I'm going by the maxim
Groups, like men, are known by their actions
This naturally leads one to ask "given groups $G, H$ which act on sets $S, T$ and the semidirect product $G \rtimes H$, how does one visualize the action of $G \rtimes H$? What does it act on? Some combination of $S$ and $T$? ($S \times T$ perhaps?)
I know some elementary examples, likr $D_n \simeq \mathbb Z_n \rtimes \mathbb Z_2$. However, given an unknown situation, I am sure I cannot identify whether it is a semidirect product that is governing the symmetry.
The best responses on similar questions like intuition about semidirect product tend to refer to this as some kind of "direct product with a twist". This is shoving too much under the rug: the twist is precisely the point that's hard to visualize. Plus, not all "twists" are allowed — only certain very constrained types of actions turn out to be semidirect product. I can justify the statement by noting that:
the space group of a crystal splits as a semidirect product iff the space group is symmorphic — this is quite a strong rigidity condition on the set of all space groups.
The closest answer that I have found to my liking was this one about discrete gauge theories on physics.se, where the answer mentions:
If the physical space is the space of orbits of $X$ under an action $H$. Ie, the physical space is $P \equiv X / H$. Then, if this space $P$ is acted upon by $G$. to extend this action of $G \rtimes H$ onto $X$ we need a connection.
This seems to imply that the existence of a semidirect product relates to the ability to consider the space modulo some action, and then some action per fiber. I feel that this also somehow relates to the short exact sequence story(though I don't know exact sequences well):
Let $1 \rightarrow K \xrightarrow{f}G \xrightarrow{g}Q \rightarrow 1$ be a short exact sequence. Suppose there exists a homomorphism $s: Q \rightarrow G$ such that $g \circ s = 1_Q$. Then $G = im(f) \rtimes im(s)$. (Link to theorem)
However, this is still to vague for my taste. Is there some way to make this more rigorous / geometric? Visual examples would be greatly appreciated.
Best Answer
There are multiple way one can construct permutation actions connected to $S$ and $T$:
Approach 1: Consider the wreath product $G \wr H:= G^H \rtimes H$, where $H$ acts on $G^H$ via pre-composition with the right multiplication, i.e. ${^h \eta} := x \mapsto \eta(xh)$. Then $(g,h) \mapsto (\eta_g,h)$ with $\eta_g(x):={^x g}$ is an embedding of $G\rtimes H$ into $G^H \rtimes H$.
Wreath products of permutation groups have two natural permutation actions, on the cartesian product of both and on the set of maps from one set to the other. In our case $G^H\rtimes H$ acts both on $S\times H$ and on $S^H$.
You can "visualise" this like so: Every element of $H$ gives you an automorphism of $G$ and therefore a way to twist (permutation) representations of $G$: For every $h\in H$ we can define $^h S$ as the $G$-set which as a copy of $S$, say $\{^h s \mid s\in S\}$, as elements and the action $({^h g})(^h s) := {^h (gs)}$. With this notation, $S\times H$ is just $\bigsqcup_{h\in H} {^h S}$ and $S^H$ is just $\prod_{h\in H} {^h S}$ where $H$ acts by permuting the ${^h S}$ amongst themselves and $G$ acts on the components individually.
If you like to think in modules: $k[S\times H]$ equals the induced module $\operatorname{Ind}_{G}^{G\rtimes H}(k[S])$ and $k[S\times H]$ equals what is called the "tensor induction" of $k[S]$.
And of course, once we have that we can combine either of those with the quotient $G^H \rtimes H \twoheadrightarrow H$ and the action of $H$ on $T$ to obtain $G^H \rtimes H$-actions on $S\times H \times T$, $(S\times H)\sqcup T$, $S^H\times T$, $S^H \sqcup T$, $(S\times H)^T$, $(S^H)^T$ etc.
Approach 1.5: Or we can use that $H$ not only has the right-multiplication-action on $H$, but also an commuting action by left-multiplication. In particular, $S\times H$ and $S^H$ not only have the $G \wr H$-action, but also a commuting $H$-action. The second construction is therefore to take $(S\times H)\times_H T$ and $S^H\times_H T$ instead of the cartesian products, i.e. to quotient out the additional $H$-action.
The first set is somewhat connected to your guess $S\times T$, but is smaller in general.
Approach 2 is to think of two actions on the same set $\Omega$. If $G$ and $H$ both act on $\Omega$ and the actions commute in the sense that $^g(^h \omega) = {^h({^g\omega})}$, then $G\times H$ acts on $\Omega$. A typical case of this is what's called a "biset", i.e. a set with an left-$G$-action and a compatible right-$H$-action. Think: $G$ acts via left-multiplication and $H$ acts via right-multiplication on $\Omega$ and we want associativity to hold. Actions of the semidirect product are "twisted bisets" in this sense. Think of them as $H$ acting by conjugation and $G$ acting by left-multiplication on $\Omega$.
What's another object on which $G$ acts by left multiplication and $H$ by conjugation? The set $G^T$. $G$ clearly acts by left multiplication in the image and since both $G$ and $T$ are $H$-sets we get a conjugation action of $H$. Together we get the $G\rtimes H$-action $(^{(g,h)}\gamma)(t) := g\cdot{^h\gamma(^{h^{-1}} t)}$.
Similar in the way $S^H$ was similar to $S\times H$ we can also combine the $H$-action on $G\times T$ with the $G$-action by left multiplication (trivial on $T$) to obtain the $G\rtimes H$-action ${^{(g,h)}(x,t)} := (ghxh^{-1},ht)$.
And of course, we can now combine all of these together to get $G\rtimes H$-actions on $G^T \times S^H$ and other funny looking things.