Find All Automorphisms of Cyclic Group of Order 10by Electromech1 Tags: abstract algebra, automorphism, cyclic groups, group 

#1
Jan2712, 09:56 AM

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1. The problem statement, all variables and given/known data
Find all the automorphisms of a cyclic group of order 10. 2. Relevant equations ψ(a)ψ(b)=ψ(ab) For G= { 1, x, x^2,..., x^9}, and some function ψ(a) = x^(a/10) 3. The attempt at a solution I know that a homomorphism takes the form Phi(a)*phi(b) = phi (ab) , and that an automorphism maps from G>G, However, I don't understand what an automorphism for a cyclic group would even look like. I suppose it should be something of the form: ψ(a) = x^(a/10) and that a should be a specific power, but I have no idea where to go from here. I appreciate any help. Thanks 



#2
Jan2712, 10:11 AM

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Think about it. An automorphism has to map a generator of the cyclic group to another generator. How many elements of your cyclic group are generators?




#3
Jan2712, 10:39 AM

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I don't have a solid understanding of generators. I understand that you need relatively prime powers to map out all the functions through multiplication. Going by this, I suppose the generators would be 1,3,5,7, & 9, all the odd numbers below 10. For what reason are would these be automorphisms though? Tell me if this is right:
[ If x^n is an automorphism, where n is an individual element of {1,3,5,7,9}, the function works by multiplying each term of G= {1,x,...,x^9} so that phi(a) = x^n *x*a = x^ (n+a) It's a homomorphism because: phi(a) * phi (b) = x^(n+a) * x^(n+b) = x^( 2n+a+b) and phi(ab) = x^(n + a +b) (I know 2n+a+b ≠ n+a+b, but it's the best I can come up with) Then the function is bijective because if phi(x) = x^n, phi(x)^1 = x^n or x^(10x) because of the cyclic nature Because it maps from G>G, it is homomorphic, and has an inverse, it is bijective automorphism for the cyclic group of order 10. ] Am I at least heading in the right direction with this? 



#4
Jan2712, 11:06 AM

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Find All Automorphisms of Cyclic Group of Order 10
I would look at it this way. x is a generator because powers of x give you all of the elements of G. x^3 is also a generator. Take all of the powers of x^3 and show you can get every element of G as a power of x^3. E.g. x^7=(x^3)^9. And yes, the reason is because 10 and 3 are relatively prime. If you are uncomfortable with the notion of generators, you should verify this by writing out all of the powers. That means if you define phi(x)=x^3 you get an automorphism. If g is ANY generator and you define phi(x)=g, you get an automorphism. BTW x^5 is NOT a generator. Why not?



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