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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!

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    $\begingroup$ A group is nil potent if and only if it is a direct product or groups of prime-power order $\endgroup$
    – ahulpke
    Commented Mar 26, 2021 at 13:51
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    $\begingroup$ See wikipedia for the classification up to order $60$, and the references given there. This requires some work and is far from "random". $\endgroup$ Commented Mar 26, 2021 at 13:51
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    $\begingroup$ And (since this seems an exercise) is there early the expectation to list all groups of order 32? That would be lots of work from scratch. $\endgroup$
    – ahulpke
    Commented Mar 26, 2021 at 13:52
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    $\begingroup$ @ahulpke I even suspect that there is a slight misunderstanding here for this homework. Indeed, listing all $51$ nilpotent groups of order $32$ without more context seems to be absurd. Also, there are $14$ different nilpotent groups of order $48$. $\endgroup$ Commented Mar 26, 2021 at 13:55
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    $\begingroup$ "All not nilpotent" is impossible, since the group $C_n$ is nilpotent for every $n\ge 1$. So for each order we have at least one nilpotent group. Conversely, for some numbers all groups are nilpotent, see this duplicate. $\endgroup$ Commented Mar 26, 2021 at 16:03

1 Answer 1

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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 $2\cdot 1=2$.

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  • $\begingroup$ What are your thoughts on this linked question? I am trying to prove a general recursion. math.stackexchange.com/questions/4084160/… $\endgroup$ Commented Mar 31, 2021 at 12:55
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    $\begingroup$ Yes, that is true exactly because the groups are direct product of their Sylows. $\endgroup$
    – ahulpke
    Commented Mar 31, 2021 at 20:09

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