Associativity of operation of finite set I'm studying Basic Algebra 1 by nathan jacobson at home.

I found that (i) implies for every element in G there exist right inverse and left inverse. I think I should show is associativity of multiplication by using (ii). I tried to find solution at google. But I couldnt find. Can you give me some hints or soultion? And where can I get solution for the book? Thank you.

I think it should suffice to say that in the rectangle we have $ac = 1, bc = x, ad = y$. Now for the fourth vertex to only depend on the pair $(x,y)$ we must have some $f(x,y)$ so that $\forall (x,y), a,b,c,d$ where $bc = x, ad = y$ we have that $bd = f(x,y)$. So to define $f$ just select some $a,b,c,d$ satisfying the above condition, now $(ii)$ implies that $f$ is independent of our choice. In which case take $1x = x, y1 = y$ so $(bc)y = xy = bd$ Now note that $d = a^{-1} y = cy$ so that $bd = b(cy)$ and we have that $(bc)y = c(by)$. This holds true for all $bc = x$. So for any pair of elements $b,c$, looking at $f(bc,y)$ shows that $bc$ associates with all other elements $y$. Since we can do that for all elements the monoid is associative.
This is probably a bit more complicated than it needs to be? The other direction can be gotten in the same exact way, association implies that we can always select $f(x,y)$ without having to make a choice by taking $a',b',c',d'$ and showing that association implies $bd = b'd'$.
