How many groups, up to isomorphism, can we find that have order (cardinality) 4? I show that there are only two groups, up to isomorphism, with 4 elements. One of them is the group of integers modulo 4, and the other is the Klein 4 group.
It's because starting with one of the three tables A, C & D, you can find a set of operations involving (a) interchanging the elements with each other, (b) interchanging rows, and (c) interchanging columns that produces a result identical to one of the others of the three. But you will never be able to arrive at table B that way.
Is there an equation or a recurrence relation that would give us the number of groups that are distinct, up to isomorphism with the general order 'n' ?
By the way, the associative property comes as a byproduct of the other desired properties in this example presented here. A question in my own mind, is why not more generally study loops and not groups? What is so dear to the soul of having an associative table?
At 5:30 when we have our 2 possible groups, A and B, don't we need to check that they are associative? My memory is not exact, but I used this Latin square method of finding all the groups of order 5, and I believe I ended up with a Latin square describing an operation that was not associative, and so the Latin square actually did not describe a group operation.
To prove a set is abelian (i.e. communicative) under an operation, just make a Cayley table and see if the table is symmetric along the main diagonal. Off the shelf proof.
Claim: Cayley tables are Latin squares. Proof: suppose there exists a pair of columns such that the element x was in row n of both the first column and the second column. Call the element that corresponds to that row w, and call the element that corresponds to the first column of the pair y, and the element that corresponds to the second column of the pair z. In other words, we have that wy = x = wz => wy = yz. Multiply by the inverse of w to get y=z. However, since the columns were different, y != z. We have a contradiction, and our most recent assumption was that there existed a pair of columns C and D such that there exists n such that the nth row of C equals the nth row of D. So in fact, the negation of the assumption is true. Therefore, there does not exist a pair of columns ... . Thus, for every pair of columns, for all n, the nth row of C doesn't equal the nth row of D. That's all there is to it. If you are good with quantifier negation, the proof basically writes itself.
This doesn't even make sense. You can set up any of the 4 in this video and see that there are 4! * 2 permutations that follow her rules because you swap rows and collumns to make a new table that still satisfies the axoims.
Yes, but most of them will be isomorphic to each other. That is to say you could relabel 'a' as 'b' and the old 'b' as 'a' and you would get another group in the permutations that you mention. However, they behave essentially the same way as relabeling does not change how the new 'a' interacts with 'c'. That is why the "up to isomorphism" clause is added.
