Find All Automorphisms of Cyclic Group of Order 10

In summary, the problem is to find the automorphisms of a cyclic group of order 10. An automorphism must map a generator of the cyclic group to another generator. In this case, the generators are the odd numbers below 10. Therefore, any function that maps one generator to another will be an automorphism. Examples of such functions include phi(x)=x^3 and phi(x)=g, where g is any generator. x^5 is not a generator because it does not give all the elements of the group when raised to different powers.
  • #1
Electromech1
2
0

Homework Statement



Find all the automorphisms of a cyclic group of order 10.

Homework Equations



ψ(a)ψ(b)=ψ(ab)

For G= { 1, x, x^2,..., x^9}, and some function

ψ(a) = x^(a/10)

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
 
Physics news on Phys.org
  • #2
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
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
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:

1. What is a cyclic group?

A cyclic group is a mathematical group that is generated by a single element, called a generator. This means that every element in the group can be written as a power of the generator. In simpler terms, it is a group where each element can be obtained by repeatedly applying a single operation to a starting element.

2. What is the order of a cyclic group?

The order of a cyclic group is the number of elements in the group. In other words, it is the number of times the generator needs to be applied to itself to generate all the elements in the group.

3. How do you find all the automorphisms of a cyclic group of order 10?

To find all the automorphisms of a cyclic group of order 10, we need to consider all possible mappings from the group to itself that preserve the group structure. In this case, as the group is cyclic, any automorphism will be uniquely determined by the image of the generator. Therefore, there will be 10 possible automorphisms, each corresponding to a different choice of the image of the generator.

4. Can you give an example of an automorphism of a cyclic group of order 10?

One example of an automorphism of a cyclic group of order 10 is the identity map, where each element is mapped to itself. Another example is the map that sends the generator to itself and all other elements to the identity element. There are 8 other possible automorphisms, each with a different image for the generator.

5. How do automorphisms of cyclic groups relate to rotations?

Automorphisms of cyclic groups can be thought of as rotations in a geometric sense. Just like how rotating a shape around a point preserves its structure, automorphisms preserve the structure of cyclic groups. In fact, the automorphisms of a cyclic group of order n are isomorphic to the group of rotations of a regular n-gon in the complex plane.

Similar threads

  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
608
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
3K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
Back
Top