examples of semigroups

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 $\{p^i\mid i\ge n\}$, 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 $2^{p-1}\equiv 1(modp)$ 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 $S-\{0\}$ 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 $S=G\cup \{0\}$.

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 $M=S\cup \{1\}$ with multiplicative identity $1$. If $S$ is already a monoid with identity $1$, then adjoining $1^{\prime }$ to $S$ and repeating the remaining step above gives us a new monoid with identity $1^{\prime }$. However, $1$ is no longer an identity, as $1^{\prime }=1\cdot 1^{\prime }$.