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Math Help - equivalence relation g~h for all g,h => group trivial

  1. #1
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    equivalence relation g~h for all g,h => group trivial

    Good morning! I have just read the following: Let G be a group and ~ be an equivalence relation in G so that g~ h \, \forall g,h \in G. Then G is the trivial group.

    I have tried to proof that but failed. Can somebody please give me a hint?
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  2. #2
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    Quote Originally Posted by Banach View Post
    Good morning! I have just read the following: Let G be a group and ~ be an equivalence relation in G so that g~ h \, \forall g,h \in G. Then G is the trivial group.

    I have tried to proof that but failed. Can somebody please give me a hint?
    that's not correct! on every set the relation, as you defined, is an equivalence relation. so i'm sure the relation that you're talking about is already defined somewhere in your book

    but you missed it. you need to find out how exactly the relation is defined, i.e. g ~ h if and only if what? for example you may define g ~ h iff gk = kh, for some k in G. then, with this

    definition, ~ becomes an equivalence relation in G and if g ~ h for all g and h in G, then G has to be trivial because g ~ 1 if and only if gk = k, for some k in G. since G is a group, we'll

    get that g = 1. so G has only one element, i.e. the identity element 1.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    that's not correct! on every set the relation, as you defined, is an equivalence relation. so i'm sure the relation that you're talking about is already defined somewhere in your book

    but you missed it. you need to find out how exactly the relation is defined, i.e. g ~ h if and only if what? for example you may define g ~ h iff gk = kh, for some k in G. then, with this

    definition, ~ becomes an equivalence relation in G and if g ~ h for all g and h in G, then G has to be trivial because g ~ 1 if and only if gk = k, for some k in G. since G is a group, we'll

    get that g = 1. so G has only one element, i.e. the identity element 1.
    Hi! Well, you are completely right. In any set A a~b iff a and b are in A defines an equivalence relation.

    It is actually strange because in the book the argument is expressed for a general equivalence relation in any group and then used for a special one.

    This is a little confusing.

    Any way, thanks for your answer.
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