Left Cancellable Elements of Semigroup form Subsemigroup

Theorem

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

Let $C_\lambda$ be the set of left cancellable elements of $\struct {S, \circ}$.

Then $\struct {C_\lambda, \circ}$ is a subsemigroup of $\struct {S, \circ}$.

Proof

Let $C_\lambda$ be the set of left cancellable elements of $\struct {S, \circ}$:

$C_\lambda = \set {x \in S: \forall a, b \in S: x \circ a = x \circ b \implies a = b}$

Let $x, y \in C_\lambda$.

Then:

\(\ds \paren {x \circ y} \circ a\)

\(=\)

\(\ds \paren {x \circ y} \circ b\)

\(\ds \leadsto \ \ \)

\(\ds x \circ \paren {y \circ a}\)

\(=\)

\(\ds x \circ \paren {y \circ b}\)

by associativity of $\circ$

\(\ds \leadsto \ \ \)

\(\ds x \circ a\)

\(=\)

\(\ds x \circ b\)

as $y \in C_\lambda$

\(\ds \leadsto \ \ \)

\(\ds a\)

\(=\)

\(\ds b\)

as $x \in C_\lambda$

\(\ds \leadsto \ \ \)

\(\ds x \circ y\)

\(\in\)

\(\ds C_\lambda\)

Thus $\struct {C_\lambda, \circ}$ is closed.

Therefore by the Subsemigroup Closure Test $\struct {C_\lambda, \circ}$ is a subsemigroup of $\struct {S, \circ}$.

$\blacksquare$