## mapping an isomorphism b/w 2 grps

Is every invertible mapping an isomorphism b/w 2 grps or does it have to be linear?
 PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
 also does an isomorphism maps connected (separated) sets to connected (separated) sets?

Blog Entries: 1
Recognitions:
Gold Member
Homework Help
 Quote by Bachelier I googled this but couldn't find a clear answer. Is every invertible mapping an isomorphism b/w 2 grps
No!
 or does it have to be linear?
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.

Blog Entries: 1
Recognitions:
Gold Member
Homework Help

## mapping an isomorphism b/w 2 grps

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

 Quote by jbunniii 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?
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?

Blog Entries: 8
Recognitions:
Gold Member