[QUOTE=topsquark;634231]The problem is this:
Prove that a non-empty finite subset S of a group G is a subgroup iff it is closed under the product in G.[/tex]
It's easy to show that if the S is a subgroup then S is closed. I am having trouble with proving that S is a subgroup if it is closed. I can easily demonstrate that this is not true for an infinite subset S, but my efforts to prove the theorem for a finite subset is running into problems. I can show, for specific examples, that if the theorem is not true then either there is more than one identity in S or that S is not associative. But I'm wasting paper on the constructed examples and I really have not been able to make headway for the general case.
For right now my question is how do I prove that S contains the group identity? (I'll tackle the existence of inverses in S on my own for now. That's what I'm planning to attack until I can understand the identity part.)
Thanks in advance.
Let be the finite closed subset of G, and define = multiplication
by (you can in fact choose any element in S that you want). Prove
1) f is 1-1 and thus onto
2) Prove now that S contains the group's unit and inverse of any element in it
3) Go celebrate since you're done.