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Math Help - Abstract Algebra: Groups

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    is up to his old tricks again! Jhevon's Avatar
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    Abstract Algebra: Groups

    honorable mathematicians,

    I was asked to verify that the set H = \{ f : f: \mathbb{R} \mapsto \mathbb{R} \mbox{ and } f(x) \ne 0 \mbox{ for all } x \in \mathbb{R} \} is a group with respect to multiplication. i did that, no problem. but then the question asks, "How does this group differ from the group of invertible (real) mappings?"

    i have no idea what this question wants from me...

    Thanks
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    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    honorable mathematicians,

    I was asked to verify that the set H = \{ f : f: \mathbb{R} \mapsto \mathbb{R} \mbox{ and } f(x) \ne 0 \mbox{ for all } x \in \mathbb{R} \} is a group with respect to multiplication. i did that, no problem. but then the question asks, "How does this group differ from the group of invertible (real) mappings?"

    i have no idea what this question wants from me...

    Thanks

    Both the set and the group operation are different.

    The first group includes things like f(x)=sin(x)+2, which is not invertible.
    While the second includes f(x)=x, which is not in the set of the first group.

    Also the group operation for the group of invertible real mappings is (usualy
    though you have not said so) function composition, that is (f o g)(x)=f(g(x)).

    RonL
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    MHF Contributor kalagota's Avatar
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    A question: does the "invertible" there the same as the existence of the inverse of f?
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    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    Both the set and the group operation are different.

    The first group includes things like f(x)=sin(x)+2, which is not invertible.
    While the second includes f(x)=x, which is not in the set of the first group.

    Also the group operation for the group of invertible real mappings is (usualy
    though you have not said so) function composition, that is (f o g)(x)=f(g(x)).

    RonL

    Oh yes nearly forgot, the first is Abelian (commutative) and the second is not.

    (and so cannot be isomorphic)

    RonL
    Last edited by CaptainBlack; February 11th 2008 at 07:20 AM.
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    Quote Originally Posted by CaptainBlank View Post
    Abelian
    For some strange reason nobody ever capitalizes and instead writes abelian. All other algebraic and not even algebraic concepts are always capitalized: Noetherian, Artinian, Lagrangian mechanics, Newtonian fluid, .... I assume it is an error that has remained through the centuries.
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    Quote Originally Posted by ThePerfectHacker View Post
    For some strange reason nobody ever capitalizes and instead writes abelian. All other algebraic and not even algebraic concepts are always capitalized: Noetherian, Artinian, Lagrangian mechanics, Newtonian fluid, .... I assume it is an error that has remained through the centuries.
    I must say I did not think twice about it, it's named after a person and therefore should be capitalised.

    RonL
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kalagota View Post
    A question: does the "invertible" there the same as the existence of the inverse of f?
    yes, i believe so. we defined the inverse of an element a to be an element b such that:

    a*b = b*a = identity element

    we call b a^{-1}

    a similar notion is applied to mappings i suppose...
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