How many of these are isomorphisms?

  • Thread starter Thread starter Phred Willard
  • Start date Start date
Phred Willard
Messages
7
Reaction score
0

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.
 
Physics news on Phys.org


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:
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...

Similar threads

Replies
1
Views
8K
Replies
8
Views
3K
Replies
3
Views
1K
Replies
6
Views
2K
Back
Top