Definition of a group

Before giving the definition of a group, it is necessary to define a binary operation on a set $S$.

Definition 2.1.1   A binary operation on a set $S$ is a mapping of $S \times S$ into $S$.

In other words a binary operation on $S$ is given when to every pair $(a,b$) of elements of $S$ another element $c \in S$ is associated. The fact that $c \in S$ is sometimes expressed by saying a binary operation, or just an operation, on $S$ is closed. This image element, $c$, is usually denoted by $ab$ or $a + b$; still other notations such as $a \circ b$ or $a * b$ are also frequently used. We will adopt for the most part the ``multiplicative'' notation $ab$ instead of the ``additive'' notation $a + b$. As a word of warning, we remind the reader that $S$ is an arbitrary set not necessarily a set of numbers and one should not give any special significance to the juxtaposition (or product) $ab$, such as the product of numbers. The elements of $S$, for example, could be mappings (functions). We shall at times speak of the ``product of $a$ and $b$'' as the image element $ab$, and we also sometimes will use ``sum of $a$ and $b$'' for $a + b$, when this notation is in use, but again the reader should not in general think of these elements as numbers. The reader should also note that a binary operation is defined on an ordered pair of set elements, so that in general, $ab$ and $ba$ are distinct.

We now proceed to the definition of a group.

Definition 2.1.2   A group is a set $G$ together with a binary operation defined on $G$ such that

1. $a(bc) = (ab)c$, for all $a,b,c \in G$ (associative law),
2. There exists an element $e \in G$, called the identity element, such that $ae = ea = a$ for all $a \in G$,
3. To each $a \in G$, there exists an element $a^{- 1}\in G$, called the inverse of $a$, such that $aa^{- 1} = a^{- 1}a = e$.

Let us remark immediately that since a group $G$ has a binary operation defined on it the operation is closed, i.e., for any $a,b \in G$ it must be true that $ab \in G$. We also note that it is customary to talk of a group $G$ in a given discussion. This is actually not precise because a group, as just defined, is a set $G$ together with a binary operation and it is possible that on a given set $G$ a number of binary operations can be introduced such that the set $G$ together with each of these operations is a group. In any discussion, however, the binary operation will be fixed and there will be no confusion in speaking just of the group $G$.

A set $G$ together with a binary operation which satisfies condition (1) of Definition 2.1.2 is called a semi-group.

Before proving some simple consequences of the axioms of a group, we shall give a number of examples of groups, semi-groups, and objects which are neither. More examples will appear during the course of our development.

Example 2.1.3   Take $G = \mathbb{Z}_+$, the set of positive integers (also denoted by $\mathbb{N}$), and let the binary operation be usual addition of integers. Clearly $G$ is a semi-group, but $G$ lacks an identity element and inverses, so that $G$ is not a group. $\Box$

Example 2.1.4   Let $G = \mathbb{Z}_+$, but now take the binary operation to be usual multiplication of integers. Again, it is clear that $G$ is a semi-group, but not a group since inverses (except for the integer $1$) are missing. $\Box$

Example 2.1.5   Let $G = \mathbb{Z}$, the set of all integers, and let the operation be addition of integers. Then $G$ is readily seen to be a group. $G$ has identity $0$, and each $x$ in $G$ has inverse $- x$. $\Box$

Example 2.1.6   Let $G = \mathbb{Z}$, and take the operation to be multiplication of integers. Then $G$ is just a semi-group. (WHY?) $\Box$

Example 2.1.7   Let $G = \mathbb{Z}$, the set of all negative integers and let the operation be multiplication of integers. This is not a binary operation on $\mathbb{Z}$ since it is not closed, or in other words it is not a mapping into $\mathbb{Z}$, and therefore, $G$ with respect to this operation is not even a semi-group. $\Box$

Example 2.1.8   Let $G$ be the set of all rotations of the plane about the origin including the rotation through $0^o$ and take the binary operation to be composition of maps. Then it is easy to see that $G$ is a group. $\Box$

