After learning about binary operations, we are ready to define the first central structure of abstract algebra. A group is a non-empty set together with a binary operation satisfying four structural conditions. These conditions are not arbitrary; they allow us to calculate, cancel, solve equations, discuss inverses, and compare algebraic systems that may look very different at first. Number systems under addition, nonzero number systems under multiplication, permutations under composition, and matrix groups all arise from this single definition. In this lesson, students will learn the formal definition of group, the roles of the identity element and inverse element, and the correct way to verify the definition in a basic example. :::definition[Group] Let $G$ be a non-empty set and let $\circ$ be a binary operation on $G$. Then $(G,\circ)$ is called a $group$ if the following conditions hold: (i) $a\circ b\in G \quad \forall\, a,b\in G$. (ii) $(a\circ b)\circ c=a\circ(b\circ c) \quad \forall\, a,b,c\in G$. (iii) There exists $e\in G$ such that $$\begin{equation*} e\circ a=a\circ e=a \quad \forall\, a\in G. \end{equation*}$$ (iv) For each $a\in G$, there exists $a^{-1}\in G$ such that $$\begin{equation*} a\circ a^{-1}=a^{-1}\circ a=e. \end{equation*}$$ The element $e$ is called the $identity element$ of $G$, and $a^{-1}$ is called the $inverse element$ of $a$. ::: A group is formed only when all four axioms work at the same time. Choose a modulus and an operation on $\{0,1,\ldots,n-1\}$. The dashboard checks closure, associativity, identity, and inverses side by side. Addition modulo $n$ passes all four tests, while multiplication modulo $n$ on the whole set usually fails because some elements do not have inverses. :::scientific-preview[Group Axiom Dashboard] ::: The notation $(G,\circ)$ must be read as one object. The set $G$ alone is not enough, because the operation decides the algebraic structure. For example, $\mathbb{Z}$ under addition behaves differently from $\mathbb{Z}$ under multiplication. Thus one should not say only that $\mathbb{Z}$ is a group without mentioning the operation. In undergraduate algebra, this habit is important because many wrong answers come from checking the right set with the wrong operation. The first condition is closure. It says that when two elements of $G$ are combined using $\circ$, the result is again an element of $G$. If $a,b\in G$, then $a\circ b$ must not leave the set. This condition is already contained in the phrase binary operation, but it is still written in the definition because it is the first condition students must verify in examples. A common mistake is to assume closure from the formula without checking the set. The second condition is associativity. It says that whenever $a,b,c\in G$, the result of multiplying three elements does not depend on how the parentheses are placed. The equality $$\begin{equation*} (a\circ b)\circ c=a\circ(b\circ c) \end{equation*}$$ does not say that $a\circ b=b\circ a$. Associativity is about grouping, not about changing order. This distinction is very important because many groups are not commutative. :::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 leaves every element unchanged from both sides. ::: The identity element is one fixed element for the entire group. It does not depend on the element $a$. Under addition, the identity element is usually $0$ because $0+a=a+0=a$. Under multiplication, the identity element is usually $1$ because $1\cdot a=a\cdot 1=a$. Students often confuse the identity element with an inverse element, but the identity is common to all elements, whereas an inverse is attached to a particular element. :::definition[Inverse Element] Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. An element $a^{-1}\in G$ is called the $inverse element$ of $a$ if $$\begin{equation*} a\circ a^{-1}=a^{-1}\circ a=e. \end{equation*}$$ Thus the inverse of $a$ is an element that combines with $a$ from both sides to give the identity element. ::: The inverse element depends on the chosen element. In $(\mathbb{Z},+)$, the inverse of $5$ is $-5$, while the inverse of $-7$ is $7$. In $(\mathbb{Q}-\{0\},\cdot)$, the inverse of $5$ is $\dfrac{1}{5}$. This is why the notation $a^{-1}$ is introduced only after choosing $a\in G$. Another common mistake is to use multiplicative inverse notation in an additive group without translating it correctly; in an additive group, the inverse of $a$ is usually written as $-a$. :::example Let $(\mathbb{Z},+)$ be the set of integers with ordinary addition. Then $(\mathbb{Z},+)$ is a group. [1] Closure: Let $m,n\in\mathbb{Z}$. Then $$\begin{equation*} m+n\in\mathbb{Z}. \end{equation*}$$ Therefore $\mathbb{Z}$ is closed under addition. [2] Associativity: Let $m,n,k\in\mathbb{Z}$. Then $$\begin{equation*} (m+n)+k=m+(n+k). \end{equation*}$$ Therefore addition is associative on $\mathbb{Z}$. [3] Identity element: There exists $0\in\mathbb{Z}$ such that $$\begin{equation*} m+0=0+m=m \quad \forall\, m\in\mathbb{Z}. \end{equation*}$$ Therefore $0$ is the identity element of $(\mathbb{Z},+)$. [4] Inverse element: Let $m\in\mathbb{Z}$. Then $-m\in\mathbb{Z}$ and $$\begin{equation*} m+(-m)=(-m)+m=0. \end{equation*}$$ Therefore every integer has an inverse in $\mathbb{Z}$ under addition. Hence, $(\mathbb{Z},+)$ is a group. ::: :::theorem Let $G$ be a non-empty set and let $\circ$ be a rule on $G$. If $(G,\circ)$ is a group, then $\circ$ is a binary operation on $G$. ::: :::proof Given that $G$ is a non-empty set, $\circ$ is a rule on $G$, and $(G,\circ)$ is a group. To prove that $\circ$ is a binary operation on $G$. Since $(G,\circ)$ is a group, by the first condition in the definition of a group, $$\begin{equation*} a\circ b\in G \quad \forall\, a,b\in G. \end{equation*}$$ Therefore the rule $\circ$ assigns to each ordered pair $(a,b)\in G\times G$ an element of $G$. Thus $\circ$ is a function from $G\times G$ into $G$. Hence, $\circ$ is a binary operation on $G$. $\square$ ::: :::example Let $(\mathbb{Q}-\{0\},\cdot)$ be the set of all nonzero rational numbers with ordinary multiplication. Then $(\mathbb{Q}-\{0\},\cdot)$ is a group. [1] Closure: Let $a,b\in\mathbb{Q}-\{0\}$. Since $a$ and $b$ are nonzero rational numbers, $ab$ is also a nonzero rational number. Therefore $$\begin{equation*} ab\in\mathbb{Q}-\{0\}. \end{equation*}$$ [2] Associativity: For all $a,b,c\in\mathbb{Q}-\{0\}$, $$\begin{equation*} (ab)c=a(bc). \end{equation*}$$ [3] Identity element: There exists $1\in\mathbb{Q}-\{0\}$ such that $$\begin{equation*} a\cdot 1=1\cdot a=a \quad \forall\, a\in\mathbb{Q}-\{0\}. \end{equation*}$$ [4] Inverse element: Let $a\in\mathbb{Q}-\{0\}$. Then $\dfrac{1}{a}\in\mathbb{Q}-\{0\}$ and $$\begin{equation*} a\cdot \frac{1}{a}=\frac{1}{a}\cdot a=1. \end{equation*}$$ Hence, $(\mathbb{Q}-\{0\},\cdot)$ is a group. ::: :::solved-problem Determine whether $(\mathbb{Z},\cdot)$ is a group. ::: :::solution Let $(\mathbb{Z},\cdot)$ be the set of integers with ordinary multiplication. To determine whether $(\mathbb{Z},\cdot)$ is a group. [1] Closure: Let $m,n\in\mathbb{Z}$. Then $$\begin{equation*} mn\in\mathbb{Z}. \end{equation*}$$ Therefore closure is satisfied. [2] Associativity: For all $m,n,k\in\mathbb{Z}$, $$\begin{equation*} (mn)k=m(nk). \end{equation*}$$ Therefore associativity is satisfied. [3] Identity element: There exists $1\in\mathbb{Z}$ such that $$\begin{equation*} m\cdot 1=1\cdot m=m \quad \forall\, m\in\mathbb{Z}. \end{equation*}$$ Therefore the identity element exists. [4] Inverse element: Take $2\in\mathbb{Z}$. If $2$ has an inverse in $\mathbb{Z}$ under multiplication, then there exists $b\in\mathbb{Z}$ such that $$\begin{equation*} 2b=1. \end{equation*}$$ Then $$\begin{align*} 2b=1 &\implies b=\frac{1}{2}. \end{align*}$$ But $\dfrac{1}{2}\notin\mathbb{Z}$. Therefore the inverse element condition is not satisfied. Hence, $(\mathbb{Z},\cdot)$ is not a group. $\square$ ::: Use the calculator below to compare two finite operations on the set $\{0,1,\ldots,n-1\}$. Choose addition modulo $n$ or multiplication modulo $n$, then check the group axioms. Try $n=5$ under addition modulo $5$ and then try multiplication modulo $5$ on the same set. This comparison shows why the operation and the set must be fixed together before deciding whether a group is formed. :::calculator[Group Axiom Test Calculator] ::: :::exercise [1] State the four conditions in the definition of a group. [2] Identify the identity element of $(\mathbb{Z},+)$. [3] Find the inverse of $-8$ in $(\mathbb{Z},+)$. [4] Determine whether $(\mathbb{N},+)$ is a group, where $\mathbb{N}=\{1,2,3,\ldots\}$. [5] Determine whether $(\mathbb{Q}-\{0\},\cdot)$ satisfies the inverse condition. ::: :::answer [1] Closure, associativity, identity element, and inverse element. [2] The identity element is $0$. [3] The inverse of $-8$ under addition is $8$. [4] No. There is no identity element in $\mathbb{N}$ under addition because $0\notin\mathbb{N}$. [5] Yes. For each $a\in\mathbb{Q}-\{0\}$, the inverse is $\dfrac{1}{a}\in\mathbb{Q}-\{0\}$. ::: :::faq Q: Is a group only a set? A: No. A group is a non-empty set together with a binary operation satisfying four conditions. Q: Can the identity element depend on $a$? A: No. The identity element is one fixed element that works for every element of the group. Q: Can the inverse element depend on $a$? A: Yes. The inverse element is chosen after choosing $a\in G$. Q: Does associativity mean commutativity? A: No. Associativity concerns parentheses, while commutativity concerns changing order. Q: Why must the inverse condition be two-sided? A: In the definition of group, the inverse element must satisfy both $a\circ a^{-1}=e$ and $a^{-1}\circ a=e$. ::: :::call-to-action[Group Axioms] subtitle: Continue by studying the four group axioms separately and learning how each axiom is checked in examples. button: Next Lesson | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/four-group-axioms button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/binary-operations-on-sets :::