Integer Multiples under Multiplication form Semigroup

Theorem
Let $n \Z$ be the set of integer multiples of $n$.

Then $\left({n \Z, \times}\right)$ is a semigroup.

Closure
Let $p, q \in n \Z$.

Then for some $p', q' \in \Z$:


 * $p = np'$


 * $q = nq'$

Hence:


 * $pq = \left({np'}\right)\left({nq'}\right)$

By the commutativity and associativity of multiplication:


 * $pq = n \left({n \left({p'q'}\right)}\right)$

Hence $pq \in n \Z$ and $n \Z$ is closed under $\times$.

Assocativity
By definition $n \Z \subseteq \Z$.

Hence $\times$ is associative on $n \Z$ as a direct results of Restriction of Associative Operation is Associative.

Also see
If $n = 1$ then $\left({n \Z, \times}\right)$ is a monoid from Integers under Multiplication form Countably Infinite Commutative Monoid.

If $n \ne 1$ then $\left({n \Z, \times}\right)$ is a semigroup but not a monoid as it is without an identity as $1 \notin n \Z$.