1. ## A subgroup problem.

Show that if H & K are subgroups of abelian group G, then {hk| h is an element of H and k is an element of K} is a subgroup of G.

2. Have you tried checking the group axioms?

Show that if H & K are subgroups of abelian group G, then {hk| h is an element of H and k is an element of K} is a subgroup of G.

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 $HK$ will be a subgroup of $G$.

This is actually a weaker form a quite a nice theorem. If one of H or K is normal then $HK$ is a group. Clearly this holds here are you are in an abelian group so all subgroups are normal.

4. 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.

...This implies that (hk)(h'k') = e.
Yes, but you need to use the fact that $G$ is abelian!

Right, you have two things to prove. You need to prove that if $hk \in HK$ then there exists an element $h_0k_0$ such that $hkh_0k_0=e$. You are right that this element is $h^{-1}k^{-1}$ but you need to mention WHY this is its inverse.

You also need to prove that if $h_1k_1 \in HK$ and $h_2k_2 \in HK$ then $h_1k_1h_2k_2 \in HK$.