After studying the order of an element, we now return to the standard examples of groups and examine them with greater confidence. The order results become meaningful only after students can recognize familiar groups quickly and correctly. Number systems provide the first reliable family of examples because ordinary addition and multiplication have well-known algebraic properties. In this lesson, students will review groups from number systems, distinguish additive and multiplicative notation, and learn why certain familiar number systems fail to be groups under some operations. :::definition[Number System Group] Let $G$ be a non-empty subset of a number system such as $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$. If $G$ is a group under ordinary addition or ordinary multiplication, then $(G,+)$ or $(G,\cdot)$ is called a $number system group$. ::: For addition, the identity element is $0$, and the inverse of $a$ is $-a$. For multiplication, the identity element is $1$, and the inverse of $a$ is $\dfrac{1}{a}$ when $a\neq0$. This is why zero is harmless in additive groups but dangerous in multiplicative groups. The set $\mathbb{Q}$ is a group under addition, but not under multiplication because $0$ has no multiplicative inverse. The set $\mathbb{Q}-\{0\}$ is a group under multiplication because every nonzero rational number has a nonzero rational reciprocal. :::theorem The structures $(\mathbb{Z},+)$, $(\mathbb{Q},+)$, $(\mathbb{R},+)$, and $(\mathbb{C},+)$ are groups under ordinary addition. ::: :::proof Given that $F$ is one of the sets $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$, and $+$ is ordinary addition. To prove that $(F,+)$ is a group. [1] To prove closure. Let $a,b\in F$. Then $$\begin{equation*} a+b\in F. \end{equation*}$$ Therefore $F$ is closed under addition. [2] To prove associativity. Let $a,b,c\in F$. Then $$\begin{equation*} (a+b)+c=a+(b+c). \end{equation*}$$ Therefore addition is associative on $F$. [3] To prove the existence of identity element. There exists $0\in F$ such that $$\begin{equation*} a+0=0+a=a \quad \forall\, a\in F. \end{equation*}$$ Therefore $0$ is the identity element. [4] To prove the existence of inverse elements. Let $a\in F$. Then $-a\in F$ and $$\begin{equation*} a+(-a)=(-a)+a=0. \end{equation*}$$ Therefore every element of $F$ has an additive inverse in $F$. Hence, $(F,+)$ is a group. $\square$ ::: :::theorem The structures $(\mathbb{Q}-\{0\},\cdot)$, $(\mathbb{R}-\{0\},\cdot)$, and $(\mathbb{C}-\{0\},\cdot)$ are groups under ordinary multiplication. ::: :::proof Given that $F$ is one of the sets $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$, and $F-\{0\}$ is the set of all nonzero elements of $F$. To prove that $(F-\{0\},\cdot)$ is a group. [1] To prove closure. Let $a,b\in F-\{0\}$. Then $a\neq0$ and $b\neq0$. Since $F$ is closed under ordinary multiplication, $ab\in F$. Also $ab\neq0$. Therefore $$\begin{equation*} ab\in F-\{0\}. \end{equation*}$$ [2] To prove associativity. Let $a,b,c\in F-\{0\}$. Then $$\begin{equation*} (ab)c=a(bc). \end{equation*}$$ Therefore multiplication is associative. [3] To prove the existence of identity element. Since $1\in F-\{0\}$ and $$\begin{equation*} a\cdot1=1\cdot a=a \quad \forall\, a\in F-\{0\}, \end{equation*}$$ the identity element is $1$. [4] To prove the existence of inverse elements. Let $a\in F-\{0\}$. Then $a\neq0$, so $\dfrac{1}{a}\in F-\{0\}$. Also $$\begin{equation*} a\cdot\frac{1}{a}=\frac{1}{a}\cdot a=1. \end{equation*}$$ Therefore every element of $F-\{0\}$ has a multiplicative inverse in $F-\{0\}$. Hence, $(F-\{0\},\cdot)$ is a group. $\square$ ::: :::example The structure $(\mathbb{Z},+)$ is an infinite group. Its identity element is $0$, and the inverse of $m\in\mathbb{Z}$ is $-m$. If $m\neq0$, then $m$ has infinite order in $(\mathbb{Z},+)$. This example connects the present lesson with the previous section on order of an element. ::: :::example The structure $(\mathbb{Z}-\{0\},\cdot)$ is not a group. The identity element $1$ belongs to the set, and multiplication is associative, but the inverse condition fails. For