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Automorphisms homework help

  1. Oct 4, 2007 #1
    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. jcsd
  3. Oct 4, 2007 #2

    morphism

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    You showed that it's a homomorphism.
     
  4. Oct 4, 2007 #3
    How would I show that it's a automorphism
     
  5. Oct 4, 2007 #4

    morphism

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    Prove that it satisfies the definition of automorphism.
     
  6. Oct 5, 2007 #5

    HallsofIvy

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    An automorphism has to be an isomorphism- you have to show that it has an inverse.
     
  7. Oct 5, 2007 #6
    You mean I have to show that its onto?
     
  8. Oct 5, 2007 #7

    morphism

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    Have you read the definition of isomorphism?
     
  9. Oct 5, 2007 #8
    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.
     
  10. Oct 5, 2007 #9

    HallsofIvy

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    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?
     
  11. Oct 5, 2007 #10
    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.
     
  12. Oct 5, 2007 #11

    morphism

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    Because an automorphism is an isomorphism from the group onto itself!
     
  13. Oct 6, 2007 #12

    HallsofIvy

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    Because to understand the definition of "automorphism" you have to understand the definition of "isomorphism". And that was precisely where your first attempt failed!
     
  14. Oct 6, 2007 #13
    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?
     
  15. Oct 6, 2007 #14

    learningphysics

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    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.
     
  16. Oct 6, 2007 #15

    HallsofIvy

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    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'.
     
  17. Oct 6, 2007 #16
    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
  18. Oct 6, 2007 #17
    Is my response unreadable again?
     
  19. Oct 6, 2007 #18

    HallsofIvy

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    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.

    " 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)."
    Your first step is correct. If y is any member of G, any positive real number, then y= phi(x)= sqrt(x) with x= y^2. Since y^2 is still a member of G, that is sufficient.
    No, you do NOT need "y= sqrt(x) and x= sqrt(y)".
     
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