# 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.

We already have that

- 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.

This theorem requires a proof.In particular: The suggestion is to use Condition for Partition between Invertible and Non-Invertible Elements to induce Congruence Relation on MonoidYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{ProofWanted}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

### Sufficient Condition

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

This theorem requires a proof.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{ProofWanted}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.18 \ \text {(a)}$