How Is the Quotient Group G/H Isomorphic to G'?

[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.

 

[LayMuon]

Thanks!

 

[LayMuon]

Homomorphism of G to G' means for any g there is a g' and because $gh \in G$ where $h \in H$ so G/H is surjective to G'.

 

[Bacle2]

O.K, let's give it another try: the statement I will use is similar to the one you had.

Here is a sketch . Fill in the details and ask if you need to:

Let h: G-->G' be a homomorphism with kernel K. Then G/K is isomorphic to h(G) ( in case h is not onto G').

Remember an isomorphism is a homomorphism that is both 1-1 and onto. We already have that h is a

homeomorphism--given-- and the map is (tautologically) onto h(G) . So we only need to show, as JBunii

pointed out :

1)h^: G/K -->h(G) is well-defined; since this is a map defined on equivalent classes, we need to show that
members of the same euivalence class have the same image.

(How do we define f^ ?)

2) h^: G/K -->h(G) is 1-1.

First, we need to define the map h^:

h^: G/K -->G' takes [ a] in G/K into h^(a).

Note that a~a' means h(a)=h(a')

Now, we need to see if it is true that a~a' implies h^(a)=h^(a'). So we have,

in the Abelian case (try the non-Abelian), : h(a)-h(a')= h(a-a')=0 .

Then show h^([a]) =h^([a'])

Then h([a]):= h(a) and h([a']):=h(a') , and h(a)-h(a')=...

We can get 2) out of the way quickly: use that G/K collapses exactly those elements with the same image in
G' . Let me do the Abelian case, and you do the non-Abelian one:

Assume g~g' in G/K . Then h(g)-h(g') is in K, which means h(g)-h(g')=...

Assume h(g)=h(g'). Then

So our map h^: G/K -->G' is 1-1 and onto.