After proving the elementary properties of groups, we now step slightly backward and study semigroups. A semigroup has only associativity as its structural law, so it is weaker than a group. This is useful because many group theorems can be understood by asking how much extra information must be added to a semigroup in order to recover a group. In this section, the central question is: when does a semigroup become a group? In this lesson, students will learn the definition of semigroup, compare it with the definition of group, and prepare for the criteria involving equations, cancellation laws, and idempotent elements. :::definition[Semigroup] Let $S$ be a non-empty set and let $\circ$ be a binary operation on $S$. Then $(S,\circ)$ is called a $semigroup$ if $\circ$ is associative on $S$, that is, $$\begin{equation*} (a\circ b)\circ c=a\circ(b\circ c) \quad \forall\, a,b,c\in S. \end{equation*}$$ Thus a semigroup is a non-empty set with an associative binary operation. ::: The difference between a semigroup and a group is important. A group requires closure, associativity, an identity element, and inverse elements. A semigroup requires only a binary operation and associativity. Since a binary operation already includes closure, the additional visible condition in a semigroup is associativity. Therefore every group is a semigroup, but not every semigroup is a group. The missing parts are the identity element and inverse elements. :::theorem Let $(G,\circ)$ be a group. Then $(G,\circ)$ is a semigroup. ::: :::proof Given that $(G,\circ)$ is a group. To prove that $(G,\circ)$ is a semigroup. Since $(G,\circ)$ is a group, $\circ$ is a binary operation on $G$. Also, by the associativity axiom of a group, $$\begin{equation*} (a\circ b)\circ c=a\circ(b\circ c) \quad \forall\, a,b,c\in G. \end{equation*}$$ Therefore $\circ$ is associative on $G$. Hence, $(G,\circ)$ is a semigroup. $\square$ ::: The converse of this theorem is false. A semigroup need not have an identity element, and even if it has an identity element, it need not have inverses. For example, $(\mathbb{N},+)$ is a semigroup because addition is associative and the sum of two positive integers is positive. But if $\mathbb{N}=\{1,2,3,\ldots\}$, then $(\mathbb{N},+)$ has no additive identity inside the set. Hence it is not a group. :::example Let $\mathbb{N}=\{1,2,3,\ldots\}$. Then $(\mathbb{N},+)$ is a semigroup. Let $a,b,c\in\mathbb{N}$. Since $a+b\in\mathbb{N}$, addition is a binary operation on $\mathbb{N}$. Also, $$\begin{equation*} (a+b)+c=a+(b+c). \end{equation*}$$ Therefore addition is associative on $\mathbb{N}$. Hence, $(\mathbb{N},+)$ is a semigroup. ::: :::example Let $S=\mathbb{Z}$ and let $\cdot$ be ordinary multiplication. Then $(\mathbb{Z},\cdot)$ is a semigroup. Let $a,b,c\in\mathbb{Z}$. Since $ab\in\mathbb{Z}$, multiplication is a binary operation on $\mathbb{Z}$. Also, $$\begin{equation*} (ab)c=a(bc). \end{equation*}$$ Therefore multiplication is associative on $\mathbb{Z}$. Hence, $(\mathbb{Z},\cdot)$ is a semigroup. ::: :::example Let $S$ be any non-empty set, and let $\mathcal{F}(S)$ be the set of all functions from $S$ to $S$. Then $\mathcal{F}(S)$ is a semigroup under composition of functions. If $f,g\in\mathcal{F}(S)$, then $f\circ g\in\mathcal{F}(S)$. Also, for all $f,g,h\in\mathcal{F}(S)$, $$\begin{equation*} (f\circ g)\circ h=f\circ(g\circ h). \end{equation*}$$ Thus composition is associative. Hence, $\mathcal{F}(S)$ is a semigroup under composition. ::: :::solved-problem Let $S=\{0,x\}$ and define $\circ$ on $S$ by $$\begin{equation*} 