0)/3), my solution has a = 1. i just don't like fractions.