Please show that Hom(Zn,Zn)=Zn.

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Homework Help Overview

The discussion revolves around the structure of the homomorphism set Hom(Z/nZ, Z/nZ) and its relationship to Z/nZ. Participants are exploring whether this set is isomorphic to Z/nZ and the implications of homomorphisms in this context.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of a homomorphism and its properties, questioning whether the mapping h(f)=f(1) is well-defined. There are inquiries about the necessity of injectivity and surjectivity in this context, as well as the distinction between homomorphisms and isomorphisms.

Discussion Status

The discussion is active, with participants offering different perspectives on the requirements for the mapping to be an isomorphism. Some participants suggest that the elements of Hom(Z/nZ, Z/nZ) do not need to be isomorphisms, while others emphasize the importance of the map being injective and surjective.

Contextual Notes

There is a noted distinction between the requirements for homomorphisms and isomorphisms, with some participants clarifying that the original question pertains to Hom(Z/nZ, Z/nZ) rather than Aut(Z/nZ).

seventson
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Please show that Hom(Z/nZ,Z/nZ) isomorphic to Z/nZ.
Thanks..
 
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Well, what are your thoughts on the problem?
 
Build a h: Hom(Z/nZ,Z/nZ) --> Z/nZ homomorphism determined by h(f)=f(1). Is it true?
 
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.
 
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.
 
He needs to check that house map from hom to Zn is injective and surjective
 
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
 
Oops, I didn't read carefully enough! Yes, the map h: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.
 
Yes the elements of hom don't need to be isomorphism just being homomorphism is enough. And i think i found proof.
 

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