# how many abelian transitive subgroups of $S_{n}$

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?

• 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 6 at 9:34
• @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 6 at 9:39
• @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 6 at 9:53
• @JackCrawford The order is precisely n because I restrict to abelian subgroup cases, see (math.stackexchange.com/questions/128098/…) – Ling Jun 6 at 11:35

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.
• 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 7 at 7:10