The discussion revolves around whether the function phi(x) = sqrt(x) serves as an automorphism of the group of positive real numbers under multiplication. Participants are exploring the properties of automorphisms and isomorphisms in the context of group theory.
The conversation is ongoing, with participants actively engaging in clarifying definitions and the necessary steps to prove that phi is an automorphism. There is no explicit consensus yet, but guidance has been provided regarding the definitions and the requirements for establishing isomorphism and automorphism.
Participants are working within the constraints of group theory definitions and the requirements set by their textbook. There is a focus on ensuring that the function phi meets the criteria for being an automorphism, including the need for clarity on the definitions involved.
morphism said:You showed that it's a homomorphism.
HallsofIvy said:An automorphism has to be an isomorphism- you have to show that it has an inverse.
morphism said:Have you read the definition of isomorphism?
morphism said:Have you read the definition of isomorphism?
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?Benzoate said: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.
Benzoate said: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.
Benzoate said: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?
HallsofIvy said: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'.
Benzoate said:Is my response unreadable again?
Pretty much unreadable! I have no idea what "the group G is onto itself to prove G is an automorphism" could mean. First, you don't want to prove "G" is an automorphism, it isn't- it's a group! you want to show that the function ph(x)= sqrt(x) is an automorphism. To do that, you must show that it is a homomorphism from G to G (which you did). You must now show that it is both one-to-one and onto. Your proof that phi is one-to-one, that if sqrt(x)= sqrt(y) then x= y, is correct.Benzoate said: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?