Examples of semigroups are numerous. This entry presents some of the most common examples.

1. The set $\mathbb{Z}$ of integers with multiplication is a semigroup, along with many of its subsets (subsemigroups):
   1. The set of non-negative integers
   2. The set of positive integers
   3. $n\mathbb{Z}$ , the set of all integral multiples of an integer $n$
   4. For any prime $p$ , the set of $\lbrace p^i\mid i\ge n\rbrace$ , where $n$ is a non-negative integer
   5. The set of all composite integers
2. $\mathbb{Z}_n$ , the set of all integers modulo an integer $n$ , with integer multiplication modulo $n$ . Here, we may find examples of nilpotent and idempotent elements, relative inverses, and eventually periodic elements:
   1. If $n=p^m$ , where $p$ is prime, then every non-zero element containing a factor of $p$ is nilpotent. For example, if $n=16$ , then $6^4=0$ .
   2. If $n=2p$ , where $p$ is an odd prime, then $p$ is a non-trivial idempotent element ($p^2=p$ ), and since $2^{p-1}\equiv 1 \pmod p$ by Fermat's little theorem, we see that $a=2^{p-2}$ is a relative inverse of $2$ , as $2\cdot a \cdot 2 = 2$ and $a\cdot 2 \cdot a=a$
   3. If $n=2^m p$ , where $p$ is an odd prime, and $m>1$ , then $2$ is eventually periodic. For example, $n=96$ , then $2^2=4$ , $2^3=8$ , $2^4=16$ , $2^5=32$ , $2^6=64$ , $2^7=32$ , $2^8=64$ , etc...
3. The set $M_n(R)$ of $n\times n$ square matrices over a ring $R$ , with matrix multiplication, is a semigroup. Unlike the previous two examples, $M_n(R)$ is not commutative.
4. The set $E(A)$ of functions on a set $A$ , with functional composition, is a semigroup.
5. Every group is a semigroup, as well as every monoid.
6. If $R$ is a ring, then $R$ with the ring multiplication (ignoring addition) is a semigroup (with $0$ ).
7. Group with Zero. A semigroup $S$ is called a group with zero if it contains a zero element $0$ , and $S-\lbrace 0\rbrace$ is a subgroup of $S$ . In $R$ in the previous example is a division ring, then $R$ with the ring multiplication is a group with zero. If $G$ is a group, by adjoining $G$ with an extra symbol $0$ , and extending the domain of group multiplication $\cdot$ by defining $0 \cdot a = a\cdot 0 =0\cdot 0:=0$ for all $a\in G$ , we get a group with zero $S=G\cup \lbrace 0\rbrace$ .
8. As mentioned earlier, every monoid is a semigroup. If $S$ is not a monoid, then it can be embedded in one: adjoin a symbol $1$ to $S$ , and extend the semigroup multiplication $\cdot$ on $S$ by defining $1\cdot a = a\cdot 1 = a$ and $1\cdot 1=1$ , we get a monoid $M=S\cup \lbrace 1\rbrace$ with multiplicative identity $1$ . If $S$ is already a monoid with identity $1$ , then adjoining $1'$ to $S$ and repeating the remaining step above gives us a new monoid with identity $1'$ . However, $1$ is no longer an identity, as $1'=1\cdot 1'$ .