Non-Cancellable Elements of Semigroup form Subsemigroup

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Theorem

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

Let $T \subseteq S$ be the subset of $S$ containing the elements of $S$ which are specifically not cancellable in $\struct {S, \circ}$.


Then $\struct {T, \circ}$ forms a subsemigroup of $S$.


Proof

Recall the definition of cancellable element:

An element $x \in \struct {S, \circ}$ is cancellable if and only if:

$\forall a, b \in S: x \circ a = x \circ b \implies a = b$
$\forall a, b \in S: a \circ x = b \circ x \implies a = b$


From the Subsemigroup Closure Test it is sufficient to demonstrate that:

$\forall x, y \in T: x \circ y \in T$


Let $x, y \in T$.

Then:

$\exists a, b \in S: a \circ y = b \circ y \text { but } a \ne b$

or:

$\exists a, b \in S: y \circ a = y \circ b \text { but } a \ne b$

Without loss of generality, suppose that:

$\exists a, b \in S: y \circ a = y \circ b \text { but } a \ne b$

Hence:

\(\ds y \circ a\) \(=\) \(\ds y \circ b\)
\(\ds \leadsto \ \ \) \(\ds x \circ \paren {y \circ a}\) \(=\) \(\ds x \circ \paren {y \circ b}\)
\(\ds \leadsto \ \ \) \(\ds \paren {x \circ y} \circ a\) \(=\) \(\ds \paren {x \circ y} \circ b\)

So we have:

$\paren {x \circ y} \circ a = \paren {x \circ y} \circ b$

but $a \ne b$.

Hence $x \circ y$ is not cancellable.

As $x$ and $y$ are arbitrary, the result follows.

$\blacksquare$


Sources

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