After studying additive groups of number systems, we now turn to multiplicative groups of nonzero numbers. This change of operation changes the identity element and the inverse notation. Under multiplication, the identity element is $1$, and the inverse of a nonzero element $a$ is $\dfrac{1}{a}$. The exclusion of zero is essential because zero has no multiplicative inverse. In this lesson, students will verify that nonzero rational, real, and complex numbers form groups under multiplication and will learn why some familiar number sets fail to be multiplicative groups. :::definition[Multiplicative Group] Let $G$ be a non-empty set with multiplication $\cdot$. Then $(G,\cdot)$ is called a $multiplicative group$ if $(G,\cdot)$ is a group. In multiplicative notation, the identity element is usually denoted by $1$, and the inverse of $a\in G$ is denoted by $a^{-1}$ or $\dfrac{1}{a}$ when the operation is ordinary multiplication. ::: The set must be chosen carefully in multiplicative examples. The set $\mathbb{Q}$ under multiplication is not a group because $0\in\mathbb{Q}$ and zero has no multiplicative inverse. Once zero is removed, every remaining rational number has a rational reciprocal. Thus $(\mathbb{Q}-\{0\},\cdot)$ becomes a group. The same principle works for $\mathbb{R}-\{0\}$ and $\mathbb{C}-\{0\}$. A multiplicative inverse is a reciprocal, so zero must be excluded. Enter a rational number $p/q$. The preview checks whether the rational number is defined and nonzero, then displays its reciprocal and verifies that the product is $1$. If the numerator is zero, the preview shows exactly why the inverse axiom fails. :::scientific-preview[Multiplicative Inverse Explorer] ::: :::theorem Let $F$ be one of the sets $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$. Then $(F-\{0\},\cdot)$ is a group 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 nonzero elements of $F$. To prove that $(F-\{0\},\cdot)$ is a group under ordinary multiplication. [1] To prove closure. Let $a,b\in F-\{0\}$. Then $a\in F$, $b\in F$, $a\neq0$, and $b\neq0$. Since $F$ is closed under ordinary multiplication, $ab\in F$. Also, since $a\neq0$ and $b\neq0$, we get $$\begin{equation*} ab\neq0. \end{equation*}$$ Therefore $ab\in F-\{0\}$. [2] To prove associativity. Let $a,b,c\in F-\{0\}$. Since ordinary multiplication in $F$ is associative, $$\begin{equation*} (ab)c=a(bc). \end{equation*}$$ [3] To prove the existence of identity element. There exists $1\in F-\{0\}$ such that $$\begin{equation*} a\cdot1=1\cdot a=a \quad \forall\, a\in F-\{0\}. \end{equation*}$$ Therefore $1$ is the identity element. [4] To prove the existence of inverse elements. Let $a\in F-\{0\}$. Then $a\neq0$. Thus $\dfrac{1}{a}\in F$ and $\dfrac{1}{a}\neq0$. Therefore $\dfrac{1}{a}\in F-\{0\}$. 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. $\square$ ::: The previous theorem covers three standard examples at once. In $(\mathbb{Q}-\{0\},\cdot)$, the inverse of $\dfrac{2}{3}$ is $\dfrac{3}{2}$. In $(\mathbb{R}-\{0\},\cdot)$, the inverse of $\sqrt{2}$ is $\dfrac{1}{\sqrt{2}}$. In $(\mathbb{C}-\{0\},\cdot)$, the inverse of $a+ib$ with $a^2+b^2\neq0$ is $$\begin{equation*} \frac{1}{a+ib}=\frac{a-ib}{a^2+b^2}. \end{equation*}$$ Each inverse remains inside the same nonzero number system. :::example Let $(\mathbb{Q}-\{0\},\cdot)$ be the set of nonzero rational numbers under multiplication. Then $(\mathbb{Q}-\{0\},\cdot)$ is a group. [1] If $a,b\in\mathbb{Q}-\{0\}$, then $ab\in\mathbb{Q}-\{0\}$. [2] If $a,b,c\in\mathbb{Q}-\{0\}$, then $(ab)c=a(bc)$. [3] The identity element is $1$. [4] The inverse of $a\in\mathbb{Q}-\{0\}$ is $\dfrac{1}{a}\in\mathbb{Q}-\{0\}$. Hence, $(\mathbb{Q}-\{0\},\cdot)$ is a group. ::: :::example Let $G=\{x\in\mathbb{R}:x>0\}$. Then $(G,\cdot)$ is a group under ordinary multiplication. If $a,b>0$, then $ab>0$, so closure holds. Real multiplication is associative. The identity element is $1$, and $1>0$. For $a>0$, the inverse $\dfrac{1}{a}$ is also positive. Hence, $(G,\cdot)$ is a group. ::: :::solved-problem Determine whether $(\mathbb{Z}-\{0\},\cdot)$ is a group. ::: :::solution Let $G=\mathbb{Z}-\{0\}$ and let $\cdot$ be ordinary multiplication. To determine whether $(G,\cdot)$ is a group. [1] Closure: Let $a,b\in G$. Then $a$ and $b$ are nonzero integers. Hence $ab$ is a nonzero integer, and so $ab\in G$. [2] Associativity: For all $a,b,c\in G$, $$\begin{equation*} (ab)c=a(bc). \end{equation*}$$ [3] Identity element: There exists $1\in G$ such that $$\begin{equation*} a\cdot1=1\cdot a=a \quad \forall\, a\in G. \end{equation*}$$ [4] Inverse element: Take $2\in G$. If $2$ has an inverse in $G$, then there exists $b\in G$ such that $$\begin{equation*} 2b=1. \end{equation*}$$ Then $$\begin{align*} 2b=1 &\implies b=\frac{1}{2}. \end{align*}$$ But $\dfrac{1}{2}\notin\mathbb{Z}-\{0\}$. Therefore the inverse element condition is not satisfied. Hence, $(\mathbb{Z}-\{0\},\cdot)$ is not a group. $\square$ ::: :::remark Zero is not removed merely for convenience. It is removed because a multiplicative group requires every element to have a multiplicative inverse. If zero were included, we would need an element $b$ such that $0\cdot b=1$, which is impossible. Thus zero is the obstruction in $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ under multiplication. ::: Use the calculator below to test multiplicative inverses in rational numbers. Enter a nonzero rational number as two integers $p$ and $q$, representing $\dfrac{p}{q}$. The calculator returns the reciprocal and explains why the number must be nonzero. This reinforces the inverse condition in multiplicative groups. :::calculator[Rational Reciprocal Calculator] ::: :::exercise [1] Prove that $(\mathbb{R}-\{0\},\cdot)$ is a group. [2] Determine whether $(\mathbb{Q},\cdot)$ is a group. [3] Determine whether $(\mathbb{Z}-\{0\},\cdot)$ is a group. [4] Find the inverse of $3-4i$ in $(\mathbb{C}-\{0\},\cdot)$. [5] Determine whether the set of positive rational numbers is a group under multiplication. ::: :::answer [1] Closure, associativity, identity $1$, and inverse $\dfrac{1}{a}$ all hold in $\mathbb{R}-\{0\}$. [2] No. The element $0$ has no multiplicative inverse. [3] No. The element $2$ has no multiplicative inverse in $\mathbb{Z}-\{0\}$. [4] The inverse is $\dfrac{3+4i}{25}$. [5] Yes. The product of two positive rational numbers is positive rational, the identity is $1$, and the reciprocal of a positive rational number is positive rational. ::: :::faq Q: Why is zero removed in multiplicative groups? A: Zero has no multiplicative inverse. Q: What is the identity element in a multiplicative group of numbers? A: The identity element is $1$. Q: Is $(\mathbb{Z}-\{0\},\cdot)$ a group? A: No. For example, $2$ has no inverse in $\mathbb{Z}-\{0\}$. Q: Is $(\mathbb{Q}-\{0\},\cdot)$ a group? A: Yes. Every nonzero rational number has a nonzero rational reciprocal. ::: :::call-to-action[Matrix Groups] subtitle: Continue to matrix groups under addition, where the same group axioms are verified for matrix spaces. button: Next Lesson | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/matrix-groups-under-addition button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/additive-groups-number-systems :::