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

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?

• 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. Aug 3 '19 at 15:40
• Does this not follow from the fact that almost all groups of order bounded by $n$ are $2$-groups? Aug 3 '19 at 20:28
• @IgorRivin It would follow from that statement, but that statement is not known. Aug 3 '19 at 20:33
• Yes sorry it was too late! Aug 4 '19 at 6:05

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