Have you tried checking the group axioms?
This is actually a weaker form a quite a nice theorem. If one of H or K is normal then is a group. Clearly this holds here are you are in an abelian group so all subgroups are normal.
I got what you two were trying to tell me by using the axioms to prove this. This is the proof that I came up.
Since H and K are subgroups of G. There exist say h an element of H, such that (h)(h') = e
=> e is an element of H.
=> (e)(h') = h' is an element of H.
Similarly, there exist say k an element of K, such that k(k') = e
=> e is an element of K
=> e(k') = k' is an element of K.
This implies that (hk)(h'k') = e.
=> e is an element of of H and K => e is also an element of G.
=> e(kh)' = (kh)' = (h'k') is an element of G.
Therefore hk is a group.
Thoughts? I think it's not yet the best proof.
Right, you have two things to prove. You need to prove that if then there exists an element such that . You are right that this element is but you need to mention WHY this is its inverse.
You also need to prove that if and then .