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Prove the set of all automorphisms of a group is a group.

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data

    An isomorphism of a group onto itself is called an automorphism. Prove that the set of all automorphisms of a group is itself a group with respect to composition.


    2. Relevant equations

    To prove that this is a group I must show that it is closed on composition, there is an identity, and each element has an inverse, but proving something is a group isn't where the trouble lies. The trouble lies in reading the problem/understanding the terms.

    3. The attempt at a solution

    First let's consider the thing called "automorphism". Is this a mapping? Say, the identity mapping? What exactly is the thing called "automorphism"?

    Second, what is the set of all automorphisms of a group? How many ways can you really list the group? Isn't there only one? I'm pretty confused about these meanings. I don't actually need help showing this is a group, I need help knowing what set looks like.
     
  2. jcsd
  3. Nov 13, 2011 #2
    Are you studying from a book? If the book has a question on automorphisms, then it should also have a definition. Have you looked in the index?
     
  4. Nov 13, 2011 #3
    The explanation given is "An isomorphism of a group onto itself is called an automorphism."
     
  5. Nov 13, 2011 #4
    What does it say that an isomorphism is?
     
  6. Nov 13, 2011 #5
    G is isomorphic to H means there is an operation preserving bijection from G to H.
     
  7. Nov 13, 2011 #6
    OK, one last question and we can get to work. What is the definition given for composition?
     
  8. Nov 13, 2011 #7
    f(g(x)) ? I don't know what you're asking here. Standard composition.
     
  9. Nov 13, 2011 #8
    That's what I was looking for. In order to prove that the set of automorphisms of a group G is itself a group under composition, you need to show 4 things. The first of these is closure. In other words, if f is an automorphism and g is an automorphism, then fg is an automorphism. In order to show that fg is an automorphism you have to show that it is an operation preserving bijection from G to G. There are four things to show:
    1. fg maps G to G.
    2. fg is one to one.
    3. fg is onto
    4. fg preserves the group operation on G.

    Start with number 1, and so on through number 4. You have a lot of facts you can use.
    1. f maps G to G.
    2. f is one to one.
    3. f is onto.
    4. f preserves the group operation on G.
    5. g maps G to G.
    6. g is one to one.
    7. g is onto.
    8. g preserves the group operation on G.

    Once you have done this, you are not finished. You will have only shown closure, but the rest is similar.
     
  10. Nov 13, 2011 #9
    Okay, that helps. Thanks :)
     
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