Subsemigroup of Monoid is not necessarily Monoid

From ProofWiki
Jump to navigation Jump to search


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

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

Then it is not necessarily the case that $\struct {T, \circ}$ has an identity.


Consider the set of integers under multiplication $\struct {\Z, \times}$.

From Integers under Multiplication form Monoid, $\struct {\Z, \times}$ is a monoid.

Let $n \in \Z$ such that $n > 1$.

Let $n \Z$ be the set of integer multiples of $n$:

$\set {x \in \Z: n \divides x}$

where $\divides$ denotes divisibility.

From Integer Multiples under Multiplication form Semigroup, $\struct {n \Z, \times}$ is a semigroup which has no identity.

By construction, $n \Z$ is a subset of $\Z$.

Hence $\struct {n \Z, \times}$ is a subsemigroup of $\struct {\Z, \times}$ which has no identity.