K, G, H are groups. Please see the 'Task 1' file for the description of the question and 'Task 2' for the description of the direct product of groups.
Question is connected with:
https://www.researchgate.net/post/Is_it_true_that_group_is_the_product_of_two_trivially_intersecting_normal_subgroups
Define a homomorphism \phi from GxH to K by taking (g,h) to \gamma(g)\delta(h). To see that this make sense, make use of the fact: If G and H are normal subgroups of K and G\cap H={e} then for any g in G and h in H gh=hg. This follows from the fact that ghg^{-1} is an element of H, so the commutator of g and h is in H; likewise it is in G.
Therefore \phi(gg',hh')=\phi(g,h)\phi(g',h'). The inverse homomorphism takes k to ( \alpha(k),\beta(k)). This needs to be checked. Obviously \alpha(\phi(g,h))=\alpha(gamma(g))\alpha(delta(h))=g. Likewise \beta(phi(g,h))=h. In the other direction, \phi(\alpha(k),\beta(k))=\gamma\alpha(k)\delta\beta(k)=gh. We need to verify that gh=k. \alpha(gh)=\alpha(k) since \alpha\delta is trivial. Therefore ghk^{-1} is in the kernel of \alpha, and so it is in H. (It is convenient to think of G and H as being subgroups of K). So ghk^{-1}=h'. Similarly, ghk^{-1} is in G, so gh=k.
If only one of G and H is normal, the argument breaks down, since \phi is no longer a homomorphism. However, by changing the multiplication on GxH you get a homomorphism from a semi-direct product of G and H to K. If you assume that every element of K can be written as a product gh, then \phi is surjective. (This was implied in the diagram, which no longer exists if one subgroup is not normal.) Injectivity is easy, since gh=g'h' implies g'^{-1}g=h'h^{-1}.
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