I'm an undergraduate who was introduced to semi-direct products of groups, $N \rtimes H$, relatively recently. At first, I found the multiplication rule for (outer) semi-direct products very annoying. But then, yesterday, I realized that there was a motivating example that really helped me to understand them, and which I believe could even be generalized to characterize semi-direct products, albeit in an unusual way.
The example is the case when $N$ is the additive group of a subspace of $\mathbb{K}^n$, and $H = GL_n(\mathbb{K})$. Then given a vector $n \in N$, and a matrix $h \in H$, we have an affine linear map of the form $v \mapsto h*v + n$, where $*$ is matrix multiplication. The set of such bijections, considered as a group under composition, is isomorphic to $N \rtimes H$.
This example led me to consider whether for arbitrary groups, $N$ and $H$, we could actually define $N \rtimes H$ in a way analogous to this example. The definition would look something like this:
Suppose we are given groups, $N$ and $H$, along with an action $H \times N \to N$ $$(h,n) \mapsto h * n $$
on $N$ as a group (meaning the action satisfies $h * (nn') = (h * n)(h * n')$).
We might consider groups $M \ge N$ where there is a faithful action $H \times M \to M$ on $M$ as a group which restricts to the original action on $N$. We can denote this action with $*$ also.
In such a case, for any $(n,h) \in N \times H$, we have a bijection $M \to M$
$$m \mapsto n(h * m) =: (n,h) \bullet m,$$
in which we first act with $h$ and then multiply by $n$. The set of such bijections forms a group, $ N \rtimes H$, under composition. We can represent an elements of this group as a pair, $(n,h)$. If $(n,h), (n',h') \in N \rtimes H$, $m \in M$, then we can see that
$$ ((n,h)(n',h'))\bullet m = (n,h) \bullet ((n',h') \bullet m)$$ $$ = (n,h) \bullet (n'(h' * m)) $$ $$ = n(h*(n'(h'* m)))$$ $$ = n(h*n')(h*(h'*m))$$ $$ = n(h*n')((hh')*m))$$ $$ = (n(h*n'),hh')\bullet m$$
So $(n,h)(n',h') = (n(h*n'),hh')$. From this explicit definition, we can see that $(n,h)^{-1} = (h^{-1} * n^{-1} ,h^{-1})$
We can also see that distinct pairs in $N \times H$ give distinct bijections in $N \rtimes H$. To see this, first suppose that we have a pair $(n,h)$ which acts as the identity. Then $(n,h) \bullet e_M = n(h * e_M)$ = $ne_M = n$, and so $n = e_N$. This implies that $(n,h) \bullet m = (e_N,h) \bullet m = h * m$, meaning $h$ must act trivially. And since the $*$ action on $M$ is faithful, $h$ can only be the identity.
So if $(n,h)$ and $(n',h')$ give the same map $M \to M$, we have
$$ (n((hh'^{-1}) * n'^{-1})),hh'^{-1}) = (n,h)(n',h')^{-1} = (e_N,e_H)$$ $$ \implies hh'^{-1} = e_H$$ $$ \implies h = h'$$ $$ \implies nn'^{-1} = n((hh^{-1}) * n'^{-1}) = e_N$$ $$ \implies n = n'$$
Anyway, I think the main thing that is unfortunate about this definition is having to consider ambient groups $M \ge N$ on which the action can be extended to be faithful. I think that such groups always exist, but introducing them in the characterization doesn't feel natural. If there are other issues with this characterization let me know!