Subsemigroup of Monoid is not necessarily Monoid
Theorem
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.
Proof
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.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets