an associative quasigroup is a group

Proposition 1.

Let $G$ be a set and $\mathrm{\cdot }$ a binary operation on $G$. Write $a\mathit{}b$ for $a\mathrm{\cdot }b$. 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 $x{e}_1=x=e_{2}x$. So $x{e}_1^2=x{e}_1=x$, 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, $ye=y{e}^2$ 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 $xy=zx=e$. Then $y=ey=(zx)y=z(xy)=ze=z$. 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 $z(xz)=(zx)z=ez=z=ze$, 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.