To prove:

Let G be the group of affine functions from R into R, as defined in A. (A = {f_m,b: R -> R | m is not equal to 0 and f_m,b(x) = mx + b}. Define phi: G -> R^x as follows: for any function f_m,b in G, let phi(f_m,b) = m. Prove that phi is a group homomorphism and find its kernel and image.

My work:

First we need to prove that A is a group.

Let we have f_m,b and f_n,v both are in A. => Composition, (f_m,b) o (f_n,v) in A => (f_m,b) o (f_n,v) = f_m,b(nx + v) = m(nx + v)+b = mnx+(mv+b) = h_mn,mv+b(x) in A. => Closure.

Let we have f_1,0 => f_1,0(x) = 1x+0 = x => Identity

Let f_m,b in A and it has inverse. Since m is not equal to 0, (f_m,b) o (f_1/m,-b/m) = f_m,b(x/m-b/m) = m(x/m-b/m)+b = x so we have inverse.

Hence A is a group.

Let we have f_m,b and f_n,v in A.

phi[(f_m,b)(f_n,v)] = phi(f_m,b o f_n,v(x)) = h_mn,mv+b(x) = mn. phi(f_m,b)*phi(f_n,v) = mn

Since they are equal, phi is a homomorphism.

And I'm stuck to prove its kernel and image..

Any help would be very appreciated.

  • 5
    $\begingroup$ please use latex commands to edit this problem so that it is readable.. I remember that almost all of your posts are being edited by somebody to make it readable.. I guess now it is your turn to do so.. $\endgroup$ – user87543 Nov 23 '13 at 5:36
  • $\begingroup$ Here is a tutorial on how to properly typeset your mathematical posts. $\endgroup$ – Omnomnomnom Nov 23 '13 at 5:47

$$ \ker(\phi) = \{f_{m,b} : m = 1\} = \{f_b : b\in \mathbb{R}\} $$ where $f_b$ denotes the map $x\mapsto x+b$.

Define a function $\ker(\phi) \to \mathbb{R}$ given by $f_b \mapsto b$. Can you check that this is an isomorphism?

Also, for any $m \in \mathbb{R}^{\times}$, define $f(x) = mx$, then what is $\phi(f)$? What can you conclude abou the image of $\phi$?


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