
## Section2.4Definitions and Properties

The integers mod $n$ and the symmetries of a triangle or a rectangle are examples of groups. A binary operation or law of composition on a set $G$ is a function $G \times G \rightarrow G$ that assigns to each pair $(a,b) \in G \times G$ a unique element $a \circ b\text{,}$ or $ab$ in $G\text{,}$ called the composition of $a$ and $b\text{.}$ A group $(G, \circ )$ is a set $G$ together with a law of composition $(a,b) \mapsto a \circ b$ that satisfies the following axioms.

• The law of composition is associative. That is,

\begin{equation*} (a \circ b) \circ c = a \circ (b \circ c) \end{equation*}

for $a, b, c \in G\text{.}$

• There exists an element $e \in G\text{,}$ called the identity element, such that for any element $a \in G$

\begin{equation*} e \circ a = a \circ e = a. \end{equation*}
• For each element $a \in G\text{,}$ there exists an inverse element in G, denoted by $a^{-1}\text{,}$ such that

\begin{equation*} a \circ a^{-1} = a^{-1} \circ a = e. \end{equation*}

A group $G$ with the property that $a \circ b = b \circ a$ for all $a, b \in G$ is called abelian or commutative. Groups not satisfying this property are said to be nonabelian or noncommutative.

###### Example2.8

The integers ${\mathbb Z } = \{ \ldots , -1, 0, 1, 2, \ldots \}$ form a group under the operation of addition. The binary operation on two integers $m, n \in {\mathbb Z}$ is just their sum. Since the integers under addition already have a well-established notation, we will use the operator $+$ instead of $\circ\text{;}$ that is, we shall write $m + n$ instead of $m \circ n\text{.}$ The identity is $0\text{,}$ and the inverse of $n \in {\mathbb Z}$ is written as $-n$ instead of $n^{-1}\text{.}$ Notice that the set of integers under addition have the additional property that $m + n = n + m$ and therefore form an abelian group.

Most of the time we will write $ab$ instead of $a \circ b\text{;}$ however, if the group already has a natural operation such as addition in the integers, we will use that operation. That is, if we are adding two integers, we still write $m + n\text{,}$ $-n$ for the inverse, and 0 for the identity as usual. We also write $m - n$ instead of $m + (-n)\text{.}$

It is often convenient to describe a group in terms of an addition or multiplication table. Such a table is called a Cayley table.

###### Example2.9

The integers mod $n$ form a group under addition modulo $n\text{.}$ Consider ${\mathbb Z}_5\text{,}$ consisting of the equivalence classes of the integers $0\text{,}$ $1\text{,}$ $2\text{,}$ $3\text{,}$ and $4\text{.}$ We define the group operation on ${\mathbb Z}_5$ by modular addition. We write the binary operation on the group additively; that is, we write $m + n\text{.}$ The element 0 is the identity of the group and each element in ${\mathbb Z}_5$ has an inverse. For instance, $2 + 3 = 3 + 2 = 0\text{.}$ Table 2.10 is a Cayley table for ${\mathbb Z}_5\text{.}$ By Proposition 2.4, ${\mathbb Z}_n = \{0, 1, \ldots, n-1 \}$ is a group under the binary operation of addition mod $n\text{.}$

###### Example2.11

Not every set with a binary operation is a group. For example, if we let modular multiplication be the binary operation on ${\mathbb Z}_n\text{,}$ then ${\mathbb Z}_n$ fails to be a group. The element 1 acts as a group identity since $1 \cdot k = k \cdot 1 = k$ for any $k \in {\mathbb Z}_n\text{;}$ however, a multiplicative inverse for $0$ does not exist since $0 \cdot k = k \cdot 0 = 0$ for every $k$ in ${\mathbb Z}_n\text{.}$ Even if we consider the set ${\mathbb Z}_n \setminus \{0 \}\text{,}$ we still may not have a group. For instance, let $2 \in {\mathbb Z}_6\text{.}$ Then 2 has no multiplicative inverse since

\begin{align*} 0 \cdot 2 & = 0 \qquad 1 \cdot 2 = 2\\ 2 \cdot 2 & = 4 \qquad 3 \cdot 2 = 0\\ 4 \cdot 2 & = 2 \qquad 5 \cdot 2 = 4. \end{align*}

