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an associative quasigroup is a group
Proposition 1.
Let $G$ be a set and $\cdot$ a binary operation on $G$. Write $ab$ for $a\cdot b$. The following are equivalent:
1. $(G,\cdot)$ is an associative quasigroup.
2. $(G,\cdot)$ is an associative loop.
3. $(G,\cdot)$ is a group.
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 $xe_{1}=x=e_{2}x$. So $xe_{1}^{2}=xe_{1}=x$, which shows that $e_{1}^{2}=e_{1}$. Let $a\in G$ be such that $e_{1}a=x$. Then $e_{2}e_{1}a=e_{2}x=x=e_{1}a$, so that $e_{2}e_{1}=e_{1}=e_{1}^{2}$, or $e_{2}=e_{1}$. Set $e=e_{1}$. For any $y\in G$, we have $ey=e^{2}y$, 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 (twosided) inverse element, which we are going to verify presently. For every $x\in G$, there are unique elements $y$ and $z$ such that $xy=zx=e$. Then $y=ey=(zx)y=z(xy)=ze=z$. This shows that $x$ has a unique twosided 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 $z(xz)=(zx)z=ez=z=ze$, so by left division (by $z$), we get $xz=e=xy$, and therefore $z=y$, again by left division (by $x$). However, $G$ may no longer be a group, because associativity may longer hold.
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