# Set of All Relations is a Monoid

## Theorem

The set of all relations $\Bbb E = \left\{{\mathcal R: \mathcal R \subseteq S \times S}\right\}$ on a set $S$ forms a monoid in which the operation is composition of relations.

The only invertible elements of $\Bbb E$ are permutations.

If $\mathcal R$ is invertible, its inverse is $\mathcal R^{-1}$.

## Proof

### Closure

A relation followed by another relation is another relation, from the definition of composition of relations.

As the domain and codomain of two relations are the same, then:

- $\forall \mathcal R_1, \mathcal R_2 \subseteq S \times S: \mathcal R_1 \circ \mathcal R_2 \subseteq S \times S$

Therefore composition of relations on a set is closed.

$\Box$

### Associativity

Composition of Relations is Associative.

$\Box$

### Identity

From Identity Mapping is Left Identity and Identity Mapping is Right Identity we have that:

- $\forall \mathcal R \subseteq S \times S: \mathcal R \circ I_S = \mathcal R = I_S \circ \mathcal R$

Hence the identity mapping is the identity element of this set of relations.

$\Box$

It is closed and associative, so it is a semigroup. It has an identity element, so it is a monoid.

### Inverses

Now if $\mathcal R \in \Bbb E$ were to be invertible, it would need to satisfy:

- $\mathcal R^{-1} \circ \mathcal R = I_S$ and
- $\mathcal R \circ \mathcal R^{-1} = I_S$.

From Inverse Relation is Left and Right Inverse iff Bijection this can only happen if $\mathcal R$ is a bijection on $S$, that is, a permutation, and its inverse is then $\mathcal R^{-1}$.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $5.8 \ \text{(g)}$