# Group theory question

• PsychonautQQ
In summary, the conversation discusses a group homomorphism with a given kernel and an element of the group. The goal is to show that the set Ka is equal to the set of elements in G that map to the same value as a under the homomorphism. The conversation also addresses a potential roadblock in the solution, which is resolved by understanding the properties of groups.

## 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)}

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!

Why do you say p(b)=0?

• PsychonautQQ
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)}

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.

• PsychonautQQ