Unique Idempotent Semigroups

After proving that every finite semigroup contains an idempotent element, we now address a natural but false guess. Since every group has exactly one idempotent element and every finite semigroup has at least one idempotent element, one might suspect that a finite semigroup with exactly one idempotent element must be a group. This is false. In this lesson, students will study a counterexample that has exactly one idempotent element but no identity element. This example is important because it shows that idempotent information alone is not enough to recover the group axioms.

Let

(S,\circ)

be a semigroup. The semigroup is said to have a

\textbf{unique idempotent}

if there exists exactly one element

a\in S

such that

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

Let

(S,\circ)

be a semigroup. The semigroup is said to have a

\textbf{unique idempotent}

if there exists exactly one element

a\in S

such that

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

In a group, the unique idempotent is the identity element. In a general semigroup, an idempotent element does not have to act as an identity. It only satisfies

a^2=a

. Acting as an identity requires

a\circ s=s\circ a=s

for every

s

in the set. These are much stronger conditions. The counterexample below separates these two ideas completely.

The statement that a finite semigroup is a group if and only if it has exactly one idempotent element is false.

Given the statement: a finite semigroup is a group if and only if it has exactly one idempotent element. To prove that the statement is false. We give a counterexample. Let

\begin{equation*} S=\{0,x\} \end{equation*}

and define

\circ

S

\begin{equation*} 0\circ0=0,\quad 0\circ x=0,\quad x\circ0=0,\quad x\circ x=0. \end{equation*}

[1] To prove that

(S,\circ)

is a finite semigroup. The set

S

is finite. For all

u,v\in S

\begin{equation*} u\circ v=0. \end{equation*}

Since

0\in S

, the operation is closed on

S

. Let

u,v,w\in S

. Then

\begin{align*} (u\circ v)\circ w &=0\circ w\\ &=0. \end{align*}

Also,

\begin{align*} u\circ(v\circ w) &=u\circ0\\ &=0. \end{align*}

Therefore

\begin{equation*} (u\circ v)\circ w=u\circ(v\circ w) \quad \forall\, u,v,w\in S. \end{equation*}

Thus

\circ

is associative. Therefore

(S,\circ)

is a finite semigroup. [2] To prove that

(S,\circ)

has exactly one idempotent element. Now

\begin{equation*} 0\circ0=0. \end{equation*}

Therefore

0

is an idempotent element. Also,

\begin{equation*} x\circ x=0\neq x. \end{equation*}

Therefore

x

is not an idempotent element. Thus

(S,\circ)

has exactly one idempotent element, namely

0

. [3] To prove that

(S,\circ)

is not a group. If

(S,\circ)

were a group, then it would have an identity element. The element

0

is not an identity element because

\begin{equation*} 0\circ x=0\neq x. \end{equation*}

The element

x

is not an identity element because

\begin{equation*} x\circ x=0\neq x. \end{equation*}

Thus no element of

S

is an identity element. Therefore

(S,\circ)

is not a group. Hence, the statement is false.

\square

The counterexample is deliberately small. Its operation sends every product to

0

. Such an operation is associative because both sides of every associative expression become

0

. The element

0

is idempotent because

0\circ0=0

, while

x

is not idempotent because

x\circ x=0

. But

0

does not act as an identity on

x

, and

x

does not act as an identity on itself. Therefore the semigroup has a unique idempotent but is not a group.

Let

(G,\circ)

be a finite group with identity element

e

. Then

G

has exactly one element

a

such that

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

Given that

(G,\circ)

is a finite group with identity element

e

. To prove that

G

has exactly one element

a

such that

a^2=a

. Since

e

is the identity element of

G

, we get

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

Therefore

e

is an idempotent element. Let

a\in G

be an idempotent element. Then

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

Since

e

is the identity element,

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

Now

\begin{align*} a^2=a &\implies a\circ a=a\circ e\\ &\implies a=e \quad [\because\, \text{left cancellation law}]. \end{align*}

Therefore every idempotent element of

G

is equal to

e

. Hence,

G

has exactly one element

a

such that

a^2=a

\square

The corollary is true, but its converse is false by the counterexample. This distinction is a valuable logical lesson. The implication

\begin{equation*} \text{finite group} \implies \text{exactly one idempotent} \end{equation*}

is true. The reverse implication

\begin{equation*} \text{exactly one idempotent} \implies \text{finite group} \end{equation*}

is false for semigroups. A single counterexample is enough to disprove the reverse implication.

For the semigroup

S=\{0,x\}

with operation

\begin{equation*} u\circ v=0 \quad \forall\, u,v\in S, \end{equation*}

determine whether

0

is an identity element.

Let

S=\{0,x\}

and suppose

\begin{equation*} u\circ v=0 \quad \forall\, u,v\in S. \end{equation*}

To determine whether

0

is an identity element. For

0

to be an identity element, we must have

\begin{equation*} 0\circ s=s\circ0=s \quad \forall\, s\in S. \end{equation*}

Take

s=x

. Then

\begin{equation*} 0\circ x=0. \end{equation*}

But

\begin{equation*} 0\neq x. \end{equation*}

Therefore

\begin{equation*} 0\circ x\neq x. \end{equation*}

Thus

0

is not an identity element. Hence,

0

is not an identity element of

S

\square

An idempotent element may look like an identity when checked only against itself. The condition

a^2=a

says only that

a

fixes itself under the operation. The identity condition says that

a

fixes every element from both sides. This is much stronger. Many wrong solutions confuse these two conditions.

Use the calculator below for the two-element zero semigroup. It displays the operation table, the idempotent elements, and whether an identity exists. The example reinforces the difference between being idempotent and being an identity element.

[1] Define a unique idempotent in a semigroup. [2] Prove that every finite group has exactly one idempotent element. [3] Give a finite semigroup with exactly one idempotent element that is not a group. [4] Explain why the element

0

in the two-element zero semigroup is not an identity element. [5] Is the converse of the statement “every group has exactly one idempotent element” true for finite semigroups?

[1] A semigroup has a unique idempotent if exactly one element

a

satisfies

a^2=a

. [2] The identity element satisfies

e^2=e

. If

a^2=a

, then

a\circ a=a\circ e

, and cancellation gives

a=e

. [3] Let

S=\{0,x\}

and define

u\circ v=0

for all

u,v\in S

. [4] Since

0\circ x=0\neq x

, the element

0

does not act as an identity. [5] No. The two-element zero semigroup has exactly one idempotent element but is not a group.

No. A finite semigroup may have exactly one idempotent element and still fail to have an identity element.

Yes. The identity element is the only idempotent element.

An idempotent element satisfies

a^2=a

, while an identity element satisfies

e\circ s=s\circ e=s

for every

s

It satisfies the hypothesis of having exactly one idempotent element but fails the conclusion of being a group.

BMLabs Mathematics

BMLabs Mathematics

Unique Idempotent Semigroups

Unique Idempotent Semigroups

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UNIQUE IDEMPOTENT TEST

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Invariants under Orthogonal Transformation