Semidirect product action and its geometry

Solution 1:

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.