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Let $G$ be a group, with $|G|=40,$ and $X$ a set with $|X|=67$ and $G$ act on $X.$ There are $5$ orbits for this action of $G$ on $X.$

How many lists of orbits lengths there are?

I know that the leght of an orbit divides the order of the group $G$ and that the sum of orbits length is equal the order of the set $X$, but I don't know how to solve the problem.

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Here's some Mathematica code to get the possible lists of orbits:

IntegerPartitions[67, {5}, Divisors[40]]

There are six such partitions, and so there are six possible actions here.

To do this by hand, it's not that bad to just try some things out (although obviously proofs by enumeration/exhaustion are annoying). First of all, note that $67/4=13.5$. This means that the list can't consist of orbits of size less than 10. So, start with a list consisting of orbits no larger than 20. Since $20+10+10+10+10 = 60 < 67$ but $20+20+10+10+10 = 70$, it's clear that there needs to be at least one orbit of size 40 or two orbits of size 20.

From there, it's a matter of trying things out. For instance, for a list with exactly two orbits of size 20 (and none of size 40), we need the three final orbits to sum to $67-40=27$. Quickly playing around with the numbers 1, 2, 4, 5, 8, and 10 yields nothing that works: since $8+8+8=24$, there needs to be at least one 10, so can we find two orbits that sum to 17? Clearly not. Etc.

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Maybe three years late, but we can get 5-orbits possibilities with GAP code:

Filtered(Partitions(67, 5), q -> ForAll(q, r -> r in DivisorsInt(40)));

which offers six orbit collections $\{20, 20, 20, 5, 2\}$, $\{40, 10, 8, 5, 4\}$, $\{40, 10, 8, 8, 1\}$, $\{40, 10, 10, 5, 2\}$, $\{40, 20, 4, 2, 1\}$, $\{40, 20, 5, 1, 1\}$.

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    $\begingroup$ I don't know why we need the one line of code in different programming languages as additional answers to this old question. $\endgroup$
    – xxxxxxxxx
    Commented Dec 13 at 13:09
  • $\begingroup$ I'm happy to remove it as it duplicates the answer (6) to the question seemingly posed. I'm not sure we've yet learned anything about the orbits though since one or more of those 6 may actually be 'impossible' - for other reasons not yet mentioned. $\endgroup$
    – Other paul
    Commented Dec 13 at 16:37
  • $\begingroup$ It's probably good to remove, it more than duplicates the answer as being 6, it duplicates the code even (just translated Mathematica to GAP). There is not much else to say about the orbits, without knowing what the group $G$ is; without restrictions on that, all of these orbit structures are possible. $\endgroup$
    – xxxxxxxxx
    Commented yesterday

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