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Math Help - Union of distinct subset in a finite group G.

  1. #1
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    Union of distinct subset in a finite group G.

    If $G$ is a finite group, $G=A \cup B$ and $A \cap B =\emptyset$ then $G\neq AB$ .

    Help me proving this exersies please !

    Thank you!
    Union of distinct subset in a finite group G.-1.gif
    Last edited by john27; November 30th 2012 at 05:59 AM.
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  2. #2
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    Re: Union of distinct subset in a finite group G.

    By AB you mean x\in G s.t x=a.b where a\in A, b\in B and the operation is the one in G?
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  3. #3
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    Re: Union of distinct subset in a finite group G.

    Yes AB={a.b|a in A and b in B} , A and B are distinct subsets of group G.
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  4. #4
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    Re: Union of distinct subset in a finite group G.

    If G = AB, then every element of G should be written as x=ab for a\in G and b\in G. for a\neq b since A\intersect B = empty. Consider the identity e of G. You cannot represent e as ab.
    Is our assumption $A \cap B =\emptyset$ correct? or is $A \cap B = identity$?
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  5. #5
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    Re: Union of distinct subset in a finite group G.

    In exercises we have that $A \cap B =\emptyset$
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  6. #6
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    Re: Union of distinct subset in a finite group G.

    If G = AB, we need to prove both inequalities. AB \in G and G \in AB

    Obviously, AB is contained in G, since every x in AB is of the form ab which is also in G since G is a group.
    Is G in AB. No. Since e \in G but e is not in AB by the assumption A \intersect B = empty. Hence G \neq AB

    I hope this answers your question.
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  7. #7
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    Re: Union of distinct subset in a finite group G.

    why you say that e is not in A.B ??...
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  8. #8
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    Re: Union of distinct subset in a finite group G.

    I am so sorry. Please ignore my answer. There is a flaw in my argument.
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