After proving the cancellation laws, we can study a useful consequence: the behaviour of idempotent elements in a group. An idempotent element is an element that remains unchanged when it is combined with itself. In many algebraic systems, there may be several idempotent elements. In a group, however, the cancellation laws force a very strong conclusion: the identity element is the only idempotent element. In this lesson, students will learn the definition of idempotent element, prove the group theorem, and compare the group case with familiar examples. :::definition[Idempotent Element] Let $(S,\circ)$ be a non-empty set with a binary operation $\circ$. 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 $$\begin{equation*} a^2=a. \end{equation*}$$ ::: The definition of idempotent element does not require a group. It can be used in semigroups, rings, matrix algebras, and many other structures. For example, under ordinary multiplication, $0$ and $1$ are idempotent real numbers because $0^2=0$ and $1^2=1$. In a group, however, the situation is more restrictive. Since every element can be cancelled, the equation $a\circ a=a$ forces $a$ to be the identity element. :::theorem Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. If $$\begin{equation*} a\circ a=a, \end{equation*}$$ then $$\begin{equation*} a=e. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a\in G$. Given that $$\begin{equation*} a\circ a=a. \end{equation*}$$ To prove that $a=e$. Since $e$ is the identity element of $G$, $$\begin{equation*} a\circ e=a. \end{equation*}$$ Now $$\begin{align*} a\circ a=a &\implies a\circ a=a\circ e \quad [\because\, a\circ e=a]\\ &\implies a=e \quad [\because\, \text{left cancellation law}]. \end{align*}$$ Hence, $$\begin{equation*} a=e. \end{equation*}$$ $\square$ ::: :::corollary Let $(G,\circ)$ be a group with identity element $e$. Then the identity element is the only idempotent element in $G$. ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$. To prove that the identity element is the only idempotent element in $G$. Since $e$ is the identity element, we get $$\begin{equation*} e\circ e=e. \end{equation*}$$ Therefore $e$ is an idempotent element. Let $a\in G$ be an idempotent element. Then $$\begin{equation*} a\circ a=a. \end{equation*}$$ By the previous theorem, $$\begin{equation*} a=e. \end{equation*}$$ Therefore every idempotent element of $G$ is equal to $e$. Hence, the identity element is the only idempotent element in $G$. $\square$ ::: Idempotent elements are solutions of the equation $a\circ a=a$. In a group, cancellation forces the only solution to be the identity. In other algebraic structures, extra idempotents can appear because cancellation may fail. Choose a modulus and an operation to list every idempotent residue. The comparison shows why extra idempotents are quick evidence that the whole structure is not a group under that operation. :::scientific-preview[Idempotent Element Explorer] ::: This corollary is a useful diagnostic tool. If an algebraic structure has more than one idempotent element, then it cannot be a group under that operation. For example, $\mathbb{R}$ under multiplication has two idempotent elements, $0$ and $1$. This is one reason $(\mathbb{R},\cdot)$ is not a group. The deeper obstruction is that $0$ has no multiplicative inverse, but the idempotent test also quickly shows that the group property cannot hold. :::example In $(\mathbb{Z},+)$, an idempotent element $a$ satisfies $$\begin{equation*} a+a=a. \end{equation*}$$ Then $$\begin{align*} a+a=a &\implies a+a=a+0\\ &\implies a=0 \quad [\because\, \text{left cancellation law}]. \end{align*}$$ Hence, the only idempotent element of $(\mathbb{Z},+)$ is $0$. ::: :::example In $(\mathbb{R}-\{0\},\cdot)$, an idempotent element $a$ satisfies $$\begin{equation*} a^2=a. \end{equation*}$$ Since $a\neq0$, cancellation in the multiplicative group gives $$\begin{align*} a^2=a &\implies a\cdot a=a\cdot1\\ &\implies a=1. \end{align*}$$ Hence, the only idempotent element of $(\mathbb{R}-\{0\},\cdot)$ is $1$. ::: :::solved-problem Let $(G,\circ)$ be a group. Prove that if $a,b\in G$ and $$\begin{equation*} a\circ b=a \end{equation*}$$ with $a=b$, then $a=e$. ::: :::solution Let $(G,\circ)$ be a group with identity element $e$, and let $a,b\in G$. Given that $$\begin{equation*} a\circ b=a \end{equation*}$$ and $a=b$. To prove that $a=e$. Since $a=b$, the equation $a\circ b=a$ becomes $$\begin{equation*} a\circ a=a. \end{equation*}$$ Thus $a$ is an idempotent element of $G$. Since the identity element is the only idempotent element in a group, $$\begin{equation*} a=e. \end{equation*}$$ Hence, $a=e$. $\square$ ::: :::solved-problem Determine all idempotent elements in the group $(\mathbb{Q}-\{0\},\cdot)$. ::: :::solution Let $(\mathbb{Q}-\{0\},\cdot)$ be the group of nonzero rational numbers under multiplication. To determine all idempotent elements in $(\mathbb{Q}-\{0\},\cdot)$. Let $a\in\mathbb{Q}-\{0\}$ be idempotent. Then $$\begin{equation*} a^2=a. \end{equation*}$$ Since the identity element of $(\mathbb{Q}-\{0\},\cdot)$ is $1$, and the identity element is the only idempotent element in a group, we get $$\begin{equation*} a=1. \end{equation*}$$ Also, $$\begin{equation*} 1^2=1. \end{equation*}$$ Hence, the only idempotent element in $(\mathbb{Q}-\{0\},\cdot)$ is $1$. $\square$ ::: :::remark The finite condition is not needed for the group result. Whether a group is finite or infinite, the identity element is the only idempotent element. The proof uses cancellation, and cancellation comes from inverse elements. This is why the result holds in all groups but does not hold in arbitrary semigroups or rings. ::: Use the calculator below to find idempotent elements modulo $n$ under addition or multiplication. Under addition modulo $n$, the only idempotent element is always $0$. Under multiplication modulo $n$, there may be more than one idempotent because the whole set $\{0,1,\ldots,n-1\}$ under multiplication modulo $n$ is not usually a group. This contrast helps students see the special role of cancellation in groups. :::calculator[Idempotent Test] @libraries: none ::: :::exercise [1] Define an idempotent element. [2] Prove that the only idempotent element of $(\mathbb{R},+)$ is $0$. [3] Determine all idempotent elements of $(\mathbb{C}-\{0\},\cdot)$. [4] Explain why $(\mathbb{R},\cdot)$ is not a group using idempotents. [5] Let $(G,\circ)$ be a group. If $a^2=a$ in multiplicative notation, prove that $a=e$. ::: :::answer [1] An element $a$ is idempotent if $a\circ a=a$. [2] If $a+a=a$, then $a+a=a+0$, so $a=0$ by cancellation. [3] The only idempotent element is $1$. [4] In $\mathbb{R}$ under multiplication, both $0$ and $1$ are idempotent, but a group has only one idempotent element. [5] Since $a^2=a$, we have $a\circ a=a\circ e$. By left cancellation, $a=e$. ::: :::faq Q: What is an idempotent element? A: It is an element $a$ such that $a\circ a=a$. Q: Which element is idempotent in every group? A: The identity element is idempotent. Q: Can a group have two idempotent elements? A: No. The identity element is the only idempotent element in a group. Q: Does the idempotent theorem require the group to be finite? A: No. It holds for every group. ::: :::call-to-action[Equations in Groups] subtitle: Continue to equations in groups and learn how inverse elements give unique solutions. button: Next Lesson | /abstract-algebra/introduction-to-groups/elementary-properties-of-groups/equations-in-groups button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/elementary-properties-of-groups/cancellation-laws-in-groups :::