Basic Group Homomorphism Proof Confusion

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.

• It is really about how you define multiplications in G. Aug 22, 2020 at 21:54

Recall that $$G$$ is the internal direct product of subgroups $$H$$ and $$K$$ if the following conditions are met:
1. $$G=\{h\cdot k:h\in H, k\in K\}$$
2. $$H\cap K=\{e\}$$ where $$e\in G$$ is identity
3. $$h\cdot k=k\cdot h$$ for all $$h\in H$$ and $$k\in K$$.
In particular, $$G$$ need not be abelian, however the elements of $$H$$ commute with the elements of $$K$$ in the way you need them to.