Finding the Right Inverses and Right Identity of a Group 
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? 
 A: 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. 
