# 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. 1.

$(G,\cdot)$ is an associative quasigroup.

2. 2.

$(G,\cdot)$ is an associative loop.

3. 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 (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.

Title an associative quasigroup is a group AnAssociativeQuasigroupIsAGroup 2013-03-22 18:28:50 2013-03-22 18:28:50 CWoo (3771) CWoo (3771) 7 CWoo (3771) Derivation msc 20N05 Group