The axiom quotient and any property of a group. 
Def. A group $G$ is a set not empty in which is definited a binary operation $G\times G\to G$ denoted with $(a,b)\mapsto ab$ such that $$(i)\quad(ab)c=a(bc)\;\text{for all}\; a,b\in G;$$ $$(ii)\quad\text{exist}\; u\in G\;\text{such that}\; ua=a=au\;\text{for all}\;a\in G;$$ $$(iii)\quad a\in G, \exists x\in G\;\text{such that}\; ax=u=xa.$$

Quotient axiom Let $G$ a not empty set with a binary operation. Let $a,b\in G$ then exist $x,y\in G$ such that $ax=b$ and $ya=b$.

Exercise Suppose true $(i)$ and the quotient axiom, we must prove that $(ii)$ is true.
My attempt. Let $a,b\in G$, then for hypothesis exists $x,y\in G$ such that $ax=b$ and $ya=b$, then we have $$bx=(ya)x=y(ax)=yb$$

Question How can I continue at this point to obtain the thesis?
Thanks!
 A: Let $a\in G$ (which exists because $G$ is not empty). We know there exists $u\in G$ such that $au=a$. I claim that for all $b\in G$, $bu=b$.
Indeed, let $b\in G$. Then there exists $y$ such that $b=ya$. Then $bu=(ya)u$. 
Can you take it from here?
Then, to show $u$ works on both sides, note that there exists $v\in G$ such that $va = a$. Show (along similar lines) that $vb=b$ for all $b\in G$. Then look at $vu$. 
A: Note that the Quotient Axiom is a "for all" statement, which means you can apply it to anything you want in place of $a$ and $b$, not just to $a$ and $b$.
What is your goal here? To prove property (ii), you need to show that there exists some $u\in G$ such that for any $a\in G$ we have $ua=a=au$.
A strategy can be as follows: (1) For any $a\in G$, with the help of the Quotient Axiom, we find its own $u_a\in G$ satisfying $u_aa=a=au_a$; (2) Then for any two $a,b\in G$, we show that $u_a=u_b$, which will automatically imply that they all are the same, and so that's the desired $u$.
Skipping some of the details, here's how we can proceed.
(1) Consider just one element $a\in G$. By the Quotient Axiom applied to $a$ and $a$, we have some elements $x,y$ such that $ax=a$ and $ya=a$. Using associativity, show that $x=y$, and so this is your $u_a$.
(2) Now consider any two elements $a,b\in G$. From the first part, we also have their corresponding $u_a$ and $u_b$. Applying the Quotient Axiom to $a$ and $u_b$, we have some $x\in G$ such that $ax=u_b$. Applying the Quotient Axiom to $b$ and $u_a$, we have some $y\in G$ such that $yb=u_a$. Now you can show that $u_a=\cdots=u_au_b$ and that $u_b=\cdots=u_au_b$, which implies they are equal to each other.
