**:**

*1. Definitions*A GROUP is a set G with one binary operation defined on it , and G satisfies the axioms:

(a) Associative law: a· (b · c) = (a· b) · c

(b) Identity element (e): there exists an element e Î G such that: e · a = a for all a Î G and a · e = a

(c) Inverse element: For any a Î G there exists an element a

**Î G such that:**

^{- 1}a · a

**= a**

^{- 1}**· a = e**

^{- 1}d) Commutativity[1]: Only if there exist elements a, b Î G such that (a · b) = (b · a), then G is said to be an

*Abelian Group*.

*2. Generating Simple Groups:*

*A) Clock Groups:*These are amongst the simplest of all, and easiest to generate – they also represent a nice introduction to modular arithmetic, the Euclidean Algorithm etc.

Let a clock group representation be denoted as shown in the diagram of Figure 1.

We seek to construct a group from the above, which is closed under addition. As we can see there are four members,

**0, 1, 2, and 3**. The process of addition is defined by adding elements – starting with 0- in a

*clockwise sense*. Doing this we should be able to find a complete closed set of addition operations for all the elements. For example, we find 0 + 1 =1, and 0 + 2 = 2 and so forth. Similarly, we find 1 + 1 = 2, 1 + 3 = 0, 2 + 3 = 1 and so on. Each result obtained by adding the portion of the cycle from the starting element. From here, we may set out the group under addition (+) (Fig. 1- right top)

The reader should easily be able to check each of these and demonstrate for himself that the table is valid. Is G+⊕ (4) a group? Yes, because

*it obeys all the properties for a group*.

*Problem:*For the same group, develop a table to show it is closed under multiplication (x). Hint: Simply extend the principle of addition to the case of multiples, and start all multiple entries counting clockwise from 0. For example, 2 + 2 + 2 amounts to three two’s counted from 0. (Three sets of two). One such counting set leads to 2, and two leads to 0 and three leads to…? Obviously, 2.

Thus: 4 + 4 = 0 and 4 + 4 + 4 = 0 so that 3 x 4 = 0

What about: 3 + 3 + 3?

By inspection and using the clock graphic we obtain: 3 + 3 + 3 = 1 so that 3 x 3 = 1

We thereby arrive at the (x) table for the group G+⊕ (4) (Bottom right in Fig. 1)

Is this group commutative? Some checks of the operation using pairs of elements will confirm that it is (E.g. 2 x 3 = 3 x 2). Hence, we can aver it is an Abelian group.

*Problem*: Set out a clock group with five elements (G+⊕ (5)), and prepare tables to show the group is closed under (+) and (x).

*Solution:*

The correct diagram is shown in Fig. 2, along with the tables for multiplication and addition.

*B) Cyclic Groups and Sub-groups*A more advanced variation on the simple clock group is what's called a "cyclic group" which we will see can also be a sub-group. To clarify definitions here - let G denote a group, and let H be a subset of G then H is a subgroup of G if: H is closed under the same operation as G, for each element x there exists the inverse element, x

**, and these are related to an identity element I such that: (x)(x**

^{- 1}**) = I.**

^{- 1}*A relevant theorem- Lagrange's Theorem*:

*Let G be a group of finite order n, then let a be any element of G. Then the order of the element divides the order of G.**Proof:*

Let H be the set of all powers of a:

H = {a, a

**, a**

^{2 }**, a**

^{3 }**...........a**

^{4}**}**

^{n}But, H must have a finite number of distinct elements.

Then: a

**= a**

^{n}**(for some m)**

^{m}a

**(a**

^{n}**)**

^{m}**= a**

^{- 1}**a**

^{n - m}**(a**

^{m }**)**

^{m}**= a**

^{- 1}***I**

^{n - m}Is the cyclic group closed under (x)? Check by consideration of C

_{4}, the cyclic group of order 4. To see an example of C

_{4}simply take the diagram for the clock group, G+⊕ (4), and re-assign it elements as follows:

0 -> 1

1 -> a

**= a**

^{1}2 -> a

^{2}3 -> a

^{3}Now, what will the multiplication table look like?

Solution: the result is shown in Fig. 3. Note especially how the table differs from the (x) table for the clock group, G+⊕ (4).

*Practice exercise*:

Using the diagram for the clock group G+⊕ (5) in Fig. 2 as a template, work out the elements for the cyclic group C

_{5}. Prepare a table for (x) applicable to its elements. Is C

_{4 }a sub-group of C

_{5}? Why or why not?

_{}

^{}

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[1] This is not a critical, or indispensable group property. If it does apply, we say the group is Abelian

_{}

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