By Proposition 2.4, every nonzero $k$ does have an inverse in ${\mathbb Z}_n$ if $k$ is relatively prime to $n\text{.}$ Denote the set of all such nonzero elements in ${\mathbb Z}_n$ by $U(n)\text{.}$ Then $U(n)$ is a group called the group of units of ${\mathbb Z}_n\text{.}$ Table 2.12 is a Cayley table for the group $U(8)\text{.}$

###### Example2.13

The symmetries of an equilateral triangle described in Section 2.3 form a nonabelian group. As we observed, it is not necessarily true that $\alpha \beta = \beta \alpha$ for two symmetries $\alpha$ and $\beta\text{.}$ Using Table 2.7, which is a Cayley table for this group, we can easily check that the symmetries of an equilateral triangle are indeed a group. We will denote this group by either $S_3$ or $D_3\text{,}$ for reasons that will be explained later.

###### Example2.14

We use ${\mathbb M}_2 ( {\mathbb R})$ to denote the set of all $2 \times 2$ matrices. Let $GL_2({\mathbb R})$ be the subset of ${\mathbb M}_2 ( {\mathbb R})$ consisting of invertible matrices; that is, a matrix

\begin{equation*} A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{equation*}

is in $GL_2( {\mathbb R})$ if there exists a matrix $A^{-1}$ such that $A A^{-1} = A^{-1} A = I\text{,}$ where $I$ is the $2 \times 2$ identity matrix. For $A$ to have an inverse is equivalent to requiring that the determinant of $A$ be nonzero; that is, $\det A = ad - bc \neq 0\text{.}$ The set of invertible matrices forms a group called the general linear group. The identity of the group is the identity matrix

\begin{equation*} I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \end{equation*}

The inverse of $A \in GL_2( {\mathbb R})$ is

\begin{equation*} A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. \end{equation*}

The product of two invertible matrices is again invertible. Matrix multiplication is associative, satisfying the other group axiom. For matrices it is not true in general that $AB = BA\text{;}$ hence, $GL_2({\mathbb R})$ is another example of a nonabelian group.

###### Example2.15

Let

\begin{align*} 1 & = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \qquad I = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}\\ J & = \begin{pmatrix} 0 & i\\ i & 0 \end{pmatrix} \qquad K = \begin{pmatrix} i & 0\\ 0 & -i \end{pmatrix}, \end{align*}

where $i^2 = -1\text{.}$ Then the relations $I^2 = J^2 = K^2 = -1\text{,}$ $IJ=K\text{,}$ $JK = I\text{,}$ $KI = J\text{,}$ $JI = -K\text{,}$ $KJ = -I\text{,}$ and $IK = -J$ hold. The set $Q_8 = \{\pm 1, \pm I, \pm J, \pm K \}$ is a group called the quaternion group. Notice that $Q_8$ is noncommutative.

###### Example2.16

Let ${\mathbb C}^\ast$ be the set of nonzero complex numbers. Under the operation of multiplication ${\mathbb C}^\ast$ forms a group. The identity is $1\text{.}$ If $z = a+bi$ is a nonzero complex number, then

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

is the inverse of $z\text{.}$ It is easy to see that the remaining group axioms hold.

A group is finite, or has finite order, if it contains a finite number of elements; otherwise, the group is said to be infinite or to have infinite order. The order of a finite group is the number of elements that it contains. If $G$ is a group containing $n$ elements, we write $|G| = n\text{.}$ The group ${\mathbb Z}_5$ is a finite group of order $5\text{;}$ the integers ${\mathbb Z}$ form an infinite group under addition, and we sometimes write $|{\mathbb Z}| = \infty\text{.}$

### Subsection2.4.1Basic Properties of Groups

Suppose that $e$ and $e'$ are both identities in $G\text{.}$ Then $eg = ge = g$ and $e'g = ge' = g$ for all $g \in G\text{.}$ We need to show that $e = e'\text{.}$ If we think of $e$ as the identity, then $ee' = e'\text{;}$ but if $e'$ is the identity, then $ee' = e\text{.}$ Combining these two equations, we have $e = ee' = e'\text{.}$

Inverses in a group are also unique. If $g'$ and $g''$ are both inverses of an element $g$ in a group $G\text{,}$ then $gg' = g'g = e$ and $gg'' = g''g = e\text{.}$ We want to show that $g' = g''\text{,}$ but $g' = g'e = g'(gg'') = (g'g)g'' = eg'' = g''\text{.}$ We summarize this fact in the following proposition.

