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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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    A group is nil potent if and only if it is a direct product or groups of prime-power order – ahulpke Mar 26 '21 at 13:51
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    See wikipedia for the classification up to order $60$, and the references given there. This requires some work and is far from "random". – Dietrich Burde Mar 26 '21 at 13:51
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    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. – ahulpke Mar 26 '21 at 13:52
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    @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 $48$. – Dietrich Burde Mar 26 '21 at 13:55
  • Is there an easy way to at least rule out orders, the groups of which are all not nilpotent? Also, some way to find (apart from what I've already done in the post) to find orders, the groups of which are all nilpotent? I'm not sure if there are more groups than the ones I've already stated. If I know, I'll only have to worry about non-abelian groups of the remaining orders. –  Mar 26 '21 at 14:45
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    "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. – Dietrich Burde Mar 26 '21 at 16:03
  • @DietrichBurde You were right, there was a slight misunderstanding. I have edited the problem with a clarification, instead of explicitly characterising the nilpotent groups, we are only required to find the number of nilpotent groups of every order. Do you think you'd have a slick way of doing that? –  Mar 26 '21 at 16:25
  • As was mentioned, a finite group is nilpotent if and only if it is a direct product of $p$-groups. But describing all $p$-groups is hopeless for anything but the smallest exponents. – Arturo Magidin Mar 26 '21 at 16:28
  • Yes, I think I have have a slick way of doing that. But it seems still too complicated to prove this without using some results. So the question remains why and how you need to know this? What is the context? – Dietrich Burde Mar 26 '21 at 16:28
  • The number of nilpotent groups of a particular order $n=\prod p_i^{e_i}$ is the product of the numbers of nilpotent groups of orders $p_i^{e_i}$ for the prime factors. But these numbers will be hard to find from scratch for $p_i^{e_i}\in{16,27,32}$ unless you were given them as part of the problem – ahulpke Mar 26 '21 at 16:29
  • @DietrichBurde What's your slick way? Since you ask for context: This is part of an optional homework exercise I received for my group theory course sometime back. Still haven't been able to complete it since it seemed like too much crazy computation. Would appreciate your help! – stoic-santiago Mar 26 '21 at 16:41
  • I suggest you ask the person who offered this exercise to give you a hint on how to classify the groups of order 16 and 32. They will probably amend the problem to exclude a few orders. – ahulpke Mar 26 '21 at 16:45
  • @epsilon-emperor It follows immediately from the nilpotent groups being direct products of their Sylow subgroups. – ahulpke Mar 26 '21 at 16:46
  • @epsilon-emperor Thanks for clarifying that. For others: epsilon-emperor and I were classmates for the group theory course! Happy to see you on MathSE! ahulpke, How about we exclude $16,32,48$? These seem like the most complicated. Still curious about finding numbers for the rest! –  Mar 26 '21 at 16:52

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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 $2\cdot 1=2$.

ahulpke
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  • What are your thoughts on this linked question? I am trying to prove a general recursion. https://math.stackexchange.com/questions/4084160/recursive-expression-for-lambdan-number-of-nilpotent-groups-of-order-n – stoic-santiago Mar 31 '21 at 12:55
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    Yes, that is true exactly because the groups are direct product of their Sylows. – ahulpke Mar 31 '21 at 20:09