Poly virtually vs virtually poly In here it is proved that a group is poly virtually abelian if and only if it is virtually poly abelian (see that page also for the definition of virtually and poly). Do you know of a property P and a group that is virtually poly P but not poly virtually P? And do you know of a property P and a group that is poly virtually P but not virtually poly P?
 A: When $1\in$ P is nonempty and passes to finite index subgroups, virtually poly-P clearly implies poly-(virtually P).
(Proof added, 29 September 2018) Indeed, suppose that $G$ is virtually poly-P, and P passes to finite index subgroups and $1\in$ P. Then poly-P also passes to finite index subgroups. So virtually-P implies (poly-P)-by-finite, which implies poly-P.
See below for counterexamples for more general P.
The converse fails, for instance, when P=(nonabelian simple) or P=(nonabelian simple or 1).
It also fails assuming that P passes to subgroups. For instance, it fails when P=torsion-free. Indeed, here poly-P = P. Virtually poly-P means virtually torsion-free, which is weaker than poly-(virtually-P). Indeed there exist finite-by-P groups that are not virtually torsion-free (=P-by-finite). I think the inverse image of some torsion-free index subgroup of $\mathrm{SL}_3(\mathbf{Z}[1/2])$ in the 2-fold covering of $\mathrm{SL}_3(\mathbf{R})$ is such an example.
Note that in both cases, already finite-by-P fails to imply virtually-(poly-P).

Here is a sufficient condition the converse to hold. Suppose that 


*

*all finite abelian groups satisfy P

*every virtually poly-P group has a characteristic subgroup of finite index that is poly-P

*virtually P passes to normal subgroups of finite index.


Then poly-(virtually P) implies virtually-(poly P).
Indeed, using induction, it is enough to show that if $G$ is ((poly P)-by-finite)-by-(virtually P) then $G$ is virtually poly-P. Using (2), $G$ is (poly P)-by-(finite-by-(virtually P)); say $N$ is normal in $G$, with $N$ poly-P and $H=G/N$ is finite-by-(virtually P). Considering the kernel of the action on the finite kernel and using (3), $H$ is virtually (finite central)-by-(virtually P). By (1), $H$ is virtually (2-step poly-P). So $G$ is virtually poly-P.
Note that (3) is clear when P is stable under taking subgroups, and (2) holds when P is stable under taking quotients. In particular, P=abelian satisfies (1),(2),(3), and more generally any if $V$ is any variety of infinite exponent, P=(belonging to $V$) satisfies all the conditions. 
Also, P=simple satisfies all the conditions (for (2), use that a poly-simple group has only finitely many subgroup of finite index).

About the implication (virtually poly-P) $\Rightarrow$ poly-(virtually P).
The naive trivial proof of this implication actually makes implicit assumptions on P (thanks to Jeremy Rickard for doubts twice about careless claims!).
First, let's be careful about the definition of poly-P: it should be understood that $1$ is poly-P (as 0-fold iterated extension). In this case, when P is the empty class, the "clear" inclusion fails: virtually poly-P means finite, while poly-(virtually P) means trivial.
Also, more seriously, when P is the singleton $\{S\}$ where $S$ is a single infinite simple group, the wreath product $(S\times S)\rtimes C_2$ is virtually poly-P, but is not poly-(virtually P), and is not even poly-(virtually (infinite simple)).
(Edit 29 Sept 2018) One more example. Let $S$ be an infinite simple group. Consider the wreath product $G_n=S^n\rtimes A_n$. Write $P=\{1,G_3\}$. Then $G_6$ is virtually poly-P but not poly-(virtually P). Indeed, if $C$ is the subgroup (3-Sylow) of $A_6$ generated by the cycles $(123)$ and $(456)$, then the subgroup of finite index $S^6\rtimes C$ of $G_6$, is isomorphic to $G_3\times G_3$. On the other hand, the subnormal groups of $G_6$ are $G_6$, its subgroup of index 2, and all possible subsums of $S^6$. In particular, the possible subquotients are isomorphic to $S^k$ for some $k\ge 6$ or are finite (and $S^k$ is not virtually P). Hence $G_6$ is not poly-(virtually P). 