Example 2.1.9   Let $G = \mathbb{Q}$, the set of all rational numbers, and let the binary operation be addition of rationals. Then $G$ is a group. Similarly, the set, $G = \mathbb{Q}^\times$, of all nonzero rationals with respect to the usual multiplication of rationals is a group. $\Box$

Example 2.1.10   Let $G = \{1,- 1\}$, i.e., the 2-element set consisting of the integers $\pm 1$, and take the binary operation to be usual multiplication. Then $G$ is a group. $\Box$

Example 2.1.11   Let $G$ be the set of all complex n-th roots of unity, i.e., $G = \{z\in \mathbb{C}\ \vert\ z^n = 1\}$, where $\mathbb{C}= \{a+ bi \ \vert\ a,b\ {\rm are\ real\ and}\ i =\sqrt{-1}\}$ is the set of all complex numbers. Let the binary operation be multiplication of complex numbers. Then $G$ is a group. This example is a generalization of the preceding one in which $n =2$. Note the order of $G$, $\vert G\vert$, is $n$.

This group shall be denoted by $\mu_n$. $\Box$

Example 2.1.12   Let $G$ be the set of all complex numbers which are roots of unity of any degree with the usual multiplication of complex numbers. Again $G$ is a group, but this time $G$ is infinite (cf. Example 2.1.11). $\Box$

Example 2.1.13   Let $G$ be the set of all $n\times n$ matrices with real entries and determinant not $0$. Take the binary operation to be matrix multiplication. Then $G$ is a group. This group is called the general linear group of $n\times n$ matrices over $\mathbb{R}$, the set of real numbers. It is denoted by $GL(n,\mathbb{R})$. (Recall that if $A$ and $B$ are $n\times n$ matrices, $\det (AB) = \det A \det B$.) This group can also be interpreted as a set of functions: The set of 1-1, onto, linear transformations from the vector space $\mathbb{R}^n$ to itself. Matrix multiplication corresponds to composition of these functions. $\Box$

Example 2.1.14   Let $G =\mathbb{C}^n$, all $n$-tuples of complex numbers, i.e., $G=\mathbb{C}\times ...\times \mathbb{C}$ ($n$ times). Let $x, y \in G$, then $x = (a_1, ..., a_n)$ and $y = (b_1, ..., b_n)$, where the $a_i$ and $b_i$ are complex numbers. Define $x + y = (a_1 + b_1, ..., a_n + b_n)$. It is easy to see that $G$ with respect to this binary operation is a group. $\Box$

Example 2.1.15   Let $A$ be any set. Then a mapping $f : A \rightarrow A$ which is both 1-1 and onto is called a permutation of $A$. To be more concrete, let $A = \{1, 2, ..., n\}$. Any 1-1, onto function, $f$, from $A$ to $A$ is a permutation (sometimes called a permutation of degree $n$) of $A$. Suppose $f$ is a permutation of degree $n$, and let $f(1) = a_1$, $f(2) = a_2$, ..., $f(n) = a_n$, where $a_1, a_2, ..., a_n$ is just some rearrangement of the set $A$ (thus the name permutation). We shall denote this situation by writing

$$f= \left( \begin{array}{cccc} 1 & 2 & ... & n\\ a_1 & a_2 & ... & a_n \end{array}\right)$$ (2.1)

i.e., the bottom entries indicate the images of the top entries under the mapping $f$. $S_n$ denotes the set of all permutations of degree $n$. Clearly ^2.1, $S_n = n!$. If $f,g \in S_n$, we take the binary operation to be composition of mapping $fg$; that this is, indeed, a binary operation follows as a special case of exercise 7 in the exercises for Section 1.1. The identity permutation, here denoted by $1$, is just

$$1= \left( \begin{array}{cccc} 1 & 2 & ... & n\\ 1 & 2 & ... & n \end{array}\right)$$

i.e., $1 = 1_A$ in previous notation. If $f$ is given by (2.1), then $f^{- 1}$ is just

$$f= \left( \begin{array}{cccc} a_1 & a_2 & ... & a_n\\ 1 & 2 & ... & n \end{array}\right)$$

Then $S_n$ is a group since the associative law is true in general for mappings (see exercise 8 in the exercises for Section 1.1). This group $S_n$ is called the symmetric group of degree $n$.

