Mapping an isomorphism b/w 2 grps

1. Feb 14, 2013

Bachelier

Is every invertible mapping an isomorphism b/w 2 grps or does it have to be linear?

2. Feb 14, 2013

Bachelier

Re: Isomorphism

also does an isomorphism maps connected (separated) sets to connected (separated) sets?

3. Feb 14, 2013

jbunniii

Re: Isomorphism

No!
It has to be invertible AND a homomorphism, meaning it must satisfy $\phi(ab) = \phi(a)\phi(b)$, where $\phi$ is the mapping and $a,b$ are arbitrary elements of the group. Here, the group operation is written multiplicatively. The additive version is $\phi(a+b) = \phi(a) + \phi(b)$.

By the way, one might think that it would also be necessary to stipulate that $\phi^{-1}$ is a homomorphism, but that turns out to be automatically true if $\phi$ is a bijection and a homomorphism.

4. Feb 14, 2013

jbunniii

Re: Isomorphism

Are we still talking about group isomorphisms? There is no notion of "connected" or "separated" for a general group. You need to impose some additional topological structure. So what kind of groups are you working with?

5. Feb 14, 2013

Bachelier

Re: Isomorphism

The whole question has to deal with analysis..

if A is connected and we have T: A ---> B an isomorphism, can we say T(A) in B is connected?

I guess one still have to show that a mapping is a homomorphism even in analysis. right?

6. Feb 14, 2013

micromass

Staff Emeritus
Re: Isomorphism

OK, but you are clearly not working with just groups. What is the structure you are working with?? What are A and B?? What kind of map is T? It's an isomorphism of what?

7. Feb 15, 2013

Bachelier

I think I was confusing the invertibilty of a Linear Mapping between 2 Vector Spaces with any function that has an inverse.

I remember in my Lin. Alg. course, we learned that if a Linear Transformation T is invertible, then it is an isomorphism between the 2 VS.

Clearly this is not the general case.