How did Kazuo Matsuzaka come to think of this? (Group Theory) I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.  
There is the following problem about group theory in this book:  
Let $G$ be a non-empty set.
"$/$" is a binary operation on $G$.
"$/$" satisfies the following 4 properties:  


*

*$a/a = b/b.$

*$a/(b/b)=a.$

*$(a/a)/(b/c)=c/b.$

*$(a/c)/(b/c)=a/b.$
We define $a*b :=a/((b/b)/b)$.  
Show that $(G, *)$ is a group.
I think this fact is very complicated.
How did Matsuzaka come to think of this?
How did Matsuzaka choose this 4 properties?
 A: These are common properties of fractions.


*

*$1 = 1$, 

*$a/1 = a$, 

*$1/(b/c) = c/b$, if $b \neq 0, c \neq 0$, and

*$\frac{a/c}{b/c} = a/b$, if $b \neq 0, c \neq 0$.


This is not all the properties of fractions of integers.  So this describes an algebraic system with division of nonzero quantities only guaranteed to have the listed properties.  If you think of another property of fractions that isn't a consequence of the above, then the algebraic system described above need not exhibit that property.
A: Let me try to guess what can have come into their mind. 
First of all, as mentioned in the comments, these were found in order to mimick the properties of  $(a,b) \mapsto ab^{-1}$ in a group.


*

*To get a group from the single operation $/$, we need to recover the neutral element. Bit $e= aa^{-1}=a/a$ for any $a$. So to get $e$ we need to have $a/a=b/b$. 

*If we call $e$ the common value of $b/b$, we need it to be a neutral element : $a/e = ae^{-1}= ae= a$. Hence $a/(b/b) = a$.

*We need $e$ to also be neutral on the left : we need to say how $e/(b/c)$ relates to $b/c$, but since there is an inversion in the process, we need to put $c/b$. 

*At this point you have recovered the neutral element, and part of the inversion. Still you have no law concerning general $a,b,c$ that can account for associativity.  
This is where 4 comes in. It relates a general $a,b,c$ and in terms of $ab^{-1}$ is : $(ac^{-1})(bc^{-1})^{-1} = ab^{-1}$. This means (thanks to 3. and to what it tells us about $(fg)^{-1}$ ) $(ac^{-1})(cb^{-1}) = ab^{-1}$. So this is a version of associativity in terms of $/$. 
Now we have recovered the neutral element, the inversion, and associativity : we've got a group. 
Note that analyzing where these laws come from tells you a great deal about how to prove that $(G, *)$ is a group, because you now have a clear route.
