# Condition on Congruence Relations for Cancellable Monoid to be Group

## Theorem

Let $\struct {S, \circ}$ be a cancellable monoid whose identity element is $e$.

Then:

$\struct {S, \circ}$ is a group
every non-trivial congruence relation on $\struct {S, \circ}$ is induced by a normal subgroup of $\struct {S, \circ}$.

### Counterexample

Let $\struct {S, \circ}$ be a monoid which is not cancellable.

Let every non-trivial congruence relation on $\struct {S, \circ}$ be induced by a normal subgroup of $\struct {S, \circ}$.

Then it is not necessarily the case that $\struct {S, \circ}$ is a group.

## Proof

### Necessary Condition

Let $\struct {S, \circ}$ be such that every non-trivial congruence relation on $\struct {S, \circ}$ is induced by a normal subgroup of $\struct {S, \circ}$.

Hence, let $\RR$ be an arbitrary non-trivial congruence relation.

From Condition for Subgroup of Monoid to be Normal, there exists a normal subgroup $\struct {H, \circ}$ such that:

the set of left cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$

and:

the set of right cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$

such that the equivalence relations induced by those partitions is $\RR$.

By the definition of normal subgroup, the set of left cosets is the same as the set of right cosets.

It remains to be shown that $\struct {S, \circ}$ is a group.

Group Axiom $\text G 0$: Closure
Group Axiom $\text G 1$: Associativity
Group Axiom $\text G 2$: Existence of Identity Element

are a priori satisfied by the fact that $\struct {S, \circ}$ is a monoid.

Hence it remains to prove Group Axiom $\text G 3$: Existence of Inverse Element.

### Sufficient Condition

Let $\struct {S, \circ}$ be a group.