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It's well-known that any abelian transitive subgroup A of a symmetric group $S_{n}$ has order $n$. Moreover, does anyone know how many abelian transitive subgroups of $S_{n}$ and what do they look like?

Santana Afton
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LQ001
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  • My immediate gut reaction is that they should all be cyclic, but I don't really know how to prove it, or whether it's really true. And just counting the number of different transitive cyclic subgroups isn't entirely trivial either. – Arthur Jun 06 '19 at 09:34
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    @Arthur The Klein-4 group is an example of an abelian, non-cyclic transitive subgroup of $S_4$. They do not need to be cyclic. – Jack Crawford Jun 06 '19 at 09:39
  • @Ling I believe you meant to say that $n$ divides its order, not that it has order $n$, right? If it has order exactly $n$, I can't believe I haven't heard of this until now! My galois theory exam was this morning! – Jack Crawford Jun 06 '19 at 09:53
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    @JackCrawford The order is precisely n because I restrict to abelian subgroup cases, see (https://math.stackexchange.com/questions/128098/show-that-any-abelian-transitive-subgroup-of-s-n-has-order-n) – LQ001 Jun 06 '19 at 11:35

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Every abelian group $A$ of order $n$ acts transitively on itself by left translation (aka Cayley action). Therefore we can view any such $A$ as a transitive subgroup of $S_n$. Furthermore, given that this kind of an action necessarily has trivial stabilizers, such an action has to be isomorphic to the translation action. In other words, a given abelian group $A$ of order $n$ has an essentially unique (up to conjugation by an element of $S_n$) transitive action of this type.

The problem of listing abelian groups of a given order $n$ (up to isomorphism) is relatively easy, but does depend on having the full factorization $n=\prod_{i=1}^kp_i^{a_i}$ with $p_i$ ranging over the prime factors of $n$. You need to list all the possible $p$-parts for $p=p_i,i=1,2,\ldots,k$. This amounts to partitioning $a_i$ in all possible ways, and the group is then a direct sum of its Sylow $p_i$-subgroups. The number of different (up to isomorphism) abelian groups of order $n$ is thus $$ \prod_{i=1}^kp(a_i), $$ where $p$ is the partition function.

Jyrki Lahtonen
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    Yes that's a nice "formula" for the number of conjugacy classes of abelian subgroups of $S_n$. The number of groups in the conjugacy class of such a group $A$ is equal to $(n-1)!/|{\rm Aut}(A)|$, so the exact number abelian subgroups of $S_n$ is more difficult to express concisely. – Derek Holt Jun 07 '19 at 07:10