Residual properties of iterated semidirect products

Consider $G^{\rtimes k} := ((G \rtimes \dots ) \rtimes G)\rtimes G$ with diagonal action by inner automorphisms, $G^{\rtimes 1} = G$. Let $\mathcal P$ be a collection of groups. Is it true that

• $G$ residually $\mathcal P$ $\Rightarrow$ $G^{\rtimes \geq 2}$ residually $\mathcal P$? (very unlikely, but I can't find a counterexample)
• $G^{\rtimes 2}$ residually $\mathcal P$ $\Rightarrow$ $G^{\rtimes \geq 2}$ residually $\mathcal P$?

I'm mostly interested in first question with $\mathcal P$ a quasivariety and $G$ finitely generated, but anything goes.

• Does $G^{\rtimes 0} \Doteq G$ belong to your sequence? – Luc Guyot Sep 21 '17 at 19:55
• (For me it seems more natural to set $G^{\ltimes 1} = G$…) I've edited question a bit. – Denis T. Sep 21 '17 at 21:43
• If $\mathcal{P}$ is not stable under taking subgroups, there's ambiguity about the meaning of "being residually $\mathcal{P}$". – YCor Sep 21 '17 at 22:09
• It's Magnus's "residual", i. e. intersection of $N \triangleleft G$ with factor in $\mathcal P$ is trivial. – Denis T. Sep 21 '17 at 22:32

The group law of $G^{\rtimes 2}$ is given by $$(g, h)(g',h') = (ghg'h^{-1}, hh')$$ and more generally for $G^{\rtimes k}$, we have $(g_1, \dots, g_{k -1}, g_k)^{(1, \dots,1, g)} = (g_1^g, \dots,g_{k - 1}^g, g_k^{g})$ where $x^y \Doteq y^{-1}xy$.
Therefore $G^{\rtimes k}$ is just $G^k$ in disguise.
Indeed, we have $G^{\rtimes 2} = (G \times \{1\}) \cdot D(G)$ with $D(G) = \left\{ (g, g^{-1}) \, \vert \, g \in G \right\} \lhd G^{\rtimes 2}$ and $[G \times \{1\}, D(G)] \subset (G \times \{1\}) \cap D(G) = 1$.
Considering the following normal subgroups of $G^{\rtimes 3}$, $$\left\{ ((g, 1), 1) \, \vert \, g \in G \right\}, \, \left\{ ((g, g^{-1}),1) \, \vert \, g \in G \right\},\, \left\{ ((1, g), g^{-1}) \, \vert \, g \in G \right\}$$ we see how the isomorphism $G^{\rtimes k} \simeq G^k$ generalizes for $k \ge 3$.
• Oh yes sure: in $G\times G$, one takes one factor and the diagonal to see the semidirect decomposition with conjugation action (no need for $G$ to be center-free as I thought) – YCor Sep 23 '17 at 11:31