Zero is Identity in Naturally Ordered Semigroup

Theorem

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

Let $0$ be the zero of $\struct {S, \circ, \preceq}$.

Then $0$ is the identity for $\circ$.

That is:

$\forall n \in S: n \circ 0 = n = 0 \circ n$

Proof

By definition of an ordering:

$0 \preceq 0$

Thus from Naturally Ordered Semigroup Axiom $\text {NO} 3$: Existence of Product:

$\exists p \in S: 0 \circ p = 0$

By the definition of zero:

$0 \preceq 0 \circ 0$ and $0 \preceq p$

Thus since $\preceq$ is compatible with $\circ$:

$0 \circ 0 \preceq 0 \circ p = 0$

Thus:

$0 \circ 0 \preceq 0$ and $0 \preceq 0 \circ 0$

Hence, as $\preceq$ is antisymmetric, it follows that:

$0 \circ 0 = 0$

Because $\struct {S, \circ, \preceq}$ is a semigroup, $\circ$ is associative.

So:

$\forall n \in S: \paren {n \circ 0} \circ 0 = n \circ \paren {0 \circ 0} = n \circ 0$

Thus from Naturally Ordered Semigroup Axiom $\text {NO} 2$: Cancellability:

$\forall n \in S: n \circ 0 = n$

Similarly:

$\forall n \in S: 0 \circ \paren {0 \circ n} = \paren {0 \circ 0} \circ n = 0 \circ n$

meaning:

$\forall n \in S: 0 \circ n = n$

Thus:

$\forall n \in S: n \circ 0 = n = 0 \circ n$

and so $0$ is the identity for $\circ$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.1$