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Possible Duplicate:
Right identity and Right inverse implies a group

Let $(G,*)$ be a binary structure that has the following properties:

1) The binary operation $*$ is associative.

2) There exists an element $e \in G$ such that for all $a \in G$, $e*a=a$ (Existence of left Identity).

3) For all $a \in G$, there exists $b \in G$ such that $b*a=e$ (existence of left inverses)

Prove that $(G,*)$ is a group.

This is how I did it.

I want to check that

$a*e=a$

$a*(a'*a)=a$ , from here I cant go on, because I only have left inverses as well. So I know something is missing. And what should I do to find the right inverse?

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    Have you looked at http://math.stackexchange.com/q/65239/17111 ?2012-03-20
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    Yea, but the truth is I don't understand what they are doing :(2012-03-20
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    Did you check out both solutions given in that page? The solution by Martin Sleziak is pretty short and easy to understand.2012-03-20
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    For some reasons He is using the right inverse, I cant use it because they are only telling me there is a left inverse.2012-03-20
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    @anilorap, once you understand the solution on that page, you will be able to figure out the answer to your own question.2012-03-20
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    Ill give it to my daughter(4 years old), she had never taken abstract algebra just like me. maybe she understand all those steps. Thanks2012-03-20
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    How do you define the operation $*$ in $G$? You can not show $(G,*)$ is a group without the definition of the operation $*$ in $G$2012-03-20
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    They are not, I don't understand the other explanation. I have more than 4 hours trying to figure out what is says. This is the first time I take this class, I dont have a clue. He is telling me when I understand the other explanation I will be able to prove mine. Thats not a nice response if I said that I dont get it. What is wrong about not getting an abstract explanation. :(2012-03-20
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    @HassanMuhammad thats what the question says and the operation is * I only have to prove there is a right identity using the left inverse and prove that there is a right inverse using its own property as a left inverse. But I keep running in circles.2012-03-20
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    Since the OP claimed that he/she already spent 4 hours, I was going to write up an answer. Then I found the answer by google for left identity, left inverse and group. Also, Page 7 of Lang's Algebra provides a complete answer. Have fun reading it!(I didn't want to take the fun away from the OP. Just like the OP would not take a toy away from his daughter)2012-03-20
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    I wrote up a solution with the left conditions instead of the right conditions.2012-03-20
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    @scaaahu.. mm good one. :)2012-03-20
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    anilorap, if I understand correctly, you had already seen the question about the right inverse when you posted yours? If so, it would have been much preferable to post a specific question with a link to the other question about what part of the answers there you have trouble understanding or transferring to the case of the left inverse, rather than to post a question that's very close to being an exact duplicate of the other one. (Two people have voted to close it as a duplicate.)2012-03-20

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Since it appears that you are very confused, I will lend you a hand. Take $a$ in your group $G$. Then by the conditions of the problem you know that it has a left inverse $b$. This means that you know that $$ba=e$$Where $e$ is the left identity. First we will show that $b$ is such that $ab=e$. This is how it goes: Let $y=ab$. Then $$yy=(ab)(ab)=a(ba)b=a(e)b=a(eb)=a(b)=ab=y$$Now since we know that we have left inverses, let us say that $q$ is the left inverse of $y$, so that $qy=e$. Multiply this at both sides and obtain: $$qyy=qy$$$$\Rightarrow ey=e$$$$\Rightarrow y=e$$$$\Rightarrow ab=e$$Hence by definition $b$ is then a $\textbf{right}$ inverse of $a$ also. Now by convention since you have seen that the right and left inverse of $a$ agree, people tend to call them $a^{-1}$ (I have not proved that it is unique, but you do not need to know this as of the moment).

Now let us prove that $e$ works as a right identity. $$ae=a(a^{-1}a)=(aa^{-1})a=ea=a$$ Above we use the fact that $a^{-1}a=e$, which we proved above, and the fact that $ea=a$ which is given as a condition of the problem.

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    Thank you very much. I got stuck in the first part i had c*c=c.. after taht I didnt know what to do... Thank you very much :)2012-03-20