After studying the four group axioms separately, the next skill is verification. To prove that a given algebraic system is a group, we must check the four axioms in an organized order. This is not a matter of guessing from familiar examples. Each verification must mention the set, the operation, closure, associativity, identity, and inverses. This method is useful throughout abstract algebra because the same pattern appears later in subgroups, matrix groups, permutation groups, cyclic groups, and quotient groups. In this lesson, students will learn a reliable verification method for proving that a structure is a group. The standard verification method has four steps. First, prove closure. Second, prove associativity. Third, identify the identity element and verify it from both sides. Fourth, for an arbitrary element of the set, find an inverse element that also belongs to the set. The word arbitrary is important: checking one or two elements is not enough unless the set is finite and all elements are checked. For infinite sets such as $\mathbb{Z}$, $\mathbb{Q}$, or $\mathbb{R}$, the proof must work for a general element. :::definition[Group Verification] Let $G$ be a non-empty set and let $\circ$ be a binary operation on $G$. To verify that $(G,\circ)$ is a group, one must prove: (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 $e\circ a=a\circ e=a \quad \forall\, a\in G$. (iv) For each $a\in G$, there exists $a^{-1}\in G$ such that $a\circ a^{-1}=a^{-1}\circ a=e$. ::: A complete group proof follows a sequence: closure, associativity, identity, and inverses. Mark the steps that have been proved in a proposed solution. The flowchart shows whether the verification can continue or where it stops. This reinforces that a missing axiom makes the proof incomplete, even when the other steps are correct. :::scientific-preview[Verification Flowchart] ::: A proof of group verification should not begin by saying that the result is known. It should show how each axiom follows from the properties of the given set and operation. When the operation is ordinary addition or multiplication, associativity may be inherited from the known associativity of numbers. However, closure and inverse conditions must still be checked with respect to the particular set. For example, integer multiplication is associative, but $(\mathbb{Z},\cdot)$ is not a group because most integers do not have multiplicative inverses in $\mathbb{Z}$. :::theorem Let $m\in\mathbb{Z}$. If $$\begin{equation*} m\mathbb{Z}=\{mk:k\in\mathbb{Z}\}, \end{equation*}$$ then $(m\mathbb{Z},+)$ is a group under ordinary addition. ::: :::proof Given that $m\in\mathbb{Z}$ and $$\begin{equation*} m\mathbb{Z}=\{mk:k\in\mathbb{Z}\}. \end{equation*}$$ To prove that $(m\mathbb{Z},+)$ is a group under ordinary addition. [1] To prove closure. Let $a,b\in m\mathbb{Z}$. Then there exist $r,s\in\mathbb{Z}$ such that $$\begin{equation*} a=mr,\quad b=ms. \end{equation*}$$ Now $$\begin{align*} a+b&=mr+ms\\ &=m(r+s). \end{align*}$$ Since $r+s\in\mathbb{Z}$, we get $a+b\in m\mathbb{Z}$. Therefore $m\mathbb{Z}$ is closed under addition. [2] To prove associativity. Let $a,b,c\in m\mathbb{Z}$. Since addition of integers is associative, $$\begin{equation*} (a+b)+c=a+(b+c). \end{equation*}$$ Therefore addition is associative on $m\mathbb{Z}$. [3] To prove the existence of identity element. Since $0=m\cdot 0$ and $0\in\mathbb{Z}$, we get $0\in m\mathbb{Z}$. For every $a\in m\mathbb{Z}$, $$\begin{equation*} 0+a=a+0=a. \end{equation*}$$ Therefore $0$ is the identity element of $(m\mathbb{Z},+)$. [4] To prove the existence of inverse elements. Let $a\in m\mathbb{Z}$. Then there exists $r\in\mathbb{Z}$ such that $a=mr$. Now $$\begin{align*} -a&=-mr\\ &=m(-r). \end{align*}$$ Since $-r\in\mathbb{Z}$, we get $-a\in m\mathbb{Z}$. Also $$\begin{equation*} a+(-a)=(-a)+a=0. \end{equation*}$$ Therefore every element of $m\mathbb{Z}$ has an inverse in $m\mathbb{Z}$ under addition. Hence, $(m\mathbb{Z},+)$ is a group. $\square$ ::: This proof illustrates the usual style of group verification. The closure part uses the internal description of the set. The identity part checks that the identity element actually belongs to the set. The inverse part starts