Let us take a look at $S_3$, i.e., all permutations of the set $\{1, 2, 3\}$:

$$\begin{array}{ccc} 1= \left( \begin{array}{ccc} 1 & 2 & 3\\ ... ...}{ccc} 1 & 2 & 3\\ 2 & 1 & 3 \end{array}\right) \end{array}.$$

The operation here is composition of functions; e.g., to find $f_1r_2$, we note that

$$\begin{array}{c} f_1r_2(1) = f_1(r_2(1)) = f_1(3) = 2,\\ ... ...1(1) = 1,\\ f_1r_2(3) = f_1(r_2(3)) = f_1(2) = 3. \end{array}$$

Thus $f_1r_2 = f_3$. Observe in $f_1r_2$, $r_2$ is applied first and $f_1$ next, so we read from right to left. We could also write

$$f_1r_2= \left( \begin{array}{ccc} 1 & 2 & 3\\ 1 & 3 & 2 \en... ...in{array}{ccc} 1 & 2 & 3\\ 2 & 1 & 3 \end{array}\right) =f_3,$$

and again reading from right to left, we begin with $\left( \begin{array}{ccc} 1 & 2 & 3\\ 3 & 1 & 2 \end{array}\right)$ to get for example that $1\longmapsto 3$ and then $3\longmapsto 2$ so $1\longmapsto 2$ under $f_1r_2$. As another example, consider

$$r_2f_1= \left( \begin{array}{ccc} 1 & 2 & 3\\ 3 & 1 & 2 \en... ...in{array}{ccc} 1 & 2 & 3\\ 3 & 2 & 1 \end{array}\right) =f_2.$$

Note that $f_1r_2\not= r_2f_1$. $\Box$

If a group $G$ contains only a finite number of elements, i.e., $\vert G\vert < \infty$, then $G$ is called a finite group; otherwise it is called an infinite group. Often when working with groups, especially finite groups, it is useful to draw a multiplication table (sometimes called a Cayley table). In general, let $G=\{g_1,...,g_n\}$ be a group with binary operation $*$, the multiplication table of $G$ is:

*	$g_1$	$g_2$	...	$g_j$	...	$g_n$
$g_1$
$g_2$
$\vdots$
$g_i$				$g_i*g_j$
$\vdots$
$g_n$

The entry in the row of $x\in G$ and column of $y\in G$ is $x * y\in G$ (in that order). The reader should check that using the notation of the previous Example 2.1.15, that $S_3$ has the Cayley table

$S_3$	$1$	$r_1$	$r_2$	$f_1$	$f_24$	$f_3$
$1$	$1$	$r_1$	$r_2$	$f_1$	$f_2$	$f_3$
$r_1$	$r_1$	$r_2$	$1$	$f_3$	$f_1$	$f_2$
$r_2$	$r_2$	$1$	$r_1$	$f_2$	$f_3$	$f_1$
$f_1$	$f_1$	$f_2$	$f_3$	$1$	$r_1$	$r_2$
$f_2$	$f_2$	$f_3$	$f_1$	$r_2$	$1$	$r_1$
$f_3$	$f_3$	$f_1$	$f_2$	$r_1$	$r_2$	$1$

If $G$ is a group and $ab = ba$ for all $a,b \in G$, then $G$ is called a commutative group or an abelian group. Note from Example 2.1.15, $S_3$ is not abelian (as a matter of fact, this implies that $S_n$ is a non-abelian for any $n \geq 3$). In the case of $G$ being abelian, it is customary to adopt an additive notation and write $a + b$ instead of $ab$, $0$ instead of $e$ (or $1$), and $-a$ instead of $a^{- 1}$.