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Group Theory

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Let ##G## be a group of order ##4, G = {e, a, b, ab}, a^2 = b^2 = e, ab = ba.## Determine Aut(G).

    2. The attempt at a solution

    How can I do these types of problems?

    When doing these types of problems is any automorphism of ##G## always determined by the images? And why would ##\text{Aut(G)}## be isomorphic to ##S_3?##

    I also have a different question. Is the mapping ##T: x\to x^2##, for positive reals under multiplication, an automorphism of its respective group?

    I know that it is a bijective but I don't know if its a homomorphism? For homomorphism, I got ##\phi(xy) = (xy)^2## and ##\phi(x)\phi(y) = x^2y^2## but does ##(xy)^2 = x^2y^2##? I know it will be true if it were an abelian group but it doesn't state it.
     
    Last edited: Nov 3, 2013
  2. jcsd
  3. Nov 3, 2013 #2

    Dick

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    If f is your automorphism then sure, you define f by saying what f(g) is for any element of the group and then showing it's an automorphism. I think you show ##\text{Aut(G)}## is isomorphic to ##S_3## by starting to work stuff out. Can you work out ANY automorphisms besides the identity? You know f(e)=e. Suppose f(a)=a, then you have two possibilities for f(b). Do they lead to automorphisms? There is a conceptual shortcut here but you aren't going to find it until you do something.
     
  4. Nov 3, 2013 #3
    Okay then, so the identity is simple, now for nontrivial elements I get ##f(a) = b , f(a) = ab, f(b) = a , f(b) = ab, f(ab) = a, f(ab) = b, ... ## so there are six automorphisms?
     
  5. Nov 3, 2013 #4

    Dick

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    Saying f(a)=b does not define an automorphism. You have to also define what f(b) and f(ab) are. IF those DID define distinct automorphisms then you would have seven (including the identity). That's wrong. You still haven't defined any nontrivial automorphisms.
     
  6. Nov 3, 2013 #5
    Thanks, Dick. I read over the definitions of homomorphism, automorphism and since as you said, "if f is your automorphism then sure, you define f by saying what f(g) is for any element of the group and then showing it's an automorphism" I think I know what to do now.

    Last question about a different issue. Is the mapping ##T: x\to x^2##, for positive reals under multiplication, an automorphism of its respective group?

    I know that it is a bijective but I don't know if its a homomorphism? For homomorphism, I got ##\phi(xy) = (xy)^2## and ##\phi(x)\phi(y) = x^2y^2## but does ##(xy)^2 = x^2y^2##? I know it will be true if it were an abelian group but it doesn't state it. I say it does not satisfy homomorphism since it is not abelian, am I right in saying that?
     
  7. Nov 3, 2013 #6

    Dick

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    When they say "for positive reals under multiplication" I think they are assuming that you know that the positive real numbers are an abelian group under multiplication. They are. ab=ba is a basic property of the real numbers. I think they were expecting you to worry more about bijective than abelian. How do you know it's bijective?
     
    Last edited: Nov 4, 2013
  8. Nov 4, 2013 #7
    I know they are bijective since it only included the positive reals under multiplication. If it wear all reals then it wouldn't be a bijection since it would not be 1-1. Am I correct to think that?
     
  9. Nov 4, 2013 #8

    Dick

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    Absolutely, but I think they want you to explain how you know it's bijective for positive reals.
     
  10. Nov 4, 2013 #9
    Thank you very much, Dick!
     
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