Proof Involving Homomorphism and Normality

  • Thread starter rlusk35
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    Proof
In summary, we need to show that if \theta is a homomorphism from G onto H and N \triangleleft G, then \theta(N) \triangleleft H. To do this, we first observe that N is a subgroup of G and \theta(G) is a subgroup of H. Since \theta is onto, we can write any element h in H as \theta(g) for some g in G. Using the fact that \theta is a homomorphism, we can show that h\theta(N) = \theta(g)\theta(N) = \theta(gN) = \theta(N)h. Therefore, we have shown that \theta(N) is normal in H.
  • #1
rlusk35
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Homework Statement


Prove that if [tex]\theta[/tex] is a homomorphism from G onto H, and N [tex]\triangleleft[/tex] G, then [tex]\theta[/tex](N) [tex]\triangleleft[/tex] H.

Homework Equations


The Attempt at a Solution


I think I have a good idea of what is going on, but I'm struggling to tie it all together.

It's given that N [tex]\triangleleft[/tex] G so I know that gng[tex]^{-1}[/tex] [tex]\in[/tex] N for all n[tex]\in[/tex]N and all g[tex]\in[/tex]G. I also know that N is a subgroup of G.

I know that [tex]\theta[/tex](G), the image of [tex]\theta[/tex], is a subgroup of H. Because of this, I would also think that [tex]\theta[/tex](N) is also a subgroup of H because of the homomorphism.

From here I need someone to lead me in the right direction. I've been trying to solve this problem for four days so any help is greatly appreciated.
 
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  • #2
You are trying to show that [tex]
\theta
[/tex](N) is normal in H, so you should start with h [tex]
\theta
[/tex](N), for some h in H. [tex]
\theta
[/tex] is onto, so h= [tex]
\theta
[/tex](g) for some g in G. Now use the fact that [tex]
\theta
[/tex] is a homomorphism and that N is normal in G.
 
  • #3
I don't think I'm following entirely. Are you saying that since I know N is in G and because the homomorphism sends elements in G to H, I can say that N is in H? If this is true, I don't understand how to show that N is normal in H.
 
  • #4
No, the whole point of the proof is to show that [tex]
\theta
[/tex](N) is normal in H. To show that, you must show that for all h in H, h[tex]
\theta
[/tex](N) = [tex]
\theta
[/tex](N)h.

Starting off with some h in H, you can observe that h = [tex]
\theta
[/tex](g) for some g in G. This is because [tex]
\theta
[/tex] is onto.

Thus you have h [tex]
\theta
[/tex](N)= [tex]
\theta
[/tex](g) [tex]
\theta
[/tex](N). Now you must use the fact that [tex]
\theta
[/tex] is a homomorphism.
 
  • #5
How do you know that [tex]\theta[/tex] is onto?
 
  • #6
because in the original problem you said ".. [tex]
\theta
[/tex] is a homomorphism from G onto H"
 
  • #7
I'm following you now. Thanks for the help.
 

1. What is a homomorphism?

A homomorphism is a mathematical function that preserves the structure of a group. In other words, it takes elements from one group and maps them to elements in another group while maintaining the operation of the group.

2. How do you prove a homomorphism?

To prove a homomorphism, you need to show that the function preserves the group operation. This means that for any two elements a and b in the original group, the function applied to their operation (f(ab)) is equal to the operation of the function on each element separately (f(a)f(b)).

3. What is normality in a group?

A normal subgroup is a subgroup of a group that is invariant under conjugation by any element of the original group. In other words, if a subgroup is normal, then conjugating any element of the subgroup by an element in the original group will still result in an element of the subgroup.

4. How does normality relate to homomorphism?

In general, a homomorphism will map a normal subgroup of one group to a normal subgroup of another group. This is because a homomorphism preserves the operation of the group, so it will also preserve the structure of normality.

5. What is the role of normality in proving a homomorphism?

In order to prove that a homomorphism exists between two groups, it is helpful to show that the kernel (the elements that map to the identity element) is a normal subgroup of the original group. This allows for a more straightforward proof, as it simplifies the structure of the group and makes it easier to show that the function preserves the operation.

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