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Thread: Abstract algebra

  1. #1
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    Abstract algebra

    Hello all,

    I have the following problem:

    Suppose a is in G.
    Let aG = {ag : g is in G}.
    Show that aG=G.

    It does not say anything about G, but I am assuming that G is a group and thus we have closure, associativity, an identity element and inverses for every element in G.

    How do I start?

    Thanks.
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  2. #2
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    Re: Abstract algebra

    Quote Originally Posted by surjective View Post
    Suppose a is in G.
    Let aG = {ag : g is in G}.
    Show that aG=G.

    It does not say anything about G, but I am assuming that G is a group and thus we have closure, associativity, an identity element and inverses for every element in G.
    Assuming that $G$ is a group, you must show $aG\subset G~\&~G\subset aG$
    1) If $x\in aG$ can you show that $x\in G~?$

    2) If $y\in G$ can you show that $y\in aG~?$
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  3. #3
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    Re: Abstract algebra

    Thanks for the answer.

    Here is what I have done:

    Assuming that $\displaystyle x \in aG$ means that x can be written on the form $\displaystyle ag$ where $\displaystyle g \in G$. Since G is a group we have closure and therefore $\displaystyle ag \in G$.

    Assuming that $\displaystyle y \in G$ means that $\displaystyle ay \in G$ due to closure. This means that ay is in aG since ay are the elements of aG. I am stuck here!!

    Assistance would be appreciated.
    Last edited by surjective; Feb 10th 2018 at 07:16 AM.
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  4. #4
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    Re: Abstract algebra

    Quote Originally Posted by surjective View Post
    Thanks for the answer.
    Here is what I have done:
    Assuming that $x \in aG$ means that x can be written on the form $ag$ where $g \in G$. Since G is a group we have closure and therefore $ag \in G$.
    Do not use the [ tex] [ /tex] tags. Instead use \$ \$ signs.
    Yes you have shown that if $\in aG$ then $x\in G$

    Now if $y\in G$ is it true that $y\in aG~?$
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  5. #5
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    Re: Abstract algebra

    If $y \in G$ then $ay \in G $ due to closure. Then $ay \in aG$?
    Last edited by surjective; Feb 10th 2018 at 07:46 AM.
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  6. #6
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    Re: Abstract algebra

    Quote Originally Posted by surjective View Post
    If $y \in G$ then $ay \in G $ due to closure.
    True. But how does that show that $y\in aG~?$
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  7. #7
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    Re: Abstract algebra

    Here I will finish this for you.
    Suppose that $y\in G$. We know that $a^{-1}\in G$ so that means $a^{-1}y\in G$ WHY?
    Thus that means $(\exists g'\in G)[a^{-1}y=g']$ But that implies that $y=ag'$ HOW?
    If that shows that $y\in aG$ how have we proven that $aG=G~?$
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  8. #8
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    Re: Abstract algebra

    Yes I see.

    Assume that $y \in G$. Assume also that $a \in G$. Since G is a group, then every element in G has an inverse which is also in G. Hence $a^{-1} \in G$.
    Due to closure then $a^{-1}y \in G$. We may call this element $g'$ and thus $a^{-1}y=g' \in G$.
    Applying the left cancellation law we get $a^{-1}y=g' \Leftrightarrow aa^{-1}y=ag' \Leftrightarrow y=ag'$.
    Hence assuming that $y \in G$ then it has the form $ag'$ which are the elements of aG. Therefore $y \in aG$.

    We have thus shown that $aG \subset G$ and $G \subset aG$ and therefore $G=aG$.

    Thanks a lot.
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