with an arbitrary element $a$ and proves that its inverse also belongs to the same set. Students often forget this membership check. For $(m\mathbb{Z},+)$, it is not enough to say that the inverse of $a$ is $-a$; one must also prove $-a\in m\mathbb{Z}$. :::example Let $G=\mathbb{Q}$ and let $+$ be ordinary addition. Then $(\mathbb{Q},+)$ is a group. [1] If $a,b\in\mathbb{Q}$, then $a+b\in\mathbb{Q}$. [2] If $a,b,c\in\mathbb{Q}$, then $(a+b)+c=a+(b+c)$. [3] There exists $0\in\mathbb{Q}$ such that $a+0=0+a=a$ for every $a\in\mathbb{Q}$. [4] If $a\in\mathbb{Q}$, then $-a\in\mathbb{Q}$ and $a+(-a)=(-a)+a=0$. Hence, $(\mathbb{Q},+)$ is a group. ::: :::solved-problem Let $G=\{x\in\mathbb{R}:x>0\}$ and let $\cdot$ be ordinary multiplication. Prove that $(G,\cdot)$ is a group. ::: :::solution Let $G=\{x\in\mathbb{R}:x>0\}$ and let $\cdot$ be ordinary multiplication. To prove that $(G,\cdot)$ is a group. [1] To prove closure. Let $a,b\in G$. Then $a>0$ and $b>0$. Therefore $$\begin{equation*} ab>0. \end{equation*}$$ Thus $ab\in G$. [2] To prove associativity. Let $a,b,c\in G$. Since multiplication of real numbers is associative, $$\begin{equation*} (ab)c=a(bc). \end{equation*}$$ [3] To prove the existence of identity element. Since $1>0$, we get $1\in G$. Also $$\begin{equation*} a\cdot 1=1\cdot a=a \quad \forall\, a\in G. \end{equation*}$$ Therefore $1$ is the identity element of $(G,\cdot)$. [4] To prove the existence of inverse elements. Let $a\in G$. Then $a>0$, and hence $$\begin{equation*} \frac{1}{a}>0. \end{equation*}$$ Thus $\dfrac{1}{a}\in G$. Also $$\begin{equation*} a\cdot \frac{1}{a}=\frac{1}{a}\cdot a=1. \end{equation*}$$ Therefore every element of $G$ has an inverse in $G$ under multiplication. Hence, $(G,\cdot)$ is a group. $\square$ ::: :::remark When verifying a group, associativity is often the easiest part for standard number systems because it is inherited from ordinary addition or multiplication. For a newly defined operation, associativity is usually the longest part of the proof. In either case, the proof must explicitly mention associativity. A group verification that skips associativity is incomplete. ::: Use the calculator below as a checklist while verifying a group. Select which axioms have been proved in a proposed solution. The calculator does not decide the mathematics for you; it reminds you that all four conditions must be established. This is especially useful when writing solutions for examples where one axiom fails and the structure is not a group. :::calculator[Group Verification Checklist Calculator] ::: :::exercise [1] Prove that $(2\mathbb{Z},+)$ is a group. [2] Prove that $(\mathbb{R},+)$ is a group. [3] Determine whether $(\mathbb{Q},\cdot)$ is a group. [4] Let $G=\{x\in\mathbb{R}:x<0\}$. Determine whether $(G,\cdot)$ is a group. [5] Explain why an inverse verification must prove membership in the set. ::: :::answer [1] If $a=2r$ and $b=2s$, then $a+b=2(r+s)\in2\mathbb{Z}$. The identity is $0$, and the inverse of $2r$ is $-2r=2(-r)\in2\mathbb{Z}$. [2] Closure, associativity, identity $0$, and additive inverse $-a$ all hold in $\mathbb{R}$. [3] No. The element $0\in\mathbb{Q}$ has no multiplicative inverse in $\mathbb{Q}$. [4] No. If $a,b<0$, then $ab>0$, so closure fails. [5] The inverse must belong to the same set because the inverse axiom requires $a^{-1}\in G$. ::: :::faq Q: Is it enough to give examples of inverses? A: No. For a general group verification, one must prove that every element has an inverse. Q: Can closure be omitted if the operation is familiar? A: No. Familiar operations may fail closure on restricted sets. Q: Why is the identity checked before inverses? A: The inverse of an element is defined relative to the identity element. Q: What is the most common missing step in group verification? A: Students often forget to prove that the inverse element belongs to the set. ::: :::call-to-action[Additive Groups] subtitle: Continue to standard additive groups of number systems, where this verification method is applied repeatedly. button: Next Lesson | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/additive-groups-number-systems button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/four-group-axioms :::