# Homework Help: Automorphisms homework help

1. Oct 4, 2007

### Benzoate

1. The problem statement, all variables and given/known data

Let R^+ be the group of positive real numbers under multiplication. Show that the mapping phi(x)=sqrt(x) is an automorphism of positive real numbers

2. Relevant equations

3. The attempt at a solution
Let x and y be real numbers.
Let phi(xy)=sqrt(xy). the sqrt(xy)=sqrt(x)*sqrt(y)=phi(x)phi(y). therefore , phi(xy) is an automorphism

2. Oct 4, 2007

### morphism

You showed that it's a homomorphism.

3. Oct 4, 2007

### Benzoate

How would I show that it's a automorphism

4. Oct 4, 2007

### morphism

Prove that it satisfies the definition of automorphism.

5. Oct 5, 2007

### HallsofIvy

An automorphism has to be an isomorphism- you have to show that it has an inverse.

6. Oct 5, 2007

### Benzoate

You mean I have to show that its onto?

7. Oct 5, 2007

### morphism

Have you read the definition of isomorphism?

8. Oct 5, 2007

### Benzoate

yes , for an isomorphism to be an automorphism, a Group g must be onto itself. When a group is onto itself , it you find the inverse of that function.

9. Oct 5, 2007

### HallsofIvy

You were asked about the definition of "isomorphism". Do you understand that what you wrote in response to that says nothing about the definition of isomophism? How does an isomorphism differ from a homomorphism?

10. Oct 5, 2007

### Benzoate

Yes. I do understand the definition of an isomorphism: a group G to a group G* is a one-to-one mapping from G onto G* that preserves the group operation. Or In symbolic terms , phi(ab)=phi(a)phi(b). Why do you think the definition of an isomorphism is more important than the definition of a automorphism in this problem.

11. Oct 5, 2007

### morphism

Because an automorphism is an isomorphism from the group onto itself!

12. Oct 6, 2007

### HallsofIvy

Because to understand the definition of "automorphism" you have to understand the definition of "isomorphism". And that was precisely where your first attempt failed!

13. Oct 6, 2007

### Benzoate

okay what should my first attempt show be to prove the isomorphism phi(xy) is an automorphism? Obviously I need to show that phi(x*y)=sqrt(xy). Before I attempt to prove that phi(xy)=sqrt(xy) is an automorphism, should I attempt to show that phi(xy) is an isomorphism first since the definition of an isomorphism says an isosomorphism from a group G onto itself is an automorphism.

Let me see if have this clear: I have to show all 4 properties of an isomorphism for phi(xy) to show that its an isomorphismand go and turn around in show that phi(xy) is onto to prove that its an automorphism?!?!?! Why do I have to prove that phi(xy) is onto twice?

14. Oct 6, 2007

### learningphysics

You haven't shown that phi is an isomorphism yet... You've shown that it is a homomorphism. Now you need to show that it is one-to-one and onto... That will show that it is an isomorphism. Then since phi is a mapping from a group to itself, it is an automorphism. An automorphism is an isomorphism that maps a group to itself... If it was a mapping from a group G to another group H, then it wouldn't be an automorphism, just an isomorphism.

The 3 steps to solve the problem are 1) show phi is a homomorphism. 2) show phi is one-to one. 3) show phi is onto

you've done the first part. now you need to do parts 2 and 3.

15. Oct 6, 2007

### HallsofIvy

You have already shown it is a homomorphism. To show it is an isomorphism, you only need to show it is one-to-one and onto. Of course, if you have already shown it is one-to-one in order to show it is an isomorphism, you don't need to do it again to show it is an automorphism! Actually, since "one-to-one" and "onto" are part of the definition of isomorphism, you could just as easily define "Automorphism on G" to be an "isomorphism from G to itself" rather than "onto'.

16. Oct 6, 2007

### Benzoate

according to my textbook, you need to show that the group G is onto itself to proved G is an automorphism . In order to prove that phi(x) is one-to-one, I need to prove that phi(x)=phi(y) => x=y right? therefore phi(x)=phi(y) => sqrt(x)=sqrt(y). After squaring both sides , x=y. There phi(x) is one to one. in ordert to prove that phi(x) is onto- I need to show that y=sqrt(x)=> x=y^2. For a group to be onto, shouldn't y= sqrt(x) and x=sqrt(y). What did I do wrong?

Last edited: Oct 6, 2007
17. Oct 6, 2007