# Nonempty associative quasigroup equals group

## History

The fact that a (nonempty) associative quasigroup is the same thing as a group is more than a mere curiosity. Many of the initial definitions of group, including Cayley's first attempted definition in 1854 and Weber's definition in 1882, defined a group as an associative quasigroup, without explicitly mentioning identity elements and inverses.

Further information: History of groups

## Statement

Let $(S,*)$ be a nonempty magma (set with binary operation) satisfying the following two conditions:

1. $S$ is a quasigroup under $*$, i.e., for any $a,b \in S$, there exist unique $x,y \in S$ such that $a * x = y * a = b$
2. $*$ is associative, or $S$ is a semigroup under $*$, i.e., for any $a,b,c \in S$, we have $a * (b * c) = (a * b) * c$

Then $S$ is a group under $*$.

Conversely, any group is an associative quasigroup. Associativity is part of the definition, and the quasigroup part is also direct. For full proof, refer: Group implies quasigroup

## Proof

Associativity is given to us, so we need to find an identity element (neutral element for the multiplication) and inverses.

### Finding the identity element (or neutral element)

Since $S$ is nonempty, we can pick $a \in S$. By the quasigroup condition, find $e \in S$ such that $a * e = a$.

Now, pick any $b \in S$. By the quasigroup condition again, there exists a unique element $y \in S$ such that $y * a = b$. Plugging in, we get:

$b * e = (y * a) * e = y * (a * e) = y * a = b$

Thus, $e$ is a right neutral element (or right identity) for the multiplication $*$.

By a similar procedure, we can find a left neutral element. Further, since any left and right neutral element are equal, we see that $e$ must be a two-sided identity element for $(S,*)$.

### Finding inverses

If $e$ is the identity element, then for any $a \in S$, we can find $x,y \in S$ such that $a * x = y * a = e$. Thus, $a$ has both a left inverse and a right inverse. By associativity, any left and right inverse must be equal, so $a$ has a two-sided inverse. Thus, every element has a two-sided inverse, so every element is invertible, and $S$ is thus a group.