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](http://www.jstor.org/stable/2589118).


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!



Generalizing the Notion of Nilpotent/Abelian/Cyclic Numbers
A nilpotent group is the direct product of its Sylow subgroups, hence the nilpotency class of a nilpotent group equals the maximum of the nilpotency classes of its Sylow subgroups. A group of order $p^n$ has nilpotency class at most n-1, since if $G/G'$ was cyclic, then $G$ itself would be cyclic. On the other hand, $p$-groups of maximal class do exist, so we obtain that for $n\geq 2$, $C\_n$ consists precisely of the nilpotent groups of class $\leq n-1$, while $C\_1$ is the class of cyclic groups.


