Proving f inverse is homomorphic

  • Thread starter Thread starter DPMachine
  • Start date Start date
  • Tags Tags
    Inverse
DPMachine
Messages
26
Reaction score
0
I'm trying to show:

If f: S -> S' is an isomorphism of <S, *> with <S', *'>, then f^(-1) is homomorphic.

My take:

So I have to show that f^(-1)(x' *' y') = f^(-1)(x') * f^(-1)(y').

Since f is bijective (onto, more precisely) I know that f^(-1)(x') = x and f^(-1)(y') = y. So f^(-1)(x') * f^(-1)(y') = xy.

How can I simplify f^(-1)(x' *' y') though?

Thanks.
 
Physics news on Phys.org
EDIT: Nevermind, that's not right.

Sorry about not using latex, by the way. I was writing from a mobile device.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top