Idempotents in Finite Semigroups

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.

Let

(S,\circ)

be a semigroup. An element

a\in S

is called an

\textbf{idempotent element}

\begin{equation*} a\circ a=a. \end{equation*}

In multiplicative notation, this condition is written as

\begin{equation*} a^2=a. \end{equation*}

a

satisfying

a^2=a

, and they will learn how repeated powers in a finite set force repetition.

Let

(S,\circ)

be a semigroup. An element

a\in S

is called an

\textbf{idempotent element}

\begin{equation*} a\circ a=a. \end{equation*}

In multiplicative notation, this condition is written as

\begin{equation*} a^2=a. \end{equation*}

Let

(S,\circ)

be a semigroup and let

x\in S

. For

n\in\mathbb{N}

, the

\textbf{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.

Let

(S,\circ)

be a finite semigroup. Then there exists an element

a\in S

such that

\begin{equation*} a^2=a. \end{equation*}

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

Let

S=\{0,x\}

and define

\circ

\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.

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.

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

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

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.

Use the calculator below 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.

[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.

[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.

Yes. Every finite semigroup contains at least one idempotent element.

No. The theorem requires finiteness.

Yes. Multiplication modulo

6

has several idempotent elements.

No. An idempotent element need not be an identity element in a semigroup.

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