2.1 Elements of group theory

A group is a set G together with a law of composition that assigns to any pair of ele ments g, $h\in G$ an element $gh\in G$ , called their product, satisfying the following three conditions:

(Gp1) The associative law holds: $g(hk)=(gh)k$ , for all $g,h,k\in G$

(Gp2) There exists an identity element $e\in G$ , such that

$$eg=ge=g\text{for all}g\in G.$$

(Gp3) Each element $g\in G$ has an inverse $g^{-1}\in G$ such that

$$g^{-1}g=g{g}^{-1}=e.$$

More concisely, a group is a semigroup with identity in which every element has an inverse. Sometimes the fact that the product of two elements is another element of G is worth noting as a separate condition, called the closure property. This is particularly relevant when $G$ is a subset of a larger set with a law of composition defined. In such cases it is always necessary to verify that G is closed with respect to this law of composition; that is, for every pair $g,h\in G$ , their product $gh\in G$ . Examples will soon clarify this point.

Condition (Gp1) means that all parentheses in products may be omitted. For example, $a((bc)d)=a(b(cd))=(ab)(cd)=((ab)c)d$ . It is a tedious but straightforward matter to show that all possible ways of bracketing a product of any number of elements are equal. There is therefore no ambiguity in omitting all parentheses in expressions such as abcd. However, it is generally important to specify the order in which the elements appear in a product.

The identity element e is easily shown to be unique. For, if $e^{\prime }$ is a second identity such that $e^{\prime }g=g{e}^{\prime }=g$ for all $g\in G$ then, setting $g=e,$ , we have $e=e^{\prime }e=e^{\prime }$ by (Gp2).

Exercise

By a similar argument, show that every $g\in G$ has a unique inverse $g^{-1}$

Exercise

Show that $(gh{)}^{-1}=h^{-1}{g}^{-1}$

A group G is called abelian if the law of composition is commutative,

$$gh=hg\text{ for all }g,h\in G.$$

The notation gh for the product of two elements is the default notation. Other possibilities are $a\cdot b,a\times b,a+b,a\circ b$ , etc. When the law of composition is written as an addition $g+h$ , we will always assume that the commutative law holds, $g+h=h+g$ . In this case the identity element is usually written as 0, so that (Gp2) reads $g+0=0+g=g$ . The inverse is then written $-g$, with (Gp3) reading $g+(-g)=0$ or, more simply, $g-g=0$ Again, the associative law means we never have to worry about parentheses in expressions such as $a+b+c+\cdots +f.$

A subgroup H of a group G is a subset that is a group in its own right. A subset $H\subseteq G$ is thus a subgroup ifit contains the identity element of G and is closed under the operations of taking products and inverses:

(a) h, $k\in H\Rightarrow hk\in H$ (closure with respect to taking products);

(b) the identity $e\in H;$

(c) $h\in H\Rightarrow h^{-1}\in H$ (closure with respect to taking inverses).

It is not necessary to verify the associative law since H automatically inherits this property from the larger group G. Every group has two trivial subgroups e and G, consisting of the identity alone and the whole group respectively.

Example 2.1

The integers Z with addition as the law of composition form a group, called the additive group of integers. Strictly speaking one should write this group as $(\mathbb{Z},+)$ , but the law of composition is implied by the word ‘additive’. The identity element is the integer 0, and the inverse of any integer $n$ is $-n$. The even integers $\{0,\pm 2,\pm 4,\dots \}$ form a subgroup of the additive group of integers.

Example 2.2

The real numbers R form a group with addition $x+y$ as the law of composition, called the additive group ofreals. Again the identity is 0 and the inverse of x is $-x$ . The additive group of integers is clearly a subgroup of R. The rational numbers $\mathbb{Q}$ are closed with respect to addition and also form a subgroup of the additive reals $\mathbb{R},$ , since the number 0 is rational and if $p/q$ is a rational number then so is $-p/q$

Example 2.3

The non-zero real numbers $\dot{\mathbb{R}}=\mathbb{R}-\{0\}$ form a group called the multiplica tive group ofreals. In this case the product is taken to be ordinary multiplication $x{y}_{;}$ , the identity is the number 1 and the inverse of x is $x^{-1}=1/x$ . The number 0 must be excluded since it has no inverse.

Exercise

Show that the non-zero rational numbers $\dot{\mathbb{Q}}$ form a multiplicative subgroup of R˙ .

Exercise

Show that the complex numbers C form a group with respect to addition, and $\dot{\mathbb{C}}=\mathbb{C}-\{0\}$ is a group with respect to multiplication of complex numbers.

Exercise

Which of the following sets form a group with respect to addition: (i) the rational numbers, (ii) the irrational numbers, (iii) the complex numbers of modulus 1? Which of them is a group with respect to multiplication?

A group G consisting of only a finite number of elements is known as a finite group. The number of elements in G is called its order, denoted G .

Example 2.4

Let k be any natural number and ${\mathbb{Z}}_k=\{[0],[1],\dots ,[k-1]\}$ the integers modulo k, defined in Example 1.3, with addition modulo k as the law of composition

$$[a]+[b]=[a+b].$$

${\mathbb{Z}}_k$ is called the additive group of integers modulo k. It is a finite group of order k, written $|{\mathbb{Z}}_k|=k$ . There is little ambiguity in writing the elements of ${\mathbb{Z}}_k$ as $0,1,\dots ,k-1$ and $[a+b]$ is often replaced by the notation $a+b$ mod k.

Exercise

Show that the definition ofaddition modulo k is independent ofthe choice ofrepresentative from the residue classes [a] and [b].

Example 2.5

If a group G has an element a such that its powers $\{a,a^2,a^3,\dots \}$ run through all of its elements, then $G$ is said to be a cyclic group and a is called a generator of the group. If G is a finite cyclic group and a is a generator, then there exists a positive integer m such that ${\mathbf{a}}^m=\mathbf{e}$ . If m is the lowest such integer then every element $g\in G$ can be uniquely written $g=a^i$ where $1\le i\le m$ , for if $g=a^i=a^j$ and $1\le i<j\le m$ then we have the contradiction $a^{j-i}=e$ with $1\le (j-i)<m$ . In this case the group is denoted $C_m$ and its order is $|C_m|=m$ . The additive group of integers modulo $k$ is a cyclic group of order $k,$ but in this case the notation $a^n$ is replaced by

$$na=\underset{n}{\underbrace{a+a+\cdots +a}}\mathrm{mod}k.$$

Example 2.6

Let $p>2$ be any prime number. The non-zero integers modulo $p$ form a group of order $p-1$ with respect to multiplication modulo $p$,

$$[a][b]=[ab]\equiv ab\mathrm{mod}p,$$

denoted $G_p$ . The identity is obviously the residue class [1], but in order to prove the existence of inverses one needs the following result from number theory: if $p$ and $q$ are relatively prime numbers then there exist integers k and m such that $kp+mq=1$ . Since $p$ is a prime number, if $[q]\ne [0]$ then $q$ is relatively prime to $p$ and for some k and m

$$[m][q]=[1]-[k][p]=[1].$$

Hence $[q]$ has an inverse $[q{]}^{-1}=[m]$

For finite groups of small order the law of composition may be displayed in the form of a multiplication table, where the $(i,j)\mathrm{th}$ entry specifies the product of the ith element and the jth element. For example, here is the multiplication table of $G_7$ :

$G_7$	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	1	3	5
3	3	6	2	5	1	4
4	4	1	5	2	6	3
5	5	3	1	6	4	2
6	6	5	4	3	2	1