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For a positive integer $n$, let $f(n)$ denote the number of transitive subgroups of the symmetric group $S_n$ up to conjugacy, i.e., the number of transitive permutation groups of degree $n$.

The OEIS sequence A002106 gives the first 47 values of $f(n)$:

1, 1, 2, 5, 5, 16, 7, 50, 34, 45, 8, 301, 9, 63, 104, 1954, 10, 983, 8, 1117, 164, 59, 7, 25000, 211, 96, 2392, 1854, 8, 5712, 12, 2801324, 162, 115, 407, 121279, 11, 76, 306, 315842, 10, 9491, 10, 2113, 10923, 56, 6

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I was surprised to see there are far more in degree $32$ than in the other degrees $\leq 47$.

A complete list of representatives for the transitive permutation groups of degree $32$ was calculated by Cannon & Holt (2008). Speculating on the prospect of extending their calculations to larger $n$, they write:

We have not yet seriously considered the problem of calculating lists of transitive permutation groups of higher degree. We would hazard a guess that for degrees between 33 and 64, 36 and 48 would be very difficult but potentially possible, degree 64 would be impossible with current techniques and resources, and the other degrees would be comparatively straightforward.

So it seems that the authors expect a similar blowup to happen in degree $64$, suggesting there is something special that occurs when the degree is a power of $2$.

Another relevant observation from the paper is the following:

Although there are $487$ distinct orders of groups in the final list, 2,737,535 of the 2,801,324 groups, that is, about 97.7% of them, are $2$-groups of order $2^n$ for $5 \leq n \leq 31$, the most popular order being $2^{15}$, of which there are 391,809 groups. Only 1051 of the groups are insoluble.

My takeaway from this quote is that the phenomenon is partially explained by the fact that $S_{32}$ has lots and lots of $2$-subgroups, though I don't immediately see why so many of them ought to be transitive.

Question. Is there a known explanation for why there are so many transitive permutation groups in degree $32$ (and conjecturally, larger powers of $2$)? Are there nontrivial lower bounds for $f(2^k)$ in general?

N. F. Taussig
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algebra learner
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    I hazard a guess. Cayley representations. And the fact that $99.99\ldots$ percent of finite groups (up to isomorrphism) are 2-groups. There is probably more to this given that the nilpotent groups have normal subgroups also, and that leads to transitive groups of larger orders. – Jyrki Lahtonen Jul 24 '25 at 14:25
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    Perhaps @Derek Holt will comment. – ancient mathematician Jul 24 '25 at 14:58
  • Are you familiar with the fact that almost all finite groups are $2$-groups? I think that (together with the fact that a necessary condition for a group $G$ to be a transitive subgroup of $S_n$ is that $n \mid |G|$) is already most of the way towards an explanation. It's not clear to me whether this is part of what you want to know about. – Qiaochu Yuan Jul 24 '25 at 16:27
  • @QiaochuYuan No, I did not know that about $2$-groups, but I can see why that would play a large role in an explanation. I'm reading this post about it now. – algebra learner Jul 24 '25 at 16:30

1 Answers1

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Most of these groups $G$ are subgroups of $C_2 \wr S_{16}$, where the projection $P$ of $G$ onto $S_{16}$ is one of the $1954$ transitive groups of degree $16$, and the intersection of $G$ with the base group of the wreath product is a submodule of the permutation module for $P$ over ${\mathbb F}_2$.

But in some cases you can get lots of distinct groups with with the same projection $P$ and the same intersection with the base group. The bulk of the calculation involved cohomology calculations to find all such groups.

Note that there are $195826352$ transitive groups of degree $48$, so if you extend your graph to the right, you will get another huge jump. See here for some details. That computation took more than $2$ CPU-years. Again most of the groups (about 98%) are subgroups of $C_2 \wr S_{24}$, and there are $25000$ transitive subgroups of $S_{24}$.

I have tried to extend the lists to higher degrees, and I have got up to degree $53$, but I am stuck on degree $54$.

Derek Holt
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