Isomorphism of quotient group

[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.

 

[Ben Niehoff]

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

 

[LayMuon]

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

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

 

[jbunniii]

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'?

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 $\text{image}(\rho)$?

 

[jbunniii]

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

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

 

[LayMuon]

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

right, sorry, $\rho'$

 

[LayMuon]

What is the most natural way to define $\phi(g+H)$ as an element of $\text{image}(\rho)$?

I don't know, how?

 

[jbunniii]

I don't know, how?

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

 

[micromass]

I don't know, how?

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

 

[LayMuon]

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

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

 

[jbunniii]

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

Yes, that's right.

 

[LayMuon]

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

Still confused.

 

[micromass]

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?

 

[LayMuon]

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?

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

 

[jbunniii]

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

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 $\rho$?

 

[LayMuon]

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 $\rho$?

$\rho(g)$

 

[jbunniii]

$\rho(g)$

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

 

[LayMuon]

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

 

[jbunniii]

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

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 $\phi(g+H) = \phi(g_1 + H)$.

 

[LayMuon]

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

 

[Bacle2]

Think of the fact that you already have an onto homomorphism (as a hypothesis/given). Show that modding out by the kernel gives you the missing ingredient --injectivity.

Modding out by the kernel collapses all elements with the same image into one coset.

 

[LayMuon]

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 $g_1 \in g + H$.

$\rho(g_1)=\rho(g_i h_n) = \rho(g_i) \rho(h_n) = \rho(g_i) e' = \rho(g_i)$

$\rho(g_2)=\rho(g_i h_k) = \rho(g_i) \rho(h_k) = \rho(g_i) e' = \rho(g_i)$

So they are mapped onto the same point.

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

I am stuck here. ((

What is the kernel of $\phi$?

 

[jbunniii]

$\rho(g_1)=\rho(g_i h_n) = \rho(g_i) \rho(h_n) = \rho(g_i) e' = \rho(g_i)$

$\rho(g_2)=\rho(g_i h_k) = \rho(g_i) \rho(h_k) = \rho(g_i) e' = \rho(g_i)$

So they are mapped onto the same point.

I think you have the right idea, but the statement is not very clear. If $g_1H = g_2H$, then $g_2^{-1} g_1 \in H = \ker(\rho)$, so what can you say about $\rho(g_2^{-1} g_1)$?

By the way, I just noticed that in several previous posts above, I wrote $g+H$ as an arbitrary element of $G/H$. Must be too much linear algebra on the brain. Should have been $gH$ as we are not assuming that $G$ is abelian.

 

[jbunniii]

I am stuck here. ((

What is the kernel of $\phi$?

If $\phi(gH) = \rho(g)$, then $\ker(\phi)$ is the set of all cosets $gH$ such that $\rho(g) = 1$.

 

[LayMuon]

I meant if $g_1 \in g_i H$ and $g_2 \in g_i H$ then they are mapped onto the same point of G'.

 

[LayMuon]

I think you have the right idea, but the statement is not very clear. If $g_1H = g_2H$, then $g_2^{-1} g_1 \in H = \ker(\rho)$, so what can you say about $\rho(g_2^{-1} g_1)$?

By the way, I just noticed that in several previous posts above, I wrote $g+H$ as an arbitrary element of $G/H$. Must be too much linear algebra on the brain. Should have been $gH$ as we are not assuming that $G$ is abelian.

$\rho(g_2^{-1} g_1) =e'$ hence $(\rho(g_2^{-1}))^{-1} =\rho(g_1)$

 

[jbunniii]

I meant if $g_1 \in g_i H$ and $g_2 \in g_i H$ then they are mapped onto the same point of G'.

OK, with that additional explanation it makes sense. Your proof is fine. So now try showing that $\phi$ is an injection.

 

[jbunniii]

$\rho(g_2^{-1} g_1) =e'$ hence $(\rho(g_2^{-1}))^{-1} =\rho(g_1)$

Right, and $\rho(g_2^{-1})^{-1} = ??$

 

[LayMuon]

Right, and $\rho(g_2^{-1})^{-1} = ??$

$\rho(g_2)$

 

[LayMuon]

If $\phi(gH) = \rho(g)$, then $\ker(\phi)$ is the set of all cosets $gH$ such that $\rho(g) = 1$.

so $g \in H$ and the kernel is H. so the kernel of gH is the ${h_i H}$ where $h_i \in H \Rightarrow ker(\phi) = H$ , hence the kernel of G/H is trivial and the mapping injective.

 

[jbunniii]

$\rho(g_2)$

Right, so once again you get $\rho(g_1) = \rho(g_2)$ and the mapping is well defined.

so $g \in H$ and the kernel is H. so the kernel of gH is the ${h_i H}$ where $h_i \in H \Rightarrow ker(\phi) = H$ , hence the kernel of G/H is trivial and the mapping injective.

I guess you mean "the kernel of $\phi$ is..."

You have the right idea, but the statement could be clearer. For example:

$\phi(gH) = 1$ if and only if $\rho(g) = 1$ if and only if $g \in H$ if and only if $gH = H$. Therefore, the kernel of $\phi$ is $\{H\}$, and $H$ is the identity element of the group $G/H$, which means that the kernel of $\phi$ is trivial, so $\phi$ is an injection.

OK next step is to show that $\phi$ is a surjection. Note that this will not be true unless you assume $\rho$ is a surjection, which is not stated in the original post.

 

[LayMuon]

Doesn't injectivity imply surjectivity and homomorphism?

 

[jbunniii]

Doesn't injectivity imply surjectivity and homomorphism?

No, it doesn't imply either one. What are the definitions of surjectivity and homomorphism?

 

[LayMuon]

No, it doesn't imply either one. What are the definitions of surjectivity and homomorphism?

Isn't injectivity the one-to-one mapping of G onto some subgroup of G', surjectivity mapping onto some subgroup which is not necessarily one-to-one, homomorphism sounds like surjectivity. Anyway I do not clearly distinguish between them. Wikipedia uses so many other terms that it would take a lot of time and distraction from physics to go through all of them. Any graphical explanation? Thanks.

 

[Bacle2]

No; injectivity means that no two elements of g have the same image in g', and surjectivity means that every element g' in G' is the image of some g in G. Sorry, I don't have any graphical sources, but,as an example, f(x)=x^2 from R to R is not injective, because f(1)=f(-1) =1 , and is not surjective, since no negative number is the image of any positive number --every square is nonnegative. But f(x)=x is both injective and surjective.