My introduction of group theory to them mainly originated with the simple groups known as 'clock groups'. 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 (which means no extraneous values enter that aren't already members of the given group). 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

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? (i.e. there exist elements a, b { G such that (a · b) = (b · a), ) 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.

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 (H ( G) - then H is a sub-group 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.

Proof:

Let H be the set of all powers of a:

H = {a, a^2, a^3, a^4...........a^n}

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

Then: a^n = a^m (for some m), therefore:

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

where I is the identity element.

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? (Note especially how the table differs from the (x) table for the clock group, G+⊕ (4).) Hint: set out the top of the table with elements: 1, a, a^2, a^3. The set out the left side of the table with elements (from top to bottom): 1, a, a^2, a^3. Your result will be a 4 x 4 array such that each side member multiplied by the corresponding top table member will yield the correct group member, e.g. 1·1 = 1, 1·a·= a, a·a= a^2 etc.

Practice exercise 2:

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.

SUB-GROUPS:

Basic definition: Let G be a group (G, •) . Let H be some subset of G. The H is a subgroup of G IF:

i) H is closed under the operation (•) on G.

ii) (H, •) satisfies the group axioms

The groups axioms are:

a) Associativity: a· (b · c) = (a· b) · c

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

(c) Inverse element: For any a { G there exists an element a^-1 {G such that: a · a^-1 = a^-1 · a = e

d) Commutativity: Only if there exist elements a, b { G such that (a · b) = (b · a), then G is said to be an Abelian Group.

We now want to test this definition for previous examples.

1. What are the proper subgroups of G+⊕ (5)?

First, a proper subgroup is not a member of itself, so that G+⊕ (5) Ë G+⊕ (5)! (Where Ë denotes 'does not belong)Second, one cannot assume any combination of a subset of elements will comprise a subgroup. For example, consider the table for the operation of (+) on Z+ (4) (e.g. the top part of the table in Fig. 2)

We perceive a number of potential subgroups at the outset:

{0, 1}, {0, 2}, {0, 3}, {0, 1, 2}, {0, 1, 3} and {0, 2, 3}

Any combination of elements that includes 4 can automatically be excluded since a group is never a proper subgroup of itself. Let’s take the extreme set first, starting with {2,3}.

A quick check of the additive-operation table discloses 2 + 2 = 4, so the set isn’t closed on {0, 2, 3}. Similar arguments can be applied to exclude {0, 1, 2} and {0, 1, 3}. For example, in the latter case, 1 + 3 = 4, so (+) is not closed for the proposed subgroup.

What about {0, 2}? Does it qualify? Again 2 + 2 = 4, so the set isn’t closed. What about {0, 1}? Does it qualify? The problem here is that 1 + 1 = 2, so the set isn’t closed, since the result (2) lies outside the group’s elements. Evidently then, there are no proper subgroups for the example. Compare this with the integer 4-element group Z4, for which the addition table is: the same as depicted in the upper portion of Fig. 1.

Are there any proper subgroups? What about {0. 3}? The problem here is, from the table: 3 + 3 = 2 and 2 Ë {0, 3}. However, if we set out the sub-table for the set of elements {0. 2} we find:

+/ ---0 -----2

--------------

0---- 0----- 2

2---- 2----- 0

By inspection, all elements are in the subgroup, so it is closed on (+).

2.: Explain why Z

_{4}(+) contains a proper subgroup within it, but not G+⊕ (5) which has one more element.

Ans. The clock group G

_{+}_{⊕ }(5) terminates at highest element 4. The clock addition (2 + 2) = 4, so the set {0, 2} is not closed under the operation, as 4 Ë {0, 2}. By the same token, {0, 3} doesn’t work because 3 + 3 = 1 and 1 Ë {0, 3}. If one now includes element 1 so that {0, 1, 3} that doesn’t work either because 3 + 1 = 4 and 4 Ë {0, 1, 3}.
Thus, Z

_{4}(+) has enough elements to allow for a small subgroup, but not so many elements that all subgroups are prohibited, as for G_{+}_{⊕ }(5). However, when one inspects the group table G_{+}_{⊕ }(4), one finds it is the same as for Z4(+).Problems for the Math Maven:

1. Show that {1,a^2} is a proper subgroup of C4, the cyclic group of order 4.

2. Sketch the cyclic graphic associated with the cyclic group C8, the cyclic group of order 8. Prepare the table for this group with all elements indicated, and thence or otherwise identify the subgroups by order and indicate which are proper, which improper and list the respective elements of each.

3. Show that the largest order subgroup in C9 is a proper subgroup of C9,.

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