an associative quasigroup is a group

Proposition 1.

Let $G$ be a set and $\cdot$ a binary operation on $G$ . Write $a b$ 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 (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
Canonical name	AnAssociativeQuasigroupIsAGroup
Date of creation	2013-03-22 18:28:50
Last modified on	2013-03-22 18:28:50
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	7
Author	CWoo (3771)
Entry type	Derivation
Classification	msc 20N05
Related topic	Group