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> Is there a way to prove, that $\lim\_{n \to \infty} \frac{\text{the number of all } 2 \text{-generated groups of order less than }n}{\text{the number of all groups of order less than } n} = 0$?
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This statement is implied by a well known conjecture:



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



Is there a way to prove, that $2$-generated groups are rare among finite groups?
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:



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> *Enumerating Boundedly Generated Finite Groups*, Journal of Algebra **238** (2001) pp 194–199. doi:[10.1006jabr.2000.8650](https://doi.org/10.1006jabr.2000.8650), [core.ac.uk version](https://core.ac.uk/display/82373534).
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