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Aren't the practical consequences it would have if P=NP highly over-rated? -- I mean, if P=NP, this would by far not guarantee that one can get any practical algorithm from it (my gut feeling is that it would be just a result of theoretical interest). On the other hand, even if P!=NP, there may still be algorithmic breakthroughs of ENORMOUS practical importance -- even if the asymptotic complexity of the new algorithms is quite boring.

@Borromean I merely reacted to a complaint of the answerer. As far as you reach agreement with them, almost anything is fine with me.

Please refrain from making substantial changes to your question which may render an answer invalid. I rolled back your question to the original version.

@markvs I checked that the assertion is true for all non-simple perfect groups of order less than 1000000.

@markvs More precisely, there are 444 perfect groups of order less than 100000. Of these, 31 are simple and 1 is trivial -- so in this range of orders, there are 412 nontrivial perfect groups which are not simple.

An immediate observation is that the class of DTI-groups is closed under taking subgroups.

@DavidESpeyer In the first line of the question, it is said "Let $r(m)$ denote the residue class $r + m\mathbb{Z}$, where $0 \leq r < m$". -- This is to exclude $2(2)$, $7(3)$ and the like. -- And that condition is also very essential to ensure that the groups indeed act on $\mathbb{N}_0$ -- i.e. don't move nonnegative integers to negative integers and vice versa.

Welcome on MathOverflow! -- Can you perhaps elaborate, to turn this into a complete answer?

@markvs Simple groups are unlikely to be counterexamples. -- For ${\rm PSL}(2,q)$, the asymptotics is just the given bound, and for other simple groups, even the bound $|G|^{\frac{3}{4}}$ seems to hold.