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 strongrigidity condition on the set of all space groups.

This question on the natural action of the semidirect product identifies one choice of natural space for the semidirect product to act on, by introducing an unmotivated (to me) equivalence relation, which "works out" magically. What's actually going on?

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.

(NOTE: this is cross posted from `math.se`

after getting upvotes but no answers)

thesemidirect product of $G\rtimes H$ as there are, in principle many (as many as there are homomorphisms $G\to Aut(H)$). So, to form your question correctly, you need THREE inputs: the action of $G$ on $S$, the action of $H$ on $T$, and the action of $G$ on $H$ described by whichever homomorphism $G\to Aut(H)$ that you choose. $\endgroup$ – Nick Gill Feb 13 '20 at 12:19