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# 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):

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:

(a) 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 $6^{4}=0$.

(b) If $n=2p$, where $p$ is an odd prime, then $p$ is a non-trivial idempotent element ($p^{2}=p$), and since $2^{{p-1}}\equiv 1\;\;(\mathop{{\rm mod}}p)$ 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 $a\cdot 2\cdot a=a$

(c) 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. 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 $1^{{\prime}}=1\cdot 1^{{\prime}}$.

## Mathematics Subject Classification

20M99*no label found*

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