Multiplicative Groups

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

\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.

Let

G

be a non-empty set with multiplication

\cdot

. Then

(G,\cdot)

is called a

\textbf{multiplicative group}

(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}

\dfrac{1}{a}

when the operation is ordinary multiplication.

1

, and the inverse of a nonzero element

a

\dfrac{1}{a}

Let

G

be a non-empty set with multiplication

\cdot

. Then

(G,\cdot)

is called a

\textbf{multiplicative group}

(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}

\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\}

Let

F

be one of the sets

\mathbb{Q}

\mathbb{R}

, or

\mathbb{C}

. Then

(F-\{0\},\cdot)

is a group under ordinary multiplication.

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}

\dfrac{3}{2}

. In

(\mathbb{R}-\{0\},\cdot)

, the inverse of

\sqrt{2}

\dfrac{1}{\sqrt{2}}

. In

(\mathbb{C}-\{0\},\cdot)

, the inverse of

a+ib

with

a^2+b^2\neq0

\begin{equation*} \frac{1}{a+ib}=\frac{a-ib}{a^2+b^2}. \end{equation*}

Each inverse remains inside the same nonzero number system.

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\}

\dfrac{1}{a}\in\mathbb{Q}-\{0\}

. Hence,

(\mathbb{Q}-\{0\},\cdot)

is a group.

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.

Determine whether

(\mathbb{Z}-\{0\},\cdot)

is a group.

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

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.

[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

(\mathbb{C}-\{0\},\cdot)

. [5] Determine whether the set of positive rational numbers is a group under multiplication.

[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.

Zero has no multiplicative inverse.

The identity element is

1

No. For example,

2

has no inverse in

\mathbb{Z}-\{0\}

Yes. Every nonzero rational number has a nonzero rational reciprocal.

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