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Math Help - automorphism of a group

  1. #1
    MHF Contributor Siron's Avatar
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    automorphism of a group

    Hi,

    Can someone help me with proving this proposition:

    Let G a group. For g \in G we define:
    \lambda_g: G \to G: x \mapsto gxg^{-1}

    Proof \lambda_g is an automorphism of G (We call it an inside automorphism of G determined by g).

    If I wan't to proof \lambda_g is an automorphism I've to demonstrate that \lambda_g is an isomorphism or an bijective homomorphism.

    1) \lambda_g is an homorphism
    But to prove if \lambda_g is an homorphism in my inion I need a composite law for the group G.

    I guess I've to proof for two arbitrary chosen numbers x,y in G that:
    f(x\perp y)=f(x)\perp f(y)

    But how do I now: \perp? Or doesn't that matter? ...

    2) \lambda_g is bijective:
    I've to prove that:
    f(x)=f(y) \Rightarrow x=y
    In this case:
    gxg^{-1}=gyg^{-1} \Rightarrow x=y
    This is true because in a group a g^{-1} is the symmetric element of g and so: gg^{-1}=1

    So the only thing I haven't proved yet is the homorphism. Does someone have a hint?

    Thanks in advance! .
    (Sorry If my English wouldn't be correct, but English mathematical terminology is not easy for me as a Belgian).
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  2. #2
    Super Member girdav's Avatar
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    Re: automorphism of a group

    You have to prove that \lambda_g(x\star y)=\lambda_g(x)\star\lambda_g( y): it's a consequence of associativity and the fact that g^{-1}\star g=e.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: automorphism of a group

    So that means I've to prove that:
    gxyg^{-1}=(gxg^{-1})\cdot(gyg^{-1})
    \Leftrightarrow xy=xy

    So I've proved that \lambda_g is an automorphism and I'm done.
    Am I right? ...
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  4. #4
    Super Member girdav's Avatar
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    Re: automorphism of a group

    You are almost done, since you have not proved yet that \lambda_g is surjective.
    Last edited by girdav; July 18th 2011 at 01:56 PM.
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: automorphism of a group

    Oh, indeed because: surjection + injection = bijection. If I've to prove that \lambda_g is surjective that means each argument x \in G has an image f(x) \in G, I think that's right because: x \mapsto gxg^{-1} so x \mapsto x (because g\cdot g^{-1}=1).

    Am I correct? ...
    I appreciate your help.
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  6. #6
    Member ModusPonens's Avatar
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    Re: automorphism of a group

    No, that's not correct. you have to find an element of the group for wich f(x)=y, for every y. If the group was abelian you could commute g and g^-1 but it's not necessarily abelian (in fact the exercise is only fun when the group may not be abelian).
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Re: automorphism of a group

    Quote Originally Posted by ModusPonens View Post
    No, that's not correct. you have to find an element of the group for wich f(x)=y, for every y. If the group was abelian you could commute g and g^-1 but it's not necessarily abelian (in fact the exercise is only fun when the group may not be abelian).
    So what you need to do, Siron, is find some element, g \in G, such that g^{-1}hg=h for all h \in G. Now, if g^{-1}hg=h then gh=hg so g\in Z(G). Can you find such an element in any given group? What is this element?...
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  8. #8
    Senior Member Tinyboss's Avatar
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    Re: automorphism of a group

    Incidentally, it's good to get into the habit of thinking of an isomorphism as a homomorphism with homomorphic inverse, rather than a bijective homomorphism. In the category of groups, the latter is a consequence of the former, and either may be taken as the definition, but in many categories it's not true. For example, in the category of topological spaces, the equivalent notions to homomorphism and isomorphism are continuous map and homeomorphism--and a bijective continuous map need not be a homeomorphism. The correct definition is a continuous map with continuous inverse.
    Last edited by Tinyboss; July 19th 2011 at 11:56 AM.
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  9. #9
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    Re: automorphism of a group

    You have already shown that \lambda_g is indeed a group homomorphism. Now we just have to show that it is both injective (one-to-one) and surjective (onto).

