If $φ:K → Aut(H)$ is the trivial homomorphism—then the semidirect product is just the ordinary direct product of groups. I'm working on the question
Show that if $φ:K → Aut(H)$ is the trivial homomorphism—i.e., $φ(k) = 1_H$ for every $k ∈ K$— then the semidirect product is just the ordinary direct product of groups.
I know that the definition of a direct product is
$$(h_1,k_1)(h_2,k_2)=(h_1h_2,k_1k_2)$$
But I'm not sure where to go from here. Thanks!
 A: For a general homomorphism $\,\phi: K\to\text{Aut}\,H\,$ , how do you perform the product of two elements $\,(h,k)\,,\,(h_1,k_1)\in H\times K$ ? Well, as follows (the following may depend on the author):
$$(h,k)\cdot (h_1,k_1):=\left(\,\,hh_1^k\,,\,kk_1\right)$$
where $\;h_1^k:=\phi(k)(h_1)\;$ . If $\;\phi\;$ is the trivial homomorphism, this means that we have
$$\phi(k)=\text{Id}_H\;,\;\;\forall\,k\in K\implies \phi(k)(h)=:h^k=\text{Id}_H(h)=h$$
and thus  we get in the product above
$$(h,k)\cdot (h_1,k_1):=\left(\,\,hh_1^k\,,\,kk_1\right)=(hh_1,kk_1)$$
and the rightmost product is just the one that we get with the direct product.
Also, note that $\;\forall\,k\in K\,,\,\,h\in H\;$ ,we'd get in this case that $\;h^k=h\;$ in the semidirect product, which of course means also $\;H\;$ is normal ( of course, we're also given $\;H\cap K=1\;$) in the product and, again, we get that the product is direct.
A: This means that $hk=kh$ for all $h\in H$ and $k\in K$. Hence $hkh^{-1}=k$ and $khk^{-1}=h$ for all $h,k$. This implies that both $H$ and $K$ are normal. Since they intersect trivially and generate the group, the group is the direct product.
