Find the number of all nilpotent groups of order $<60$, up to isomorphism - i.e. for every $n \in \{1,2,\ldots,59\}$, find the number of nilpotent groups up to isomorphism.
We know that
Result 1: Every abelian group is nilpotent.
Result 2: Every finite $p$-group is nilpotent.
Hence, all abelian groups of orders $1$ to $59$ are nilpotent. Additionally, all groups of orders $2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,37,41,43,47,49,53,59$ are nilpotent, due to Result 2.
So the only groups left to worry about, are non-abelian groups of non-prime (or prime power) order. This feels really random, and cumbersome. Is there a systematic approach to listing all the desired nilpotent groups? Thanks!
For a prime $p$ there is one group of order $p$, and two (cyclic and $p\times p$) of order $p^2$. Also (and this is the part that is hard in that hand calculation will require days or more) there are $5$ groups each of orders $8$ and $27$, $14$ groups of order $16$ and $51$ groups of order $32$. With this information you can calculate the numbers of nilpotent groups of these orders as products of the numbers for the dividing prime powers. For example, the number of nilpotent groups of order $12=4\cdot 3$ is the number of groups of order $4$ (which is $2$ because of the $p^2$ rule) times the number of groups of order $3$ (which is one) for a total of $2\cdot 1=2$.