Tuesday, April 18, 2017

Math Revisited: Abstract Algebra - Groups

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 - 1  Î G such that:
a · a - 1 = a - 1 · a = e

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).


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 - 1, and these are related to an identity element I such that: (x)(x - 1) = I.

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.


Let H be the set of all powers of a:

H = {a, a2 , a3 , a4...........a n}

But, H must have a finite number of distinct elements.

Then: a n = a m (for some m)

a n(a m- 1 = a n - m   a m (a m- 1 = a n - m *I

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 1 = a

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?


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

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