Proving Find isomorphism: What Else Needed?

[Shackleford]

Is what I did all I need to do? Is there anything else I need to prove?

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110728_175901.jpg?t=1311905852

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110728_175917.jpg?t=1311905865

 

[micromass]

Hi Shackleford!

You're probably not going to like this, but you're not done yet. You have yet to show that this is an isomorphism.

Your function f is an isomorphism if it is bijective (this is trivial) and if

$$f(xy)=f(x)f(y)$$

for all x and y. You have to show that this holds for your groups.

For example, you'll need to show that

$$f(a^2)=f(a)^2,~f(ba)=f(b)f(a)$$

and so on for every element. This is a lot of work, but maybe you can find some shortcuts that could reduce the calculations a bit? Or maybe you won't bother with checking all of these things

 

[Shackleford]

Hi Shackleford!

You're probably not going to like this, but you're not done yet. You have yet to show that this is an isomorphism.

Your function f is an isomorphism if it is bijective (this is trivial) and if

$$f(xy)=f(x)f(y)$$

for all x and y. You have to show that this holds for your groups.

For example, you'll need to show that

$$f(a^2)=f(a)^2,~f(ba)=f(b)f(a)$$

and so on for every element. This is a lot of work, but maybe you can find some shortcuts that could reduce the calculations a bit? Or maybe you won't bother with checking all of these things

Am I correct in making the transformation by assigning the matrices e, a, b, and ab respectively? If so, then I already showed ab = [ ] [ ] = ba. I need to do a², too?

 

[micromass]

Am I correct in making the transformation by assigning the matrices e, a, b, and ab respectively? If so, then I already showed ab = [ ] [ ] = ba. I need to do a², too?

Yes, you're matrices are assigned perfectly! The assignment you wrote down is an homomorphism, but you just need to prove it.

 

[Robert1986]

One thing to consider is that both groups are abelian. This means that you don't need to check ab and ba. You just need to check one. So I think there's really only 6 mappings you need to check and it is better to do it in a certain order, I think.

 

[Shackleford]

One thing to consider is that both groups are abelian. This means that you don't need to check ab and ba. You just need to check one. So I think there's really only 6 mappings you need to check and it is better to do it in a certain order, I think.

I need to check six mappings?

The properties for isomorphism are one-to-one correspondence and, generally, if φ(a*b) = φ(a) x φ(b).

Clearly it has the one-to-one correspondence, and I've already shown φ(ab) = φ(a)φ(b). As mentioned previously, I suppose I need to do a² and b².

 

[Robert1986]

Don't you need to show it for a*ab b*ab and ab*ab. These will be trivial, but I don't see how you can get around it.

 

[Shackleford]

Don't you need to show it for a*ab b*ab and ab*ab. These will be trivial, but I don't see how you can get around it.

I suppose you're right, since the set is so small.