After learning the definition and first examples of groups, we now begin the elementary properties that follow from the group axioms. These properties are not extra assumptions; they are consequences of closure, associativity, identity, and inverse. The first such property is the uniqueness of the identity element. This result is important because the group definition says that there exists an identity element, but it does not initially say that there is only one. In this lesson, students will learn why a group cannot have two different identity elements and how this uniqueness is used in later proofs involving inverses, cancellation, equations, and powers. :::definition[Identity Element] Let $(G,\circ)$ be a group. An element $e\in G$ is called the $identity element$ of $G$ if $$\begin{equation*} e\circ a=a\circ e=a \quad \forall\, a\in G. \end{equation*}$$ The identity element must leave every element of $G$ unchanged from both sides. ::: The word identity means that the element does not change any other element under the group operation. In $(\mathbb{Z},+)$, the identity element is $0$, because $0+a=a+0=a$ for every integer $a$. In $(\mathbb{Q}-\{0\},\cdot)$, the identity element is $1$, because $1\cdot a=a\cdot1=a$ for every nonzero rational number $a$. These examples may make the identity look familiar, but in an abstract group the identity is not guessed from appearance. It is determined by the operation. :::theorem Let $(G,\circ)$ be a group. Then the identity element of $G$ is unique. ::: :::proof Given that $(G,\circ)$ is a group. To prove that the identity element of $G$ is unique. Let $e$ and $e'$ be two identity elements of $G$. Since $e$ is an identity element, we get $$\begin{equation*} e\circ e'=e'. \end{equation*}$$ Since $e'$ is an identity element, we get $$\begin{equation*} e\circ e'=e. \end{equation*}$$ Therefore $$\begin{equation*} e=e'. \end{equation*}$$ Hence, the identity element of $G$ is unique. $\square$ ::: An identity element must work on both sides of every element. In a finite modular example, it is possible to test this condition row by row. Choose an operation and a candidate, then compare the left product with the right product for each element. A true two-sided identity leaves every row unchanged. The table also shows why a one-sided check is not enough in the abstract definition. :::scientific-preview[Identity Element Explorer] ::: This proof is short, but it contains an important method. When two objects are claimed to have the same universal property, apply one object to the other. Since $e$ behaves as an identity, it leaves $e'$ unchanged. Since $e'$ behaves as an identity, it leaves $e$ unchanged. The same product $e\circ e'$ is forced to be both $e'$ and $e$, and so the two elements are equal. Students often try to prove uniqueness by assuming formulas from familiar number systems, but the proof must use only the group axioms. :::example In the additive group $(\mathbb{Z},+)$, the identity element is $0$. Let $e\in\mathbb{Z}$ be any identity element under addition. Then $$\begin{equation*} e+a=a \quad \forall\, a\in\mathbb{Z}. \end{equation*}$$ Taking $a=0$, we get $$\begin{align*} e+0&=0\\ e&=0. \end{align*}$$ Hence, the identity element of $(\mathbb{Z},+)$ is $0$. ::: :::example In the multiplicative group $(\mathbb{R}-\{0\},\cdot)$, the identity element is $1$. Let $e\in\mathbb{R}-\{0\}$ be any identity element under multiplication. Then $$\begin{equation*} e\cdot a=a \quad \forall\, a\in\mathbb{R}-\{0\}. \end{equation*}$$ Taking $a=1$, we get $$\begin{align*} e\cdot1&=1\\ e&=1. \end{align*}$$ Hence, the identity element of $(\mathbb{R}-\{0\},\cdot)$ is $1$. ::: :::solved-problem Let $(G,\circ)$ be a group. Suppose $u\in G$ satisfies $$\begin{equation*} u\circ a=a \quad \forall\, a\in G. \end{equation*}$$ Suppose $v\in G$ satisfies $$\begin{equation*} a\circ v=a \quad \forall\, a\in G. \end{equation*}$$ Prove that $u=v$. ::: :::solution Let $(G,\circ)$ be a group, and let $u,v\in G$ satisfy $$\begin{equation*} u\circ a=a \quad \forall\, a\in G \end{equation*}$$ and $$\begin{equation*} a\circ v=a \quad \forall\, a\in G. \end{equation*}$$ To prove that $u=v$. Since $u\circ a=a$ for every $a\in G$, taking $a=v$, we get $$\begin{equation*} u\circ v=v. \end{equation*}$$ Since $a\circ v=a$ for every $a\in G$, taking $a=u$, we get $$\begin{equation*} u\circ v=u. \end{equation*}$$ Therefore $$\begin{equation*} u=v. \end{equation*}$$ Hence, $u=v$. $\square$ ::: :::remark The previous problem shows why a left identity and a right identity cannot be different in a group-like setting when both act on all elements. If $u$ acts as a left identity and $v$ acts as a right identity, then the product $u\circ v$ is forced to equal both of them. This is the same idea used in the proof of uniqueness of identity. In later proofs, this uniqueness allows us to write the identity element simply as $e$ without ambiguity. ::: Use the calculator below to test identity candidates in modular arithmetic. Choose a modulus $n$, an operation, and a candidate identity. The calculator tests the identity condition on the set $\{0,1,\ldots,n-1\}$. This is a finite illustration of the two-sided condition $e\circ a=a\circ e=a$. :::calculator[Identity Test] @libraries: none ::: :::exercise [1] Prove that $0$ is the identity element of $(\mathbb{Q},+)$. [2] Prove that $1$ is the identity element of $(\mathbb{C}-\{0\},\cdot)$. [3] Let $(G,\circ)$ be a group. If $p\in G$ satisfies $p\circ a=a$ for every $a\in G$, prove that $p=e$. [4] Let $(G,\circ)$ be a group. If $q\in G$ satisfies $a\circ q=a$ for every $a\in G$, prove that $q=e$. [5] Explain why the identity element cannot depend on $a$. ::: :::answer [1] For every $a\in\mathbb{Q}$, $0+a=a+0=a$. [2] For every $z\in\mathbb{C}-\{0\}$, $1\cdot z=z\cdot1=z$. [3] Since $e$ is the identity element, $p\circ e=p$. Since $p$ is a left identity, $p\circ e=e$, and also $e\circ p=p$. By uniqueness of identity, $p=e$. [4] Since $e$ is the identity element, $e\circ q=q$. Since $q$ is a right identity, $e\circ q=e$. Therefore $q=e$. [5] The identity element must satisfy the identity equation for every element of the group, so it is one fixed element of $G$. ::: :::faq Q: Does every group have an identity element? A: Yes. The existence of an identity element is one of the group axioms. Q: Can a group have two identity elements? A: No. The identity element of a group is unique. Q: Is the identity element always $0$? A: No. In additive notation it is usually $0$, while in multiplicative notation it is usually $1$. Q: Why must the identity be two-sided? A: The group definition requires $e\circ a=a\circ e=a$ for every $a\in G$. ::: :::call-to-action[Inverse Element] subtitle: Continue to the uniqueness of inverse elements, the next elementary property derived from the group axioms. button: Next Lesson | /abstract-algebra/introduction-to-groups/elementary-properties-of-groups/inverse-element-in-groups button-ghost: Previous Section | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/first-nonexamples-of-groups :::