Number of finite groups: is $\operatorname{gnu}(4n) \geq 2 \operatorname{gnu}(n)$?

Question

Let $\operatorname{gnu}(n)$ be the number of finite groups of order $n$.

Question: Is it true that $\operatorname{gnu}(4n) \geq 2 \operatorname{gnu}(n)$ for all $n \geq 1$?

Surely this must be true, and it is clear if $n$ is odd.

For general $n$ one might try to prove that among all semidirect products $C_4 \rtimes A$, $(C_2 \times C_2) \rtimes A$, $C_4 \ltimes A$, $(C_2 \times C_2) \ltimes A$ with $|A| = n$, there are at least $2 \operatorname{gnu}(n)$ different groups up to isomorphism. Is it clear that this is the case?

Intuitively there should not be so many coincidences among these groups, but I am not sure how to make this precise.

(This is a special case $a = 4$ of a math.SE question, which asks if $\operatorname{gnu}(ab) \geq \operatorname{gnu}(a)\operatorname{gnu}(b)$ for all $a,b \geq 1$.)

Answer 1

This is not a complete answer but a reduction to $2$-groups (for example) which I think might be more amenable.

Let $\mathcal{S}$ be a class of groups that is closed under taking direct products and direct factors (and so contains the trivial group). Let $\mathcal{S'}$ be the class of groups with no nontrivial $\mathcal{S}$-group as a direct factor.

Let $G$ be a finite group. Write $G=S_G\times S'_G$ where $S_G\in \mathcal{S}$ and $S'_G\in\mathcal{S'}$. Note that $S_G$ and $S'_G$ are uniquely determined up to isomorphism. Note that $G\in\mathcal{S}$ if and only if $S'_G=1$.

Now, for $T\in\mathcal{S'}$ let $\operatorname{gnu}_T(n)$ be the number of groups $G$ (up to isomorphism) of order $n$ such that $S'_G\cong T$.

Note that, for every $T\in\mathcal{S'}$, we have that $\operatorname{gnu}_T(n)=\operatorname{gnu}_1\left(\frac{n}{|T|}\right)$.

Now, suppose that the result you want holds for groups in $\mathcal{S}$, that is $\operatorname{gnu}_1(4n)\geq2\operatorname{gnu}_1(n)$ for every $n$. Then, for every $T\in\mathcal{S'}$, we have $$\operatorname{gnu}_T(4n)=\operatorname{gnu}_1\left(\frac{4n}{|T|}\right)\geq2\operatorname{gnu}_1\left(\frac{n}{|T|}\right)=2\operatorname{gnu}_T(n)$$ and the result you want follows. (We can partition groups according to $S'_G$ and if the result holds in every part, then it holds for the set of all groups.)

So it remains to pick a class $\mathcal{S}$ so that we can prove your result in that class. Note that $\mathcal{S}$ must include $C_4$ and $C_2^2$ so that the result holds at $n=1$.

A promising attempt is to take $\mathcal{S}$ to be the class of $2$-groups, where the asymptotics are better understood. The best result I know is that $\operatorname{gnu}(2^m)=2^{(2/27)m^3+O(m^{8/3})}$, which is just short of being strong enough here (I think the error term is too big). But we can also restrict $\mathcal{S}$ to a smaller family. It is tempting to try $\mathcal{S}$ being abelian $2$-groups but there are not enough of these. Perhaps $2$-groups of nilpotency class at most $2$ (and of exponent at most $4$)? I'm not sure if the error term is better understood in this case?