    It suffices to show that \lambda_g has an inverse. When you look further into this then you will notice that \lambda_{g^{-1}} is indeed the inverse of \lambda_g. You can see this by noting that if \lambda_{g^{-1}}(x) = g^{-1}x (g^{-1})^{-1} = g^{-1}xg then it follows that \lambda_g(\lambda_{g^{-1}}(x)) = \lambda_{g^{-1}}(\lambda_g(x)) = x for any element x\in G (because g^{-1}g = gg^{-1} = e for any element g\in G).

    Does this make sense?
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  10. #10
    MHF Contributor Swlabr's Avatar
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    Re: automorphism of a group

    Quote Originally Posted by obd2 View Post
    You have already shown that \lambda_g is indeed a group homomorphism. Now we just have to show that it is both injective (one-to-one) and surjective (onto).

    It suffices to show that \lambda_g has an inverse. When you look further into this then you will notice that \lambda_{g^{-1}} is indeed the inverse of \lambda_g. You can see this by noting that if \lambda_{g^{-1}}(x) = g^{-1}x (g^{-1})^{-1} = g^{-1}xg then it follows that \lambda_g(\lambda_{g^{-1}}(x)) = \lambda_{g^{-1}}(\lambda_g(x)) = x for any element x\in G (because g^{-1}g = gg^{-1} = e for any element g\in G).

    Does this make sense?
    This isn't entirely true - you have shown that every element in the image has an inverse. That is to say, you have shown that the function is 1-1. You still need to show that it is surjective!

    (For example, if we take the function \phi: \mathbb{N}\rightarrow \mathbb{N}, a\mapsto a^2, then if we define \varphi: \mathbb{N}\rightarrow \mathbb{N}, a\mapsto \sqrt{a} then \phi^{-1}=\varphi, but clearly \phi isn't a bijection...(what I have just written abuses notation horrendously, but my point is that my abuse is the same as your abuse...))
    Last edited by Swlabr; July 20th 2011 at 05:47 AM.
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  11. #11
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    Re: automorphism of a group

    Quote Originally Posted by Swlabr View Post
    This isn't entirely true - you have shown that every element in the image has an inverse. That is to say, you have shown that the function is 1-1. You still need to show that it is surjective!

    (For example, if we take the function \phi: \mathbb{N}\rightarrow \mathbb{N}, a\mapsto a^2, then if we define \varphi: \mathbb{N}\rightarrow \mathbb{N}, a\mapsto \sqrt{a} then \phi^{-1}=\varphi, but clearly \phi isn't a bijection...(what I have just written abuses notation horrendously, but my point is that my abuse is the same as your abuse...))
    Well... I think I might have shown that it is surjective. Given an element x\in G then I have shown that the element \lambda_{g^{-1}}(x) maps to x with \lambda_g. That is to say set y = \lambda_{g^{-1}}(x) then \lambda_g(y) = x.
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  12. #12
    MHF Contributor Swlabr's Avatar
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    Re: automorphism of a group

    Quote Originally Posted by obd2 View Post
    Well... I think I might have shown that it is surjective. Given an element x\in G then I have shown that the element \lambda_{g^{-1}}(x) maps to x with \lambda_g. That is to say set y = \lambda_{g^{-1}}(x) then \lambda_g(y) = x.
    Ah yes, sorry, you're right.
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  13. #13
    MHF Contributor Swlabr's Avatar
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    Re: automorphism of a group

    Quote Originally Posted by Swlabr View Post
    Ah yes, sorry, you're right.
    Actually...doesn't this only work in a finite group? You are saying that both \lambda_g and \lambda_g^{-1} are injections, but that doesn't really mean that they are bijections, unless your group is infinite...
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  14. #14
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    Re: automorphism of a group

    fog = id say that g is an injection and f a surjection.

    So showing fog = id = gof show that f and g are both bijections (and that f = g^{-1})
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