# Proportion of nonabelian $2$-groups of a certain order whose exponent is $4$

Let

$$\displaystyle A(n)=\frac{\text{number of nonabelian 2-groups of order n whose exponent is }4}{\text{total number of nonabelian 2-groups of order n}}.$$

Using GAP, I could observe the following:

$$A(16)=\frac{5}{9}=0.5556, A(32)=\frac{21}{44}=0.4773, A(64)=\frac{93}{256}=0.3633, A(128)=\frac{820}{2313}=0.3545, A(256)=\frac{30446}{56070}=0.5430 \text{ and } A(512)=\frac{8791058}{10494183}=0.8377.$$

Can one prove that if $n>4$, then $A(n)>\frac{1}{3}$?

• @ Alex, I was trying to classify nonabelian 2-groups that satisfy a certain property "say Q" (which I don't want to talk about here). However, I predicted that the number of nonabelian 2-groups whose exponent is 4 is at least a third of the total no of nonabelian 2-groups of a certain order. I have proved the Q property for nonabelian 2-groups whose exponent is 4. I posed the above question to get other researchers view on the proportion of nonabelian 2-groups of exponent 4. – Chuks Dec 22 '15 at 18:11
• Thanks. BTW I've once computed a file with exponents of all 10494213 groups of order 512 and there are 8791062 of exponent 4. This gives a rough idea about A(512), only not taking into account whether the groups are abelian or not... Oh wait a minute, I have also the data for the order of the centre of all these groups. I will give you exact A(512) shortly. – Alexander Konovalov Dec 22 '15 at 21:40
• @ Alex, thanks a lot for your attempts. It is now left to determine the bound of $|A(n)|$. Can we say something like $\frac{1}{3}<A(n)\leq1$ for $n>4$? If yes, any idea on the lower bound proof? [Lest I forget, I use the GAP code "Length(AllSmallGroups(Size,256,IsAbelian,false,Exponent,4));" to get the number of nonabelian groups of order 256 whose exponent is 4, and "Length(AllSmallGroups(Size,256,IsAbelian,false));" to get the number of nonabelian groups of order 256. But my computer can't give me $A(2^m)$ for $m>9$ in less than an hour.] – Chuks Dec 22 '15 at 23:10
• # meant $m\geq 9$; but you have already provided the value of $A(2^9)$. – Chuks Dec 23 '15 at 0:10
• It may be faster for order 256 to call Length(AllSmallGroups(Size,256,IsAbelian,true)); since this is precomputed (see SmallGroupsInformation(256);). Also, $m=9$ is the maximal $m$ you can do with GAP's Small Groups Library. I don't have any ideas about the proof, though. – Alexander Konovalov Dec 23 '15 at 13:25

It is not hard to prove a weaker result based using the seminal work of Sims (1961) and Higman (1964). Proving your result, I fear, will require a deeper understanding of $$p$$-groups than currently exists. Let $$f(n,p)$$ be the number of groups of order $$p^n$$ and let $$f_2(n,p)$$ be the number of such groups with $$\Phi(G)=G'G^p$$ central and elementary abelian. It follows from Sims that $$f(n,p)\leqslant p^{cn^3+dn^{5/2}}$$ (see [1]) and from Higman that $$f_2(n,p)\geqslant p^{cn^3+en^2}$$ where $$c=\frac{2}{27}$$ and $$d$$, $$e$$ are constants. Note that $$d>0$$. Higman had $$e=-\frac{2}{9}$$.
How does this relate to the ratio $$A(n)$$ in the above question? The groups of exponent dividing $$p^2$$ (and order $$p^n$$) are precisely counted by $$f_2(n,p)$$. Since a group of exponent 2 is elementary, we have $$A(2^n)=\frac{f_2(n,2)-1}{f(n,2)}\geqslant\frac{2^{cn^3+en^2}-1}{2^{cn^3+dn^{5/2}}}\to 0\qquad{\rm as}\ d>0.$$ Taking logarithms of the numerator and denominator above (and ignoring the $$-1$$), this ratio approaches 1, that is $$\lim_{n\to\infty}\frac{cn^3+en^2}{cn^3+dn^{5/2}}=1$$. Even if there were a constant $$d'$$ such that $$f(n,p)\leqslant p^{cn^3+d'n^2}$$ as Sims conjectured on p. 153. We must have $$e as $$f_2(n,p) and then $$\frac{f_2(n,2)}{f(n,2)}\geqslant\frac{2^{cn^3+en^2}}{2^{cn^3+d'n^{5/2}}}=2^{(e-d')n^2}\to 0\qquad{\rm as}\ e-d'<0.$$