# Condition for Cosets of Subgroup of Monoid to be Partition

## Theorem

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

Let $\struct {H, \circ}$ be a subgroup of $\struct {S, \circ}$.

Let the identity element of $\struct {H, \circ}$ be $e$.

Then:

- 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 $S$.

## Proof

Because of the fact that it is not necessarily the case that $x \in S$ has an inverse, we cannot invoke the result Left Congruence Modulo Subgroup is Equivalence Relation.

Instead we prove partitionhood directly.

First we recall the definition of left coset and right coset:

- The
**left coset of $H$ by $x$**, is:

- $x \circ H = \set {y \in S: \exists h \in H: y = x \circ h}$

- The
**right coset of $H$ by $x$**, is:

- $H \circ x = \set {y \in S: \exists h \in H: y = h \circ x}$

Next we recall the definition of partition:

- A
**partition**of $S$ is a set of non-empty subsets $\Bbb S$ of $S$ such that each element of $S$ lies in exactly one element of $\Bbb S$.

Testing each of the criteria for a partition as follows:

### No coset is empty, and each element in at least one coset

\(\ds \forall k \in S: \, \) | \(\ds k \circ e\) | \(=\) | \(\ds k\) | Monoid Axiom $\text S 2$: Identity | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \forall k \circ H \subseteq S: \, \) | \(\ds k\) | \(\in\) | \(\ds k \circ H\) | Definition of Left Coset: as $e \in H$ |

Thus:

- no left coset of $\struct {H, \circ}$ in $\struct {S, \circ}$ is empty.
- every element of $S$ is in at least one left coset of $\struct {H, \circ}$.

Also:

\(\ds \forall k \in S: \, \) | \(\ds e \circ k\) | \(=\) | \(\ds k\) | Monoid Axiom $\text S 2$: Identity | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \forall H \circ k \subseteq S: \, \) | \(\ds k\) | \(\in\) | \(\ds H \circ k\) | Definition of Right Coset: as $e \in H$ |

Thus:

- no right coset of $\struct {H, \circ}$ in $\struct {S, \circ}$ is empty.
- every element of $S$ is in at least one right coset of $\struct {H, \circ}$.

$\Box$

### Each element in no more than one coset

Let $r, s \in S$ such that:

- $r \in s \circ H$

for some $s \in S$.

It will be shown that $r \circ H$ and $s \circ H$ are the same left coset.

We have that:

- $r \in r \circ H$

from above.

By definition of left coset:

- $r \in \set {y \in S: \exists h \in H: y = s \circ h}$

That is:

- $r = s \circ h$

and so:

- $r \circ h^{-1} = s \circ \paren {h \circ h^{-1} }$

leading to:

- $r \circ h^{-1} = s$

That is:

- $s \in \set {y \in S: \exists h^{-1} \in H: y = r \circ h^{-1} }$

and so:

- $s \in r \circ H$

That is:

- $r \circ H = s \circ H$

Similarly, if $r \in H \circ s$, then $H \circ r$ and $H \circ s$ are the same right coset.

Thus:

- each element of $S$ is in no more than one left coset of $\struct {H, \circ}$ in $\struct {S, \circ}$

and

- each element of $S$ is in no more than one right coset of $\struct {H, \circ}$ in $\struct {S, \circ}$.

$\Box$

Thus:

- 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 $S$

as we were to prove.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.15$