How many of these are isomorphisms?

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Homework Help Overview

The discussion revolves around the number of group homomorphisms and isomorphisms from the cyclic group Zn to itself. The original poster seeks to understand the relationship between homomorphisms and isomorphisms in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the concept that the number of homomorphisms from Zn to Zn is equal to n, based on the gcd of the groups. There is a discussion about how homomorphisms are determined by the image of the generator of the group. Questions arise regarding which elements generate Zn and how many of these correspond to isomorphisms.

Discussion Status

Participants are actively engaging with the problem, with some providing insights into the nature of homomorphisms and isomorphisms. There is a mix of interpretations regarding the number of homomorphisms, particularly in specific cases like Z6, and some participants express uncertainty about the correctness of their reasoning.

Contextual Notes

There is mention of specific examples, such as Z6 and Z8, and the need to consider elements that are relatively prime to n. The original poster's confusion about the factorial representation of homomorphisms suggests a potential misunderstanding of the definitions involved.

Phred Willard
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Homework Statement


Show that the number of group homomorphisms from Zn to Zn is equal to n. How many of these are isomorphisms?

The Attempt at a Solution


It has been shown by other proofs that the number of homomorphisms from Zm to Zn is the gcd(m,n), but here m=n, so the gcd(n,n)=n so that is the number of homomorphisms. (Correct?) and I have no idea how to determine how many would be isomorphisms.
 
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Since 1 generates Zn, any homomorphism is determined by where 1 gets sent. So the image of a homomorphism is generated by the image of 1. What elements of Zn generate Zn?
 


Wouldn't it be all elements relatively prime to n? In the case of Z6:1,3,5; In the case of Z8:1,3,5,7
 


Yep.
 


Does this necessarily mean that there are only 3 homomorphisms in the case of Z6->Z6?
I was under the impression that the number of homomorphisms in this example would be 6! = 720 and only 6 of those would be isomorphisms.

Here was my thought process:
The mappings cycle through, i.e. Z1 maps to Z1-6, then Z2 maps to Z1-6, on till Z6 maps to Z1-6, giving 6! homomorphisms. Each isomorphism occurs at Zn->Zn, so Z1->Z1, Z2->Z2, ..., Z6->Z6.

Is that not correct?
 
Last edited:

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