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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!

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Since the paper you cite also characterizes solvable numbers, maybe you can extend your filtration and numbering scheme accordingly. And perhaps even incorporate "p-group numbers". –  Michael Zieve Oct 1 '13 at 6:46

1 Answer 1

up vote 7 down vote accepted

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.

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Thank you :) In retrospect, that should have been pretty obvious! –  Alex Youcis Oct 2 '13 at 23:54
    
See also math.stackexchange.com/a/67469/589. –  lhf Jul 17 at 10:56

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