0\circ0=0,\quad 0\circ x=0,\quad x\circ0=0,\quad x\circ x=0. \end{equation*}$$ Prove that $(S,\circ)$ is a semigroup. ::: :::solution Let $S=\{0,x\}$ and let $\circ$ be defined by $$\begin{equation*} 0\circ0=0,\quad 0\circ x=0,\quad x\circ0=0,\quad x\circ x=0. \end{equation*}$$ To prove that $(S,\circ)$ is a semigroup. [1] To prove closure. For all $a,b\in S$, the operation gives $$\begin{equation*} a\circ b=0. \end{equation*}$$ Since $0\in S$, we get $a\circ b\in S$ for all $a,b\in S$. [2] To prove associativity. Let $a,b,c\in S$. Since $a\circ b=0$, we get $$\begin{align*} (a\circ b)\circ c&=0\circ c\\ &=0. \end{align*}$$ Also, since $b\circ c=0$, we get $$\begin{align*} a\circ(b\circ c)&=a\circ0\\ &=0. \end{align*}$$ Therefore $$\begin{equation*} (a\circ b)\circ c=a\circ(b\circ c) \quad \forall\, a,b,c\in S. \end{equation*}$$ Hence, $(S,\circ)$ is a semigroup. $\square$ ::: :::remark This example is a useful warning. The operation is associative, so the structure is a semigroup. But it is not a group because there is no identity element. Indeed, $0\circ x=0\neq x$, $x\circ0=0\neq x$, and $x\circ x=0\neq x$. Thus neither $0$ nor $x$ can act as an identity element. A semigroup may be very far from being a group. ::: A semigroup table on the set $\{0,1\}$ is determined by four products. We observe how each triple $(a,b,c)$ compares the two values $(a\circ b)\circ c$ and $a\circ(b\circ c)$. Change the table and watch the number of associative triples update. If all eight triples agree, the table defines a semigroup on the two-element set. :::scientific-preview[Associativity Triple Counter] @libraries: plotly ::: Use the calculator to check associativity for a small operation table on the set $\{0,1\}$. Enter the four products $0\circ0$, $0\circ1$, $1\circ0$, and $1\circ1$. The calculator tests all triples from $\{0,1\}$. This illustrates that a semigroup condition is entirely about associativity once closure on the set is built into the table. :::calculator[Semigroup Test] ::: :::exercise [1] Define a semigroup. [2] Prove that every group is a semigroup. [3] Prove that $(\mathbb{N},+)$ is a semigroup when $\mathbb{N}=\{1,2,3,\ldots\}$. [4] Determine whether $(\mathbb{Z},-)$ is a semigroup. [5] Give an example of a semigroup that is not a group. ::: :::answer [1] A semigroup is a non-empty set with an associative binary operation. [2] Every group has an associative binary operation, so every group is a semigroup. [3] Addition is closed and associative on $\mathbb{N}$. [4] No. Subtraction is not associative because $(5-3)-1=1$, while $5-(3-1)=3$. [5] $(\mathbb{N},+)$ is a semigroup but not a group when $\mathbb{N}=\{1,2,3,\ldots\}$. ::: :::faq Q: Is every group a semigroup? A: Yes. A group has an associative binary operation, so it is a semigroup. Q: Is every semigroup a group? A: No. A semigroup may fail to have an identity element or inverse elements. Q: What is the main condition in a semigroup? A: Associativity of the binary operation. Q: Why study semigroups in group theory? A: Semigroups help us understand which extra conditions force a weaker algebraic structure to become a group. ::: :::call-to-action[Group Criterion by Equations] subtitle: Continue to the first major criterion showing how unique solvability of semigroup equations forces a semigroup to be a group. button: Next Lesson | /abstract-algebra/introduction-to-groups/semigroup-conditions-for-groups/semigroup-equation-criterion-for-groups button-ghost: Previous Section | /abstract-algebra/introduction-to-groups/elementary-properties-of-groups/properties-of-inverses :::