# Find All Automorphisms of Cyclic Group of Order 10

1. Jan 27, 2012

### Electromech1

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. Jan 27, 2012

### Dick

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. Jan 27, 2012

### Electromech1

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^(10-x) 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. Jan 27, 2012

### Dick

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?

Last edited: Jan 27, 2012