1. Apr 7, 2012

### seventson

Please show that Hom(Z/nZ,Z/nZ) isomorphic to Z/nZ.
Thanks..

2. Apr 7, 2012

### morphism

Well, what are your thoughts on the problem?

3. Apr 8, 2012

### seventson

Build a h: Hom(Z/nZ,Z/nZ) --> Z/nZ homomorphism determined by h(f)=f(1). Is it true?

4. Apr 8, 2012

### conquest

Go from here and check whether it is well defined (i.e. actually a homomorphism), whether it is in fact injective and surjective (sometimes called one-to-one and onto) then determine whether the inverse is also a homomorphism.

5. Apr 8, 2012

### A. Bahat

Actually, it is not necessary to check that the homomorphism is injective or surjective: he's asking for Hom(Z/nZ,Z/nZ), not Aut(Z/nZ), so the maps need not be isomorphisms.

6. Apr 8, 2012

### Office_Shredder

Staff Emeritus
He needs to check that house map from hom to Zn is injective and surjective

7. Apr 8, 2012

### conquest

Exactly the map h mentioned above does actually need to be an isomorphism thus injective and surjective. Also the inverse map needs to be a homomorphism.

The elements of Hom(Z/nZ, Z/nZ) do not need to be isomorphisms.

The trick here is of course to see that homomorphism are uniquely defined by their image of 1 To show this you need to use the fact that they are homomorphisms

8. Apr 8, 2012

### A. Bahat

Oops, I didn't read carefully enough! Yes, the map hom(Z/nZ,Z/nZ)--->Z/nZ must be an isomorphism; I was thinking of the elements of Hom(Z/nZ,Z/nZ), which of course don't have to be isomorphisms.

9. Apr 8, 2012

### seventson

Yes the elements of hom dont need to be isomorphism just being homomorphism is enough. And i think i found proof.