Prove the following set endowed with the binary operation is an abelian group

Question

Let $∗$ be a commutative and associative binary operation on a set $S$. Assume that for every $x$ and $y$ in $S$, there exists $z$ in $S$ such that $x ∗ z = y$. Show that this set with the operation $∗$ is an abelian group. The original question is problem A-2 in the 2012 Putnam Competition which asks to prove that if $a,b,c$ are in $S$ and $a ∗ c = b ∗ c$, then $a = b$ but I'm interested in why it's an abelian group. I'd like to see your colorful creative answers, please.

Answer 1

We have to show that

1. there exists the identity element $1_S$
2. for each $x\in S$ there exists the inverse $x^{-1}$

For each $x,y\in S$ let $e_{x,y}\in S$ be such that $x*e_{x,y} = y$, and let $e_x = e_{x,x}$. For $x\in S$ consider the nonempty set $$ E_x := \{e_x\in S: ~x*e_x = x\} $$ of inverses of $x$ (which are both right and left inverses, in that $*$ is abelian).

The intuition is that $e_x\in E$ should be the identity $1_S$, but to prove this we have to show that $E_x = E_y$ for every $y\in S$.

Using associativity and commutativity of $*$, $$ y*e_x = (x*e_{x,y})*e_x = (x*e_x) * e_{x,y} = x * e_{x,y} = y $$ which proves that $E_x \subseteq E_y$. Similarly we can show that the converse inclusion holds and therefore that $E_x=E_y$ for all $x,y$ as expected. This proves the existence of $1_S$.

As to the inverse, trivially, for $x,1_S\in S$ by hypothesis there exists $z$ s.t. $x*z = 1_S$, and hence such $z$ is $x^{-1}$.