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. :::definition[Unique Idempotent] Let $(S,\circ)$ be a semigroup. The semigroup is said to have a $unique idempotent$ 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. :::theorem The statement that a finite semigroup is a group if and only if it has exactly one idempotent element is false. ::: :::proof 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 $\circ$ on $S$ by $$\begin{equation*} 0\circ0=0,\quad 0\circ x=0,\quad x\circ0=0,\quad x\circ x=0. \end{equation*}$$ [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. :::corollary Let $(G,\circ)$ be a finite group with identity element $e$. Then $G$ has exactly one element $a$ such that $$\begin{equation*} a^2=a. \end{equation*}$$ ::: :::proof 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. :::solved-problem 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. ::: :::solution 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 $S$. $\square$ ::: :::remark 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. ::: The two-element zero semigroup separates idempotent behavior from identity behavior. We observe that every product becomes $0$, so $0\circ0=0$ makes $0$ idempotent. But the same rule gives $0\circ x=0$, not $x$, so $0$ cannot be an identity element. This is the counterexample that disproves the false converse. :::scientific-preview[Zero Semigroup Property Chart] @libraries: plotly ::: Use the calculator 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. :::calculator[Unique Idempotent Test] ::: :::exercise [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? ::: :::answer [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. ::: :::faq Q: Does a unique idempotent make a finite semigroup a group? A: No. A finite semigroup may have exactly one idempotent element and still fail to have an identity element. Q: Does every finite group have exactly one idempotent element? A: Yes. The identity element is the only idempotent element. Q: What is the difference between idempotent and identity? A: An idempotent element satisfies $a^2=a$, while an identity element satisfies $e\circ s=s\circ e=s$ for every $s$. Q: Why does the counterexample disprove the converse? A: It satisfies the hypothesis of having exactly one idempotent element but fails the conclusion of being a group. ::: :::call-to-action[Powers of Elements] subtitle: Continue to powers of elements in a group, where repeated products are developed systematically. button: Next Section | /abstract-algebra/introduction-to-groups/powers-of-elements-in-groups/positive-and-negative-powers button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/semigroup-conditions-for-groups/idempotents-in-finite-semigroups :::