# Group implies quasigroup

## Statement

### Verbal statement

Any group is a quasigroup.

### Statement with symbols

Let $G$ be a group, and $a,b \in G$ be (not necessarily distinct) elements. Then, there exist unique $x,y \in G$ satisfying $ax = b$ and $ya = b$ respectively.

## Definitions used

### Quasigroup

Further information: Quasigroup

A magma $(S,*)$ (a set $S$ with binary operation $*$) is termed a quasigroup if for any $a,b \in S$, there exist unique $x,y \in S$ such that $a * x = y * a = b$.

## Proof

Given: A group $G$, elements $a,b \in G$

To prove: There exist unique solutions to $ax = b$ and $ya = b$

Proof: We have:

$ax = b \implies a^{-1}(ax) = a^{-1}b \implies x = a^{-1}b$

Conversely:

$x = a^{-1}b \implies ax = a(a^{-1}b) \implies ax = b$

Thus:

$ax = b \iff x = a^{-1}b$

So, $ax = b$ has a unique solution.

Similarly:

$ya = b \implies (ya)a^{-1} = ba^{-1} \implies y = ba^{-1}$

Conversely:

$y = ba^{-1}\implies ya = (ba^{-1})a \implies ya = b$

Thus:

$ya = b \iff y = ba^{-1}$

So, $ya = b$ as a unique solution.