Let $a, b \in G\text{.}$ Then $abb^{-1}a^{-1} = aea^{-1} = aa^{-1} = e\text{.}$ Similarly, $b^{-1}a^{-1}ab = e\text{.}$ But by the previous proposition, inverses are unique; hence, $(ab)^{-1} = b^{-1}a^{-1}\text{.}$

Observe that $a^{-1} (a^{-1})^{-1} = e\text{.}$ Consequently, multiplying both sides of this equation by $a\text{,}$ we have

\begin{equation*} (a^{-1})^{-1} = e (a^{-1})^{-1} = a a^{-1} (a^{-1})^{-1} = ae = a. \end{equation*}

It makes sense to write equations with group elements and group operations. If $a$ and $b$ are two elements in a group $G\text{,}$ does there exist an element $x \in G$ such that $ax = b\text{?}$ If such an $x$ does exist, is it unique? The following proposition answers both of these questions positively.

Suppose that $ax = b\text{.}$ We must show that such an $x$ exists. We can multiply both sides of $ax = b$ by $a^{-1}$ to find $x = ex = a^{-1}ax = a^{-1}b\text{.}$

To show uniqueness, suppose that $x_1$ and $x_2$ are both solutions of $ax = b\text{;}$ then $ax_1 = b = ax_2\text{.}$ So $x_1 = a^{-1}ax_1 = a^{-1}ax_2 = x_2\text{.}$ The proof for the existence and uniqueness of the solution of $xa = b$ is similar.

This proposition tells us that the right and left cancellation laws are true in groups. We leave the proof as an exercise.

We can use exponential notation for groups just as we do in ordinary algebra. If $G$ is a group and $g \in G\text{,}$ then we define $g^0 = e\text{.}$ For $n \in {\mathbb N}\text{,}$ we define

\begin{equation*} g^n = \underbrace{g \cdot g \cdots g}_{n \; \text{times}} \end{equation*}

and

\begin{equation*} g^{-n} = \underbrace{g^{-1} \cdot g^{-1} \cdots g^{-1}}_{n \; \text{times}}. \end{equation*}

We will leave the proof of this theorem as an exercise. Notice that $(gh)^n \neq g^nh^n$ in general, since the group may not be abelian. If the group is ${\mathbb Z}$ or ${\mathbb Z}_n\text{,}$ we write the group operation additively and the exponential operation multiplicatively; that is, we write $ng$ instead of $g^n\text{.}$ The laws of exponents now become

1. $mg + ng = (m+n)g$ for all $m, n \in {\mathbb Z}\text{;}$

2. $m(ng) = (mn)g$ for all $m, n \in {\mathbb Z}\text{;}$

3. $m(g + h) = mg + mh$ for all $n \in {\mathbb Z}\text{.}$

It is important to realize that the last statement can be made only because ${\mathbb Z}$ and ${\mathbb Z}_n$ are commutative groups.

### Subsection2.4.2Historical Note

Although the first clear axiomatic definition of a group was not given until the late 1800s, group-theoretic methods had been employed before this time in the development of many areas of mathematics, including geometry and the theory of algebraic equations.

Joseph-Louis Lagrange used group-theoretic methods in a 1770–1771 memoir to study methods of solving polynomial equations. Later, Évariste Galois (1811–1832) succeeded in developing the mathematics necessary to determine exactly which polynomial equations could be solved in terms of the polynomials'coefficients. Galois' primary tool was group theory.

The study of geometry was revolutionized in 1872 when Felix Klein proposed that geometric spaces should be studied by examining those properties that are invariant under a transformation of the space. Sophus Lie, a contemporary of Klein, used group theory to study solutions of partial differential equations. One of the first modern treatments of group theory appeared in William Burnside's The Theory of Groups of Finite Order [1], first published in 1897.