instance, $2\in\mathbb{Z}-\{0\}$, but its multiplicative inverse $\dfrac{1}{2}$ does not belong to $\mathbb{Z}-\{0\}$. ::: :::solved-problem Determine whether $(\mathbb{Q},\cdot)$ is a group. ::: :::solution Let $\mathbb{Q}$ be the set of rational numbers under ordinary multiplication. To determine whether $(\mathbb{Q},\cdot)$ is a group. The element $0$ belongs to $\mathbb{Q}$. If $(\mathbb{Q},\cdot)$ were a group, then every element of $\mathbb{Q}$ would have a multiplicative inverse in $\mathbb{Q}$. In particular, there would exist $q\in\mathbb{Q}$ such that $$\begin{equation*} 0\cdot q=1. \end{equation*}$$ But $$\begin{equation*} 0\cdot q=0 \quad \forall\, q\in\mathbb{Q}. \end{equation*}$$ Therefore no such $q$ exists. Thus the inverse axiom fails. Hence, $(\mathbb{Q},\cdot)$ is not a group. $\square$ ::: :::solved-problem Prove that the set of positive real numbers is a group under multiplication. ::: :::solution Let $$\begin{equation*} G=\{x\in\mathbb{R}:x>0\}. \end{equation*}$$ 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. For all $a,b,c\in G$, $$\begin{equation*} (ab)c=a(bc). \end{equation*}$$ [3] To prove identity element. Since $1>0$, we get $1\in G$. Also $$\begin{equation*} a\cdot1=1\cdot a=a \quad \forall\, a\in G. \end{equation*}$$ [4] To prove inverse elements. Let $a\in G$. Then $a>0$, so $$\begin{equation*} \frac{1}{a}>0. \end{equation*}$$ Therefore $\dfrac{1}{a}\in G$. Also $$\begin{equation*} a\cdot\frac{1}{a}=\frac{1}{a}\cdot a=1. \end{equation*}$$ Hence, $(G,\cdot)$ is a group. $\square$ ::: :::remark When checking number system examples, students should first identify the operation. The same set may be a group under one operation and not under another. For instance, $\mathbb{Q}$ is a group under addition, but $\mathbb{Q}$ is not a group under multiplication. Removing zero changes the multiplicative example into a group. ::: Number-system examples are easiest to compare by checking the four group axioms in order. We observe how changing the operation changes the identity and inverse requirements. Choose an example and notice which axiom fails when the structure is not a group. This makes the obstruction visible instead of hiding it inside a long proof. :::scientific-preview[Number System Axiom Chart] @libraries: plotly ::: Use the calculator to test standard number-system examples. Choose the set and operation, then read the identity element, inverse rule, and group verdict. The calculation is a conceptual checklist that points to the exact axiom responsible for success or failure. :::calculator[Number Group Test] ::: :::exercise [1] Prove that $(\mathbb{R},+)$ is a group. [2] Prove that $(\mathbb{C}-\{0\},\cdot)$ is a group. [3] Determine whether $(\mathbb{N},+)$ is a group when $\mathbb{N}=\{1,2,3,\ldots\}$. [4] Determine whether $(\mathbb{Z},\cdot)$ is a group. [5] Find the identity and inverse of $\dfrac{3}{5}$ in $(\mathbb{Q}-\{0\},\cdot)$. ::: :::answer [1] Closure, associativity, identity $0$, and inverse $-a$ all hold in $\mathbb{R}$. [2] Closure, associativity, identity $1$, and inverse $\dfrac{1}{z}$ all hold in $\mathbb{C}-\{0\}$. [3] No. The identity element $0$ is not in $\mathbb{N}$. [4] No. For example, $2$ has no multiplicative inverse in $\mathbb{Z}$. [5] The identity is $1$, and the inverse of $\dfrac{3}{5}$ is $\dfrac{5}{3}$. ::: :::faq Q: Is every number system a group under every operation? A: No. The set and operation must be checked together. Q: Why is $\mathbb{Q}$ not a group under multiplication? A: The element $0$ has no multiplicative inverse. Q: Why is $\mathbb{Q}-\{0\}$ a group under multiplication? A: Every nonzero rational number has a nonzero rational reciprocal. Q: What is the additive inverse of $a$? A: It is $-a$. ::: :::call-to-action[Roots of Unity] subtitle: Continue to finite groups formed by complex roots of unity under multiplication. button: Next Lesson | /abstract-algebra/introduction-to-groups/standard-examples-of-groups/roots-of-unity-groups button-ghost: Previous Section | /abstract-algebra/introduction-to-groups/order-of-an-element/orders-of-products-and-conjugates :::