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In the paper 'Nilpotent Numbers' by Pakianathan and Shankar (http://www2.math.ou.edu/~shankar/papers/nil2.pdf), it was proven that every group of order $n$ is nilpotent if and only if $p^k\not\equiv 1\mod q$ whenever $p,q$ are distinct primes with $p^k$ and $q$ dividing $n$. I am interested in finding conditions on $n$ such that every group of order $n$ is nilpotent of class at most $c$; has there been any work on this problem already?

(At least the case of $c=1$ has been done, as these are the abelian numbers.)

Francis Begbie
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2 Answers2

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The numbers $n$ for which every group of order $n$ is nilpotent of class at most $c$ are the nilpotent numbers that are not divisible by any $(c+1)^\mathrm{th}$ power.

This is because a finite nilpotent group is the direct product of its Sylow subgroups, the class of the group is bounded by the classes of its direct factors, and a $p$-group of order $p^{c+1}$ has class at most $c$.

See Wikipedia for the existence of $p$-groups of maximal class.

James
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For a reference, see the following article by Müller.

Müller, Thomas W. An arithmetic theorem related to groups of bounded nilpotency class. J. Algebra 300 (2006), no. 1, 10–15. DOI

The solution is basically the one given by James in his answer.

Mikko Korhonen
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