Group Theory

From ZuluNotes - Free Leaving Cert Notes

What is Group Theory?

A 'group' is a combination of a set and an operator that together obey certain properties (listed below). A set is simply a list of objects, usually numbers. For example $\{1,2,3,4\}$ . An operator is something that takes two elements and produces a third element. For example, if we perform the addition operator on the numbers $1$ and $5$ , the number $6$ is produced. Dozens of operators appear on the Leaving Certificate course - from addition to division to nCr.

In fact, not all operators take two elements. Factorial, for example, takes only one - we can perform the factorial 'operator' on $5$ to produce $120$ . Operators that take two elements are called binary operators. When we deal with groups we will only be concerned with binary operations.

Group Theory is an area of mathematics concerned with deciding if a certain set and operation form a group, and examining certain properties of groups. At its hardest, Leaving Certificate Group Theory involves investigating similarities between different groups. Group Theory is not difficult, but may appear so due to the amount of definitions involved, coupled with the fact that it is unlike almost all other areas of Leaving Certificate maths. However, if you develop a good understanding of these definitions, the questions asked become quite repetitive and straight-forward.

Properties of Groups

There are four properties that groups must have: closure, associativity, identity and inverses.

Closure

Let's represent some operation by $*$ , and a let the set $H=\{a,b,...,z\}$ . Then, if for every two elements, $p$ and $q$ , $p*q \in H$ , then $H$ forms a group under the operation $*$ (we could also write this as " $H,*$ is a group").

For example, the set of natural numbers under the addition operator is closed, as if we add any two natural numbers, we get another natural number.

Associativity

$H,*$ is associative if $(a*b)*c=a*(b*c)$ for all $a,b,c \in H$ .

Addition is associative: $(a+b)+c=a+(b+c)$ . However, subtraction is not, as $(a-b)-c \not= a-(b-c)$ . Composition of permutations and multiplication (modulo and matrix) are associative

Identity

There must exist some element in the set, let's call it $e$ , such that $e*a=a*e=a$ for all $a \in H$ .

In addition, $0$ is the identity, as $a+0=0+a=a$ . In multiplication $1$ is the identity.

Inverses

For every element $a$ in the set H, there is another element $a^{-1}$ such that $a*a^{-1}=e$ , where $e$ is the identity.

Note: $a^{-1}$ is not necessarily the same as $a$ to the power of $-1$ in this context. For example, if we take the integers under addition, then the inverse of $3$ is $-3$ as $(3)+(-3)=0$ . So we could write $3^{-1}=-3$ ! This is just notation, don't take it literally.

If a set and operation have these four properties, then they form a group.

Examples of Groups

Let's take the integers ( $...,-2,-1,0,1,2,...$ ) under the addition operation. Do they form a group? Let's check each of the properties.

First of all, is $a+b$ always an integer when $a$ and $b$ are integers? Yes, of course (we could never have $1532+3452=4984.125$ !). So we have closure.

We know that $(a+b)+c=a+(b+c)$ is true (the associativity of addition can be assumed at Leaving Cert level). So we have associativity.

We also know that $a+0=0+a=a$ , so we know that the identity is $0$ .

Finally, for every integer $a$ , we know that $-a$ is also an integer, and so because $(a)+(-a)=0$ , we have inverses.

So, they do form a group! Now, something to consider, do the Naturals ( $0,1,2,3,...$ ) form a group under addition?

Sample Problems

Believe it or not, we have already covered enough to tackle some LC questions (well, parts of questions anyway).

2006. Paper 2.

  10.(a) For each of the following, give a reason why it is not a group.
     (i) The set of natural numbers under subtraction.
     (ii) The set of real numbers under multiplication.

(i) Firstly, subtraction is not associative. Eg. $(a-b)-c \not= a-(b-c)$ . Secondly, we do not have closure. Eg. $1-5=-4$ , which is not a natural number.
(ii) Note that $1$ is the identity under multiplication. But there is no number such that $(a)(0)=1$ , so $0$ has no inverse.

2004. Paper 2.

  10(a) The binary operation  is identified by , where  are real numbers and .
    (i) Find the identity element.
    (ii) Calculate , the inverse of .
    (iii) Find  in terms of .
    (iv) Show that .
    (v) Show that  for all .</b>

(i) Let's call the identity $e$ . Then:
$a*e=a$
=> $a+e-ae=a$
=> $e=ea$
=> $e=0$ or $e=1$ .
But $e\not=1$ , therefore $e=0$ .

(ii) $3*3^{-1}=0$
=> $3+3^{-1}=(3)(3^{-1})$
=> $3=(2)(3^{-1})$
=> $3^{-1}=\frac{3}{2}$

(iii) $x*x^{-1}=0$
=> $x+x^{-1}=(x)(x^{-1})$
=> $x=(x-1)(x^{-1})$
=> $x^{-1}=\frac{x}{x-1}$

(iv) $a*b=a+b-ab$

=> $(a*b)*c = (a+b-ab)*c = a+b-ab+c-ac-cb+abc$ ... equation I
$b*c=b+c-bc$ </ br> => $a*(b*c) = a*(b+c-bc) = a+b+c-ab-ac-bc+abc$ ... equation II
Equation I = equation II, therefore $(a*b)*c=a*(b*c)$ .