Product of Quotient Groups Isomorphism

tylerc1991
Messages
158
Reaction score
0

Homework Statement



I have attached the problem below.

Homework Equations





The Attempt at a Solution



I have tried to use the natural epimorphism from G x G x ... x G to (G x G x ... x G)/(K1 x K2 x ... x Kn), but I do not believe that this is an injective function. Then I tried to use a function f from G to (G/K1) x (G/K2) x ... x (G/Kn) defined by f(g) = (k1g, k2g, ..., kng) for all g in G. I found that this function was injective (and a homomorphism) but not surjective. I've also been trying to find a way to use the first isomorphism theorem, but with no luck. Any hints would be greatly appreciated!
 

Attachments

  • herstein_problem.jpg
    herstein_problem.jpg
    10.7 KB · Views: 507
Physics news on Phys.org
How can something be a homeomorphism and not be subjective? It was my understanding that a homeomorphism is a bi-continuous function. That is, it is a continuous function with a continuous inverse. If its invertible then it is both surjective and injective (both the function and its inverse). Was that a typo?

As for the problem I think (G/K1) x (G/K2) x ... x (G/Kn) defined by f(g) = (k1g, k2g, ..., kng) for all g in G is a good start. But what theorems (if any) have you developed so far in this context?
 
stephenkeiths said:
How can something be a homeomorphism and not be subjective? It was my understanding that a homeomorphism is a bi-continuous function. That is, it is a continuous function with a continuous inverse. If its invertible then it is both surjective and injective (both the function and its inverse). Was that a typo?

What do homeomorphisms and topology have to do with this? :confused:
This is a group theory question, not topology. You do know what a homomorphism is, right? If not, then maybe you should not answer questions like this.
 
tylerc1991 said:

Homework Statement



I have attached the problem below.

Homework Equations





The Attempt at a Solution



I have tried to use the natural epimorphism from G x G x ... x G to (G x G x ... x G)/(K1 x K2 x ... x Kn), but I do not believe that this is an injective function. Then I tried to use a function f from G to (G/K1) x (G/K2) x ... x (G/Kn) defined by f(g) = (k1g, k2g, ..., kng) for all g in G. I found that this function was injective (and a homomorphism) but not surjective. I've also been trying to find a way to use the first isomorphism theorem, but with no luck. Any hints would be greatly appreciated!

I think there is something missing in the problem. If you take K_1=K_2=\{e\}, then this gives G\sim G\times G. This is surely not true in general.
 
i misread homo as homeo. That confused me as much as my post confused you. And questioning whether or not I know a homomorphism is a function f:(G1,*)->(G2,~) such that f(g1*g2)=f(g1)~f(g2) for any g1,g2 in G1 hardly seems helpful.

And its my intuition that you'll have to use the 2nd isomorphism theorem also (or some theorem about products). Do you have any more specific theorems about products?
 
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