I am not sure how you have defined semi-direct product, but any way, I'll write a complete answer, recalling the definitions, in the process, we'll also establish notation.
I learnt this from Rotman, so, there'll certainly be a flavour of his notation here.
Definition 1
A group $G$ is a semi-direct product of $K$ by $Q$, if the following is true:
- $K \lhd G$ (Atleast one of them is good.)
- $KQ=G$ (The subgroups are fat enough!)
- $K \cap Q=\{e_G\}$ (They are fat in a nice way.)
This is the important abstract thing going on here:
Theorem 2
Given a group $K$, $Q$ and a homomorphism $\theta:K \to\operatorname{Aut} H$, there is a way to build a new group, which is a semi-direct product of $K$ and $Q$ in the following way:
The set $G=\{(h,k)|h \in H; k \in K\}$ and the operation in the group is twisted by the element that $\theta$ points to.
(h,k) \ast (h',k')=(h,[\theta(k)](h),kk') $(h,k)^{-1}=([\theta(k^{-1})](h),k^{-1})$
Prove that the above structure is a group. (Or so you'd have done this in your class.)
The difficulty will then lie in understanding the following:
What is a trivial action? All elements of $K$ are so good (bad?) that they do nothing to any element of $H$. Then, all elements in $K$ fix all elements of $H$. The, $K \to \operatorname{Aut} H$ is the trivial map.
So, no new twist, then what happens?
This means $\theta(k)$ is the identity permutation for all $k \in K$. Then, $[\theta(k)](h)=h \forall h \in H$. Then, the operations just become that of the usual direct product.
I know, I have not used mathematical language, but this is how I keep this in my head! And, it is not difficult to put this into Mathematical formalism.
P.S.:
Think of group actions as following:
Suppose $G$ is a group, and $S$ is a set. Let us say $G$ acts on $S$. Then, this is what that means: Every element in $G$ takes an element of $S$ to some other element of $S$. And, identity, "is identity" on the elements of $S$, I mean it does not take any element in $S$ to a new element. And, if $g$ takes $s$ to $s_1$ and $h$ takes $s_1$ to $s_2$, then $gh$ takes $s$ to $s_2$.
This means that every element in $G$, in short, permutes $S$.