Maybe the statement has an error? If , then I think
...Just a thought.
Define as . Show this map is well-defined, note, this is were you are going to use the fact that . Now argue that this is a homomorphism. This mapping is onto, and so . The kernel of the homomorphism will be the set (group) , I want to mention here another notation, we can write this also as , which looks a little neater. And so by fundamental homomorphism theorem .
i just want to clarify something before i go storming off to do this. we did a theorem that says a homomorphism is one-to-one if and only if the kernel is the set with only the identity. that does not seem to be happening here, but we need one-to-one for the isomorphism
Ok, thanks. does this work?
Define by for all . This function is well-defined if and only if by homework problem 21.6 (you don’t expect me to re-invent the wheel, do you?).
We claim that this defines a homomorphism. For if , then .
This mapping is onto, since for any we can find a such that because (should I explain this part more?). So that we have . Clearly the kernel of this homomorphism is .
Thus we can define a mapping, by for each .
This mapping is an isomorphism of onto by the Fundamental Homomorphism Theorem. So that
This can also be proved using the third isomorphism theorem. Note that . And *.
Thus, .
*)Where .
that is very interesting. it seems you can treat it just like a quotient and actually "cancel" the . very nice.
thanks for the link as well. i'll read through it when i have the time, or maybe we'll get to this in class and i can use it as supplementary material