Let the group $G = H \otimes K$ be the result of the internal direct product of subgroups $H$ and $K$. Then show that the map $$H \times K \rightarrow G : \phi(h,k) \rightarrow hk$$ is an isomorphism.
Here's my attempt. We want to show that $$\phi ((h_1, k_1) \cdot (h_2,k_2)) = \phi ((h_1,k_1)) \cdot \phi ((h_2, k_2)).$$ Since multiplication of the pairs will occur "coordinate-wise" (I know this isn't quite proper lingo, it just helps me think about the operation this way, sorry), $$\phi ((h_1, k_1) \cdot (h_2,k_2)) = \phi ((h_1h_2, k_1k_2)) = h_1h_2 k_1k_2.$$ Since we aren't given that $G$ is abelian, we can't infer much more from here. On the other hand, $$\phi ((h_1, k_1)) \phi ((h_2,k_2)) = h_1k_1h_2k_2...$$ I'm not sure how to get through this point, however. I'm completely stumped. As for the bijective part of this problem, I'm pretty sure I have that down. Could someone please help by clearing up the homomorphism proof above? I have a feeling this theorem implicitly assumes that $G$ is abelian but I'm not sure. Thank you.