Identity of Cancellable Monoid is Identity of Submonoid

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {S, \circ}$ be a monoid, all of whose elements are cancellable.

Let $\struct {T, \circ}$ be a submonoid of $\struct {S, \circ}$.


Then the identity of $\struct {T, \circ}$ is the same element as the identity of $\struct {S, \circ}$.


Proof

By Identity of Monoid is Unique:

there is only one identity element of $\struct {S, \circ}$

and:

there is only one identity element of $\struct {T, \circ}$.


Let $e_S$ be the identity of $\struct {S, \circ}$, and $e_T$ the identity of $\struct {T, \circ}$.

From Identity is only Idempotent Cancellable Element, $e_S$ is the only cancellable element of $\struct {S, \circ}$ which is idempotent.

But all elements of $\struct {S, \circ}$ are cancellable.

Thus $e_S$ is the only idempotent element of $\struct {S, \circ}$.


Again, all elements of $\struct {S, \circ}$ are cancellable.

Thus from Cancellable Element is Cancellable in Subset, all elements of $\struct {T, \circ}$ are cancellable.

Thus, $e_T$ is the only element of $\struct {T, \circ}$ which is idempotent.

Thus, as $e_T \in S$, we have $e_S \circ e_T = e_T = e_T \circ e_T$ and thus $e_S = e_T$.

$\blacksquare$


Also see