# Isomorphism of quotient group

1. Apr 4, 2013

### LayMuon

How can one prove that for homomorphism $G \xrightarrow{\rho} G'$ and H as kernel of homomorphism, quotient group G/H is isomorphic to G'?

Thanks.

2. Apr 4, 2013

### Ben Niehoff

Think about what happens to H under the mapping, and how this relates to the definition of G/H.

3. Apr 4, 2013

### LayMuon

I understand that it suffices to prove that $Ker[G/H \xrightarrow{\rho} G'] =e$. How do you prove this?

4. Apr 5, 2013

### jbunniii

This is only true if $\rho$ is surjective. In general, $G/\ker(\rho)$ is isomorphic to $\text{image}(\rho)$, which may be a proper subgroup of $G'$.

Hint: you need to find an isomorphism $\phi : G/H \to \text{image}(\rho)$. Every element of $G/H$ is a coset of the form $g + H$ for some $g \in G$. What is the most natural way to define $\phi(g+H)$ as an element of $\text{image}(\rho)$?

5. Apr 5, 2013

### jbunniii

But you already defined $\rho$ as a map from $G$ to $G'$. It can't also be a map from $G/H$ to $G'$.

6. Apr 5, 2013

### LayMuon

right, sorry, $\rho'$

7. Apr 5, 2013

### LayMuon

I don't know, how?

8. Apr 5, 2013

### jbunniii

What is an element of $\text{image}(\rho)$ which is in some way related to $g$?

9. Apr 5, 2013

### micromass

Staff Emeritus
Make an educated guess and prove that your guess is correct.

10. Apr 5, 2013

### LayMuon

By image(rho) you mean the subgroup of G' onto which G is mapped?

11. Apr 5, 2013

### jbunniii

Yes, that's right.

12. Apr 5, 2013

### LayMuon

Still confused.

13. Apr 5, 2013

### micromass

Staff Emeritus
So $\rho$ takes in an element of $G$ and gives you an element of $G^\prime$.

You want to define some map $G/H\rightarrow G^\prime$. are given the point $g+H$. You know $g\in G$. You want to use $\rho$ somehow to get a point in $G^\prime$. What are you going to try?

14. Apr 5, 2013

### LayMuon

bundle all $\rho$s to get a $\rho'$?

15. Apr 5, 2013

### jbunniii

You are overthinking this. Given an element $g\in G$, and the mapping $\rho: G \to G'$, can you write down a formula for an element of $G'$ which involves $g$ and $\rho$?

16. Apr 5, 2013

### LayMuon

$\rho(g)$

17. Apr 5, 2013

### jbunniii

Yes, that's what you need. So now how can you define $\phi : G/H \to G'$?
$$\phi(g+H) = ???$$

18. Apr 5, 2013

### LayMuon

$\phi(g+H)=\rho(g)$ but how do you prove the isomorphism?

19. Apr 5, 2013

### jbunniii

OK good, you have found the right mapping, that's half the battle.

Now you need to prove the following:

1. $\phi$ is well defined. This is important because you defined the map $\phi$ in terms of a particular element, $g$, of the coset $g + H$. But that coset may have other elements. You need to show that $\phi$ gives you the same result if you choose a different element $g_1 \in g + H$.

2. $\phi$ is an injection. i.e. $\ker(\phi) = \{e\}$.

3. $\phi$ is a surjection.

4. $\phi$ is a homomorphism.

I recommend starting with 1, so you can be sure you understand why the definition of $\phi$ makes sense. Formally, you need to show that if $g + H = g_1 + H$, then $\phi(g+H) = \phi(g_1 + H)$.

20. Apr 5, 2013

### LayMuon

thanks, let me think and report back. this is an important thing to thoroughly understand.

Last edited: Apr 5, 2013
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