Suppose we have a group $N$ acting on a set $X$, and a group $Q$ acting on $Y$. Moreover, we are given $\phi: Q \to \mathrm{Aut}(N)$.

My question is, is there a "natural" action on a set that can be constructed from $X$ and $Y$ (maybe $X \times Y$?) on which the semidirect product $N \rtimes_{\varphi} Q$ acts.

For example, when $\phi$ is trivial (so we are considering direct product of groups), then the answer to this question is trivial (our set is $X \times Y$ and the action is coordinatewise).

PS In general I am interested in a case when $X$ and $Y$ have more structure, but let us start with sets.

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    $\begingroup$ I don't think you can get an action on $X\times Y$ unless $\varphi$ is particularly well-behaved with respect to the action. You probably want to take a quotient of $X$ modulo something that "equalizes" the actions of all elements $\varphi(q)(n)$, as $q$ ranges over all elements of $Q$. $\endgroup$ – Arturo Magidin Nov 23 '15 at 20:12
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    $\begingroup$ In the case that $X=Y$ you can extend the actions to the semi direct product iff they semi directly commute. (i.e. $q(nx)= \phi(q)(n)(qx)$) $\endgroup$ – Daniel Valenzuela Nov 23 '15 at 21:26
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    $\begingroup$ On one hand the (unrestricted) wreath product $N \;\mathrm{WR}_\Omega\; Q$ acts canonically on $X\times Y$ (see wikipedia), on the other hand by the Krasner-Kaloujnine embedding theorem any extension of $N$ by $Q$ embeds in this unrestricted wreath product. So you can rephrase your question to "Is the restriction of the action to the semidirect product independent of the chosen embedding?" $\endgroup$ – j.p. Nov 24 '15 at 10:07
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    $\begingroup$ @j.p.: If $N$ acts on $X$ and $Q$ acts on $Y$, then $N\wr_Y Q$ acts on $X\times Y$; the Kaloujnine-Krasner embedding, though, maps the extension into $N\wr_Q Q$, so you would need $Q$ acting on itself by left multiplication to get that working, no? $\endgroup$ – Arturo Magidin Nov 24 '15 at 15:17
  • $\begingroup$ @ArturoMagidin: Oh, this can well be true! I'll leave my wrong comment and upvote yours in the hope that in future nobody else gets this wrong idea! $\endgroup$ – j.p. Nov 24 '15 at 17:25

Suppose you define some equivalence relation on $X$ so that the quotient $\overline{X}$ has the property that $N\rtimes Q$ acts on $\overline{X}\times Y$ coordinatewise, with the action of $n$ on $[x]$ being $[nx]$ (here, $[x]$ is the equivalence class of $x$).

Writing ${}^qn$ for $\varphi(q)(n)$, and writing the elements of $N\rtimes Q$ as pairs $(n,q)$, we would need $[nx] = [{}^qnx]$ for all $q\in Q$ (with fixed $x$ and $n$). For $({}^qn,1) = (1,q)(n,1)(1,q^{-1})$, and looking at the actions on a pair $([x],y)$ readily gives the need for $[nx]=[{}^qnx]$.

Conversely, if $\sim$ is any equivalence relation on $x$ that satisfies this property, then the action on $\overline{X}\times Y$ given by $(n,q)([x],y) = ([nx],y)$ is readily seen to be an action: $$\begin{align*} (n_1,q_1)\Bigl((n_2,q_2)([x],y)\Bigr) &= (n_1,q_1)([n_2x],q_2y)\\\ &= ([n_1(n_2x)],q_1(q_2y))\\\ \Bigl((n_1,q_1)(n_2,q_2)\Bigr)([x],y) &= (n_1{}^{q_1}n_2,q_1q_2)([x],y)\\\ &= ([n_1{}^{q_1}n_2 x],(q_1q_2)y). \end{align*}$$ The assumption on $\sim$ gives that $[{}^{q_1}n_2 x] = [n_2x]$, and hence that $[n_1{}^{q_1}n_2 x] = [n_1n_2x]$.

If you let $\sim$ be the transitive closure of the relation with $x\sim x'$ if and only if there exists $x_0\in X$, $n\in N$, and $q\in Q$ with $nx_0 = x$, ${}^qnx_0 =x'$, then any equivalence relation that is coarser than $\sim$ will do. (I may have my "coarser" and "finer" mixed up here; I mean that the equivalence classes are unions of equivalence classes under $\sim$, so that if you think of it as if it were a "topology", you would have a coarser topology, with fewer open sets).

In particular, taking $\sim$ to be the total relation ($x\sim x'$ for all $x,x'\in X$), you get the natural action on $Y$ given by mapping to the quotient (well, technically an action on $\{\bullet\}\times Y$, but that is naturally isomorphic to an action on $Y$). If the action of $Q$ on $N$ is trivial, so that ${}^qn = n$, then $\sim$ is the identity relation, so $\overline{X}=X$ and you get the natural action on $X\times Y$.

  • $\begingroup$ Of course, there are other things on which $G=N\rtimes Q$ acts naturally, such as $N$ (on whom it acts by conjugation)... $\endgroup$ – Arturo Magidin Nov 23 '15 at 20:48

In general, $N$ (acting on itself) is a natural answer. Let us consider the example of the affine group of a vector space $V$: the affine group is the semi-direct product of $Aut(V)=Gl(V)$ and the additive group of $V$. This is "the" natural space where $Aff(V)$ acts.


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