If isomorphic groups are regarded as being the same, prove, for each positive integer , that there are only finitely many distinct groups with exactly elements.

Thanks in advance.

Printable View

- April 14th 2009, 03:23 PMBiscaimisomorphic groups
If isomorphic groups are regarded as being the same, prove, for each positive integer , that there are only finitely many distinct groups with exactly elements.

Thanks in advance. - April 14th 2009, 03:41 PMNonCommAlg
- April 14th 2009, 04:09 PMBiscaim
- April 14th 2009, 06:59 PMNonCommAlg
another way: let and let be a group of order define multiplication in by where then is a group and

so every group of order defines a map note that if two groups define the same on then thus the number of groups of order is at

most the number of maps which can be defined from to which is

an easier way to get this upper bound (this just came to my mind!) is to look at the multiplication table of a group of order each of entries have at most pssibilities!! (Nod) - April 14th 2009, 09:26 PMThePerfectHacker
Another (not so nice) approach.

Given a finite set we need to define .

Therefore, .

Thus, I get as an upper bound.

Which is a horrible upper bound (Puke).