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Is there a way to prove, that $\lim_{n \to \infty} \frac{\text{the number of all } 2 \text{-generated groups of order less than }n}{\text{the number of all groups of order less than } n} = 0$?

This statement is implied by a well known conjecture:

$\lim_{n \to \infty} \frac{\text{the number of all groups of nilpotency class } 2 \text{, exponent } 4 \text{ and order less than }n}{\text{the number of all groups of order less than } n} = 1$

(because the $2$-generated relatively free group of nilpotency class $2$ and exponent $4$ is finite)

However, the problem with aforementioned statement is, that it is...

...well, just a conjecture.

Is there a way to strictly prove the main statement of the question?

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  • $\begingroup$ Do I remember right that the conjecture you mention is equivalent to the same conjecture but with "p-group of order less than n" in the numerator? If so, that seems discouraging because 2-generated is a powerful condition if you already know the group is a p-group (Burnside basis theorem, etc.), but seems pretty useless for groups that aren't p-groups. $\endgroup$ Aug 3 '19 at 15:40
  • $\begingroup$ Does this not follow from the fact that almost all groups of order bounded by $n$ are $2$-groups? $\endgroup$
    – Igor Rivin
    Aug 3 '19 at 20:28
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    $\begingroup$ @IgorRivin It would follow from that statement, but that statement is not known. $\endgroup$
    – verret
    Aug 3 '19 at 20:33
  • $\begingroup$ Yes sorry it was too late! $\endgroup$
    – Derek Holt
    Aug 4 '19 at 6:05
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Lubotzky proved that the number of $2$-generator groups of order at most $n$ (or order exactly $n$) is bounded by $n^{A \log n}$ for some constant $A$. The number of $2$-groups of order $2^m$ with $n \ge 2^m > n/2$ is (corrected, see Will Sawin's comment) $n^{B \log^2 n}$ for some explicit constant $B$. So you win! Using Lubotzky's theorem, you still win even if $2$-generator groups are replaced by $d$-generator groups for any fixed $d$.

Here is Lubotzky's paper:

Enumerating Boundedly Generated Finite Groups, Journal of Algebra 238 (2001) pp 194–199. doi:10.1006jabr.2000.8650, core.ac.uk version.

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    $\begingroup$ There's a cube in the exponent for $p$-groups, so in fact we win by a mile. $\endgroup$
    – Will Sawin
    Aug 4 '19 at 3:03
  • $\begingroup$ $\log^2 n=(\log n)^2$? Or $\log(\log n)$? I'm guessing the former, but I initially thought the latter on reading it. $\endgroup$ Aug 4 '19 at 8:26

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