After proving that finite cancellative semigroups are groups, we now study a different feature of finite semigroups: idempotent elements. In a group, the identity element is the only idempotent element. In a finite semigroup, an identity element may not exist, but an idempotent element must exist. This is a remarkable consequence of finiteness and associativity. In this lesson, students will prove that every finite semigroup contains an element $a$ satisfying $a^2=a$, and they will learn how repeated powers in a finite set force repetition. :::definition[Idempotent Element] Let $(S,\circ)$ be a semigroup. An element $a\in S$ is called an $idempotent element$ if $$\begin{equation*} a\circ a=a. \end{equation*}$$ In multiplicative notation, this condition is written as $$\begin{equation*} a^2=a. \end{equation*}$$ ::: :::definition[Power in a Semigroup] Let $(S,\circ)$ be a semigroup and let $x\in S$. For $n\in\mathbb{N}$, the $positive power$ $x^n$ is defined by $$\begin{equation*} x^n=\underbrace{x\circ x\circ\cdots\circ x}_{n\text{ times}}. \end{equation*}$$ Associativity ensures that this expression is unambiguous. ::: In a semigroup, positive powers are meaningful because the operation is associative. We do not need an identity element to define $x^1,x^2,x^3,\ldots$. We only need a repeated product of positive length. If the semigroup is finite, then the infinite sequence $x,x^2,x^3,\ldots$ must repeat. This repetition is the key to finding an idempotent element. :::theorem Let $(S,\circ)$ be a finite semigroup. Then there exists an element $a\in S$ such that $$\begin{equation*} a^2=a. \end{equation*}$$ ::: :::proof Given that $(S,\circ)$ is a finite semigroup. To prove that there exists an element $a\in S$ such that $a^2=a$. Let $x\in S$. Consider the sequence $$\begin{equation*} x,x^2,x^3,\ldots. \end{equation*}$$ Since $S$ is finite, there exist positive integers $r$ and $s$ such that $$\begin{equation*} r<s \end{equation*}$$ and $$\begin{equation*} x^r=x^s. \end{equation*}$$ Let $$\begin{equation*} d=s-r. \end{equation*}$$ Then $d\in\mathbb{N}$. For every $k\geq r$, multiplying $x^r=x^s$ on the right by $x^{k-r}$ gives $$\begin{equation*} x^k=x^{k+d}. \end{equation*}$$ Choose $t\in\mathbb{N}$ such that $$\begin{equation*} td\geq r. \end{equation*}$$ Let $$\begin{equation*} a=x^{td}. \end{equation*}$$ Since $td\geq r$, repeated use of $x^k=x^{k+d}$ for $k\geq r$ gives $$\begin{equation*} x^{td}=x^{td+d}=x^{td+2d}=\cdots=x^{td+td}=x^{2td}. \end{equation*}$$ Therefore $$\begin{align*} a^2 &=(x^{td})^2\\ &=x^{2td}\\ &=x^{td}\\ &=a. \end{align*}$$ Hence, there exists an element $a\in S$ such that $a^2=a$. $\square$ ::: The proof uses only finiteness and associativity. Finiteness forces repetition among the powers of $x$. Associativity allows us to manipulate powers consistently. The element $a=x^{td}$ is chosen far enough along the repeating part of the sequence so that doubling its exponent does not change the value. This is why $a^2=a$. :::example Let $S=\{0,x\}$ and define $\circ$ by $$\begin{equation*} 0\circ0=0,\quad 0\circ x=0,\quad x\circ0=0,\quad x\circ x=0. \end{equation*}$$ Then $0$ is an idempotent element because $$\begin{equation*} 0\circ0=0. \end{equation*}$$ Also, $x$ is not idempotent because $$\begin{equation*} x\circ x=0\neq x. \end{equation*}$$ Hence, this finite semigroup contains an idempotent element. ::: :::example Let $S=\{1,2,3,4\}$ under multiplication modulo $5$. This is a finite semigroup. Since $$\begin{equation*} 1^2\equiv1 \pmod{5}, \end{equation*}$$ the element $1$ is idempotent. In fact, this semigroup is a group, and the identity