Have you tried checking the group axioms?
As Roninpro said, just check the group axioms. In fact, you just need to check the two conditions for something to be a subgroup, as will be a subgroup of .
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.
Yes, but you need to use the fact that is abelian!
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 .