Proof. We will prove this in the following direction $(1)\Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$

$(1)\Rightarrow (2)$

Let $x\in G$ and $e_1,e_2\in G$ such that $xe_1=x=e_2x$ So $xe_1^2=xe_1=x$ which shows that $e_1^2=e_1$ Let $a\in G$ be such that $e_1a=x$ Then $e_2e_1a=e_2x=x=e_1a$ so that $e_2e_1=e_1=e_1^2$ or $e_2=e_1$ Set $e=e_1$ For any $y\in G$ we have $ey=e^2y$ so $y=ey$ Similarly, $ye=ye^2$ implies $y=ye$ This shows that $e$ is an identity of $G$

$(2)\Rightarrow (3)$

First note that all of the group axioms are automatically satisfied in $G$ under $\cdot$ except the existence of an (two-sided) inverse element, which we are going to verify presently. For every $x\in G$ there are unique elements $y$ and $z$ such that $x y=z x=e$ Then $y = e y = (z x) y = z (x y)= z e = z$ This shows that $x$ has a unique two-sided inverse $x^{-1}:=y=z$ Therefore, $G$ is a group under $\cdot$

$(3)\Rightarrow (1)$

Every group is clearly a quasigroup, and the binary operation is associative.

This completes the proof.

Remark. In fact, if $\cdot$ on $G$ is flexible, then every element in $G$ has a unique inverse: for $z (x z) = (z x) z = e z = z = z e$ so by left division (by $z$ , we get $x z=e=x y$ and therefore $z=y$ again by left division (by $x$ . However, $G$ may no longer be a group, because associativity may longer hold.