element is the only idempotent element. ::: :::solved-problem Let $(S,\circ)$ be a finite semigroup and let $x\in S$. Suppose $$\begin{equation*} x^3=x^7. \end{equation*}$$ Construct an idempotent element from powers of $x$. ::: :::solution Let $(S,\circ)$ be a finite semigroup and let $x\in S$. Given that $$\begin{equation*} x^3=x^7. \end{equation*}$$ To construct an idempotent element from powers of $x$. Here $r=3$ and $s=7$. Thus $$\begin{equation*} d=s-r=4. \end{equation*}$$ Choose $t=1$. Then $$\begin{equation*} td=4\geq3. \end{equation*}$$ Let $$\begin{equation*} a=x^4. \end{equation*}$$ Since $x^k=x^{k+4}$ for all $k\geq3$, we get $$\begin{equation*} x^4=x^8. \end{equation*}$$ Now $$\begin{align*} a^2 &=(x^4)^2\\ &=x^8\\ &=x^4\\ &=a. \end{align*}$$ Hence, $a=x^4$ is an idempotent element. $\square$ ::: :::remark A finite semigroup may have one idempotent element, many idempotent elements, or, if it is a group, exactly one idempotent element. The theorem guarantees existence, not uniqueness. A common mistake is to think that because finite groups have exactly one idempotent element, finite semigroups must behave the same way. They do not. ::: Multiplication modulo $n$ gives a finite associative operation, so it is a useful place to search for idempotents. We observe the values of $a^2-a$ modulo $n$. An element is idempotent exactly when this value is $0$. Change $n$ and notice that some moduli have several idempotents, showing that idempotents need not be unique in a finite semigroup. :::scientific-preview[Idempotent Residue Explorer] @libraries: plotly ::: Use the calculator to find idempotent elements in multiplication modulo $n$ on the set $\{0,1,\ldots,n-1\}$. This operation is associative and the set is finite, so at least one idempotent element will appear. Try $n=6$, where more than one idempotent occurs. :::calculator[Finite Semigroup Idempotents] ::: :::exercise [1] Define an idempotent element in a semigroup. [2] Prove that every finite semigroup contains an idempotent element. [3] Find all idempotent elements under multiplication modulo $6$. [4] If $x^2=x^5$ in a semigroup, construct an idempotent power of $x$. [5] Explain why associativity is needed to discuss powers in a semigroup. ::: :::answer [1] An element $a$ is idempotent if $a^2=a$. [2] The sequence $x,x^2,x^3,\ldots$ must repeat in a finite semigroup. The repetition gives a periodic tail from which an idempotent power can be chosen. [3] The idempotent elements modulo $6$ under multiplication are $0,1,3,4$. [4] Here $r=2$, $s=5$, and $d=3$. Choose $t=1$, so $td=3\geq2$. Then $a=x^3$ satisfies $a^2=a$. [5] Associativity ensures that products such as $x^n$ are unambiguous. ::: :::faq Q: Does every finite semigroup have an idempotent element? A: Yes. Every finite semigroup contains at least one idempotent element. Q: Does every semigroup have an idempotent element? A: No. The theorem requires finiteness. Q: Can a finite semigroup have more than one idempotent element? A: Yes. Multiplication modulo $6$ has several idempotent elements. Q: Is the idempotent element always an identity element? A: No. An idempotent element need not be an identity element in a semigroup. ::: :::call-to-action[Unique Idempotent Semigroups] subtitle: Continue to the counterexample showing that a unique idempotent in a finite semigroup does not guarantee a group. button: Next Lesson | /abstract-algebra/introduction-to-groups/semigroup-conditions-for-groups/unique-idempotent-semigroups button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/semigroup-conditions-for-groups/finite-cancellative-semigroups :::