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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.

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    $\begingroup$ It is really about how you define multiplications in G. $\endgroup$ Aug 22, 2020 at 21:54

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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.

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  • $\begingroup$ Yep, I forgot to use number 3. Thank you for your response. $\endgroup$ Aug 22, 2020 at 22:02

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