A group comprises a set (G) and an operator (*) (I don't think it needs to be binary) and observes the three property below:
1. Associativity
For a, b, c elements of set G
$ a*(b*c) = (a*b)*c $
2. Existence of identity, e
$ a*e = a $
3. Existence of inverse, $ a^{-1} $, for all elements in the set
$ a * a^{-1} = e $
A step towards abstraction suits the mathematical study of groups, and I enjoy the part where I stop caring about what symbols I use to illustrate an idea. Fraleigh suggests that I think about an alien civilization using entirely different language structure than our own world; only in abstraction will we be able to identify the similarities in the algebraic structure of our mathematical system and theirs. (but I wonder if what we find rational is irrational to the alien society, and vice versa)
For instance, consider a group with one element in its set. We can set up a multiplication table for such set to illustrate what happens when its operator (binary) acts upon different pair of the sets:
| * | e |
| e | e |
So, from the multiplication table, we can conclude $ e * e = e $. We can further deduce that that one element is an identity element that is also an inverse of itself. So the three conditions of a group are met (I imply the associativity), and a set with identity element is rightfully a group with the above operation. Note that I could have written,
| + | 0 |
| 0 | 0 |
but we can easily see that the algebraic structure of the group is no different than the former. It is not interesting to discover different ways of representing what is essentially the same.
Consider a set with two elements; how must the elements "interact"? Now, since an identity element must always exist, we keep the "e" and consider a, the other element of the set.
| * | e | a |
| e | ||
| a |
We know that whatever element that pairs with 'e' results in producing itself.
| * | e | a |
| e | e | a |
| a | a |
We also know that for this particular arrangement to be qualified as a group, a must have an inverse. We are left with only a single option for the choice to fill in the remaining square- that a is also an inverse of itself.
| * | e | a |
| e | e | a |
| a | a | e |
Let's do it with three elements. I need to prove two theorems to chart the interactions of this proposed group.
1. if a * b = c, b is unique.
Proof
If not, and conisder b1 and b2 satisfying the above condition. Then
$ a * b1 = a * b2 $
$ a^{-1} * a * b1 = a^{-1} * a * b2 $
$ e * b1 = e * b2 $
$ b1 = b2 $
I think I cheated just a little bit (on step 2), but I hope the idea is illustrated.
2. inverse of each element in the set is unique
Proof
It is the same as the above proof, really; substitute b1 and b2 with $ a^{-1}1 $ and $ a^{-1}2 $.
Now, with the above two fact, I can say that each element of the set must only appear in a given row/column no more than once. This is because if a * b = a * c = d, b = c. But b does not equal c- that is why we gave each element a unique row/column in the first place.
So let append an element 'b' to the group of order 2.
| * | e | a | b |
| e | |||
| a | |||
| b |
Apply the obvious result of operations with the identity element
| * | e | a | b |
| e | e | a | b |
| a | a | ||
| b | b |
Now because we know that each element can only appear once in a given row or column, we can fill in the rest.
| * | e | a | b |
| e | e | a | b |
| a | a | b | e |
| b | b | e | a |
For a little, I tinkered with the table
| * | e | a | b |
| e | e | a | b |
| a | a | e | b |
| b | b | a | e |
and wondere what was wrong with it since commutativity is not a requirement for a group (but it is a condition for an abelian group). But then, theorem 1.
Group of order 4 is where it should get more interesting. I append the set in the similar manner and apply the obivous
| * | e | a | b | c |
| e | e | a | b | c |
| a | a | |||
| b | b | |||
| c | c |
Now, there isn't only one way to fill in this table, but two.
| * | e | a | b | c |
| e | e | a | b | c |
| a | a | b | c | e |
| b | b | c | e | a |
| c | c | e | a | b |
and
| * | e | a | b | c |
| e | e | a | b | c |
| a | a | e | c | b |
| b | b | c | e | a |
| c | c | b | a | e |
TBC!
