Group Theory Question: Ker(p) and Homomorphisms Explained in Detail

PsychonautQQ
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Homework Statement


Let p: G-->M be a group homomorphism with ker(p) = K. If a is an element of G, how that Ka = {g in G | p(g) = p(a)}

Homework Equations


none needed

The Attempt at a Solution


Okay, I've been struggling with this problem for awhile and I've ran into a problem:

-Let g be an element of Ka
-Let b be an element of K such that ba = g.

Since g is an element of Ka and the intersection of Ka and K is {1}, p(g) does not equal zero.

But then if ba = g then:
p(ba) = p(g)
p(b)p(a) = p(g)
0p(a) = 0, but p(g) can't be zero!

Someone want to shed some light perhaps? I guess I need help on understanding this road block I've run into as well as the actual problem >.<. Thanks!
 
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Why do you say p(b)=0?
 
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Nevermind. Lol I'm a noob i had an error in how I was thinking.
 
PsychonautQQ said:

Homework Statement


Let p: G-->M be a group homomorphism with ker(p) = K. If a is an element of G, how that Ka = {g in G | p(g) = p(a)}

Homework Equations


none needed

The Attempt at a Solution


Okay, I've been struggling with this problem for awhile and I've ran into a problem:

-Let g be an element of Ka
-Let b be an element of K such that ba = g.

Since g is an element of Ka and the intersection of Ka and K is {1}, p(g) does not equal zero.

What is zero? You're dealing with groups. 1 is the identity and all elements are invertible.
 
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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...
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