Let us call a number $n\in\mathbb{N}$ *nilpotent* if

$$n=p_1^{e_1}\cdots p_m^{e_m}$$

with $p_i^k\not\equiv 1\mod p_j$ for $i,j\in\{1,\ldots,m\}$ and $1\leqslant k\leqslant e_i$.

A cute theorem says the following:

Theorem:Every group of order $n$ is nilpotent, if and only if $n$ is a nilpotent number.

But, the interesting thing is that the theorem doesn't stop there. It also says the following:

Theorem:Every group of order $n$ is abelian, if and only if $n$ is a cubefree nilpotent number.

Theorem:Every group of order $n$ is cyclic, if and only if $n$ is a squarefree nilpotent number.

For this reason squarefree and cubefree nilpotent numbers are called *cyclic* and *abelian* respectively.

The proofs of these theorems can be found here.

There is, of course, after hearing these last two theorems an obvious question: what does being a nilpotent number indivisible by an $\ell^{\text{th}}$ power, for $\ell\geqslant 4$ correspond to? In particular, one might guess that there is some filtration of classes of groups

$$C_2\subseteq C_3\subseteq C_4\subseteq\cdots\subseteq\{\text{nilpotent groups}\}$$

where all groups of order $n$ are in $C_\ell$ if and only if $n$ is a nilpotent number not divisible by an $\ell^{\text{th}}$-power.

For example, we have already remarked that

$$C_1=\{\text{cyclic groups}\}\qquad C_2=\{\text{abelian groups}\}$$

Also, if we can find such classes $C_j$, is there some unifying, governing statistic of groups naturally indexed by $\mathbb{N}$, for which $C_j$ is just the class of groups satisfying the $j^{\text{th}}$ statistic?

Thanks!