Find Aut(Z n ) for cases n = 2,3,4, and 5

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In summary, Aut(Zn) is the set of automorphisms of the cyclic group Zn. For n = 2,3,4, and 5, the order of Aut(Zn) is equal to the Euler totient function of n. The only functions that satisfy Aut(Z) are f(x) = x and f(x) = -x, and for Aut(Z2), f(x) = kx and f(x) = x+k, where k is an integer. The set of automorphisms forms a group and the identity element is mapped to an identity in Z. It is important to check that the order of each element is the same before committing to an answer.
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zcdfhn
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Find Aut(Zn) for cases n = 2,3,4, and 5

My question is to find Aut(Zn) for cases n = 2,3,4, and 5

First, I am a little confused about the case Aut(Z), the integers. I know that the only functions that satisfy Aut(Z) are f(x) = x, and f(x) = -x, but what is wrong with f(x) = x+1?

It would be great if you explain to me how to find automorphic functions, and why the previous example works and doesn't work.

My attempts at Aut(Z2): I think that f(x) = kx and f(x) = x+k, k [tex]\in[/tex] Z are the only functions that apply in this case.

The rest, I believe, are similar, but I just need help before I commit to an answer.

Thank you for your time.
 
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zcdfhn said:
My question is to find Aut(Zn) for cases n = 2,3,4, and 5

First, I am a little confused about the case Aut(Z), the integers. I know that the only functions that satisfy Aut(Z) are f(x) = x, and f(x) = -x, but what is wrong with f(x) = x+1?

It would be great if you explain to me how to find automorphic functions, and why the previous example works and doesn't work.

My attempts at Aut(Z2): I think that f(x) = kx and f(x) = x+k, k [tex]\in[/tex] Z are the only functions that apply in this case.

The rest, I believe, are similar, but I just need help before I commit to an answer.

Thank you for your time.

[tex] f \in Aut (Z)[/tex] implies [tex]f:Z \rightarrow Z[/tex] and [tex]f^{-1}:Z \rightarrow Z[/tex] should be injective. Thus, both the kernels of [tex]f, f^{-1}[/tex] should be identity. That implies the identity is mapped to an identity in Z.

Generally speaking, the set of automorphisms forms a group and the order of [tex]Aut(Z_n)[/tex] is [tex]\phi(n)[/tex] (Euler totient function).
Once you find an automorphism, you need to make sure that the order of each element is same with [tex]x \in Z_n[/tex] and [tex]f(x) \in Z_n[/tex], where f is an automorphism of [tex]Z_n[/tex].
 

What is Aut(Zn)?

Aut(Zn) is the automorphism group of the group Zn, also known as the integers modulo n. It consists of all bijective homomorphisms from Zn to itself.

What are the possible values of n for Aut(Zn)?

The possible values of n for Aut(Zn) are 2, 3, 4, and 5, as specified in the question. These correspond to the groups Z2, Z3, Z4, and Z5, respectively.

How do you find Aut(Zn)?

To find Aut(Zn), we first need to determine the elements of the group Zn. These are the integers from 0 to n-1. Then, we need to find all possible bijective homomorphisms (functions that preserve the group structure) from Zn to itself. Finally, we can combine these functions using composition to form the automorphism group Aut(Zn).

What is the size of Aut(Zn)?

The size of Aut(Zn) is equal to the number of bijective homomorphisms from Zn to itself. This can be calculated using the formula (n-1)!, where n is the order of the group Zn. Therefore, the sizes of Aut(Zn) for n=2,3,4, and 5 are 1, 2, 6, and 24, respectively.

What is the significance of Aut(Zn)?

Aut(Zn) is an important concept in group theory as it helps us understand the structure of the group Zn. It also has applications in cryptography, where the automorphism group of a finite abelian group is used to generate keys for encryption.

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