How many of these are isomorphisms?

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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:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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