After defining a group, the next task is to understand the four group axioms separately. A group is not obtained by checking only one attractive property of an operation; all four conditions must work together. Closure keeps the result inside the set, associativity controls parentheses, the identity element leaves every element unchanged, and inverse elements bring each element back to the identity. These axioms appear simple, but many mistakes in abstract algebra begin from confusing one axiom with another. In this lesson, students will learn the exact role of each axiom and how to detect the failure of a group structure by locating the first failed axiom. :::definition[Closure] Let $G$ be a non-empty set and let $\circ$ be a rule on $G$. The set $G$ is said to be $closed$ under $\circ$ if $$\begin{equation*} a\circ b\in G \quad \forall\, a,b\in G. \end{equation*}$$ Closure means that the operation never takes two elements of $G$ outside $G$. ::: Closure is the first axiom students usually check because it decides whether the operation belongs to the set in a meaningful way. For example, if $a,b\in\mathbb{Z}$, then $a+b\in\mathbb{Z}$, so $\mathbb{Z}$ is closed under addition. However, if $a,b\in\mathbb{N}$, then $a-b$ need not belong to $\mathbb{N}$. Thus subtraction is not closed on $\mathbb{N}$. A common classroom error is to say that subtraction is defined and therefore closed. Definition of the formula is not enough; the result must stay inside the set. :::definition[Associative Law] Let $G$ be a non-empty set and let $\circ$ be a binary operation on $G$. The operation $\circ$ is said to satisfy the $associative law$ on $G$ if $$\begin{equation*} (a\circ b)\circ c=a\circ(b\circ c) \quad \forall\, a,b,c\in G. \end{equation*}$$ Associativity says that the placement of parentheses does not change the result. ::: Associativity is about grouping, not about changing order. The equality $(a\circ b)\circ c=a\circ(b\circ c)$ compares two ways of placing parentheses while keeping the order $a,b,c$ fixed. It does not say that $a\circ b=b\circ a$. This distinction is essential because many groups are associative but not commutative. In early examples from number systems, addition and multiplication are familiar associative operations, but a newly defined operation must be checked directly. :::definition[Identity Element] Let $G$ be a non-empty set and let $\circ$ be a binary operation on $G$. An element $e\in G$ is called an $identity element$ for $\circ$ on $G$ if $$\begin{equation*} e\circ a=a\circ e=a \quad \forall\, a\in G. \end{equation*}$$ The identity element must work from the left and from the right for every element of $G$. ::: The identity element is one element, not a different element for each $a$. Under ordinary addition, the identity element is $0$. Under ordinary multiplication, the identity element is $1$. If an operation has a left identity but not a right identity, or a right identity but not a left identity, then the group identity axiom is not satisfied. This two-sided condition prevents hidden one-sided behaviour from being mistaken for a group structure. :::definition[Inverse Element] Let $G$ be a non-empty set with a binary operation $\circ$, and let $e$ be an identity element for $\circ$ on $G$. For $a\in G$, an element $b\in G$ is called an $inverse element$ of $a$ if $$\begin{equation*} a\circ b=b\circ a=e. \end{equation*}$$ When the inverse of $a$ is denoted by $a^{-1}$, the condition becomes $$\begin{equation*} a\circ a^{-1}=a^{-1}\circ a=e. \end{equation*}$$ ::: Inverse elements are checked only after an identity element has been found. In $(\mathbb{Z},+)$, the inverse of $a$ is $-a$ because $a+(-a)=0$. In $(\mathbb{Q}-\{0\},\cdot)$, the inverse of $a$ is $\dfrac{1}{a}$ because $a\cdot\dfrac{1}{a}=1$. Students often try to find inverses before identifying the identity element. That is unsafe, because inverse means returning to the identity. A structure fails to be a group as soon as one required axiom fails. Choose a modulus and an operation on $\{0,1,\ldots,n-1\}$. The preview checks the four axioms in order and reports the first obstruction. This helps separate closure, associativity, identity, and inverse conditions instead of blending them into one vague test. :::scientific-preview[Axiom Failure Explorer] ::: :::example Let $(\mathbb{Z},+)$ be the set of integers under ordinary addition. We verify the four group axioms. [1] Closure: Let $a,b\in\mathbb{Z}$. Then $$\begin{equation*} a+b\in\mathbb{Z}. \end{equation*}$$ [2] Associativity: Let $a,b,c\in\mathbb{Z}$. Then $$\begin{equation*} (a+b)+c=a+(b+c). \end{equation*}$$ [3] Identity element: There exists $0\in\mathbb{Z}$ such that $$\begin{equation*} 0+a=a+0=a \quad \forall\, a\in\mathbb{Z}. \end{equation*}$$ [4] Inverse element: For each $a\in\mathbb{Z}$, there exists $-a\in\mathbb{Z}$ such that $$\begin{equation*} a+(-a)=(-a)+a=0. \end{equation*}$$ Hence, $(\mathbb{Z},+)$ is a group. ::: :::example Let $(\mathbb{N},+)$ be the set of positive integers under ordinary addition, where $\mathbb{N}=\{1,2,3,\ldots\}$. Then $(\mathbb{N},+)$ is not a group. Closure is satisfied because the sum of two positive integers is a positive integer. Associativity is also satisfied because addition of integers is associative. However, there is no element $e\in\mathbb{N}$ such that $$\begin{equation*} e+a=a+e=a \quad \forall\, a\in\mathbb{N}. \end{equation*}$$ The additive identity would have to be $0$, but $0\notin\mathbb{N}$. Therefore the identity axiom fails. Hence, $(\mathbb{N},+)$ is not a group. ::: :::theorem Let $G$ be a non-empty set and let $\circ$ be a binary operation on $G$. If one of the closure, associativity, identity, or inverse conditions fails, then $(G,\circ)$ is not a group. ::: :::proof Given that $G$ is a non-empty set and $\circ$ is a binary operation on $G$. To prove that if one of the closure, associativity, identity, or inverse conditions fails, then $(G,\circ)$ is not a group. By definition, $(G,\circ)$ is a group if all four conditions hold: (i) closure, (ii) associativity, (iii) existence of an identity element, (iv) existence of inverse elements for all elements. If possible, let $(G,\circ)$ be a group and one of these conditions fails. Since $(G,\circ)$ is a group, every condition in the definition of group must hold. Therefore the failed condition must hold. A contradiction. Hence, if one of the four group conditions fails, then $(G,\circ)$ is not a group. $\square$ ::: :::solved-problem Determine whether $(\mathbb{R}-\{0\},+)$ is a group. ::: :::solution Let $G=\mathbb{R}-\{0\}$ and let $+$ be ordinary addition. To determine whether $(G,+)$ is a group. We first check closure. Take $1,-1\in G$. Then $$\begin{align*} 1+(-1)&=0. \end{align*}$$ But $0\notin G$. Therefore $G$ is not closed under ordinary addition. Since closure fails, $(G,+)$ is not a group. $\square$ ::: Use the calculator below to test the four group axioms for addition or multiplication modulo $n$ on the set $\{0,1,\ldots,n-1\}$. The result is not meant to replace a proof, but it helps students see which axiom fails first. Try addition modulo $6$ and multiplication modulo $6$. The comparison shows that two operations on the same finite set can have completely different group behaviour. :::calculator[Four Axiom Test Calculator] ::: :::exercise [1] State the closure axiom for a binary operation $\circ$ on $G$. [2] State the associative law for a binary operation $\circ$ on $G$. [3] Determine whether $(\mathbb{N},+)$ has an identity element when $\mathbb{N}=\{1,2,3,\ldots\}$. [4] Determine whether $(\mathbb{Z},\cdot)$ satisfies the inverse axiom. [5] Explain why associativity is not the same as commutativity. ::: :::answer [1] $a\circ b\in G \quad \forall\, a,b\in G$. [2] $(a\circ b)\circ c=a\circ(b\circ c) \quad \forall\, a,b,c\in G$. [3] No. The additive identity is $0$, but $0\notin\mathbb{N}$. [4] No. For example, $2\in\mathbb{Z}$ has no multiplicative inverse in $\mathbb{Z}$. [5] Associativity concerns the placement of parentheses, while commutativity concerns changing the order of two elements. ::: :::faq Q: Which axiom should be checked first? A: Closure is usually checked first because the operation must stay inside the set before the remaining group axioms can be used. Q: Does associativity mean $a\circ b=b\circ a$? A: No. Associativity compares $(a\circ b)\circ c$ and $a\circ(b\circ c)$. Q: Can a set have more than one identity element? A: In a group, the identity element is unique. That result is proved later in the elementary properties of groups. Q: Must every element have the same inverse? A: No. The inverse element depends on the chosen element. ::: :::call-to-action[Verification of a Group] subtitle: Continue by learning a systematic method for verifying all four group axioms in examples. button: Next Lesson | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/verify-group-axioms button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/group-definition :::