Are Hall subgroups of symmetric groups determined up to isomorphism by their order? (For example,) if $k$ is coprime to $n$ and there is a Hall subgroup of order $k$ in $S_n$, then it is clear, that it isomorphic to the Hall subgroup of order $k$ contained with $S_{n-1}$. But does this extend, up to determining all Hall subgroups of symmetric groups just by their order? Edit: Immediately after typing this question, it became clear that this works by induction. The Hall subgroup of $S_{n-1}$ is in turn isomorphic to that of $S_{n-2}$, and so on, as long as it's coprime to each of $n,\,n-1\,\dots$. As soon as it's not coprime, there is no Hall subgroup of that order any more. So, by induction, they are all isomorphic. Does this sound correct?

The results of the papers

P. Hall, "Theorems like Sylow's", Proc. London Math. Soc. (3) **6** (1956), 286--304,

which deals with soluble Hall subgroups, and

J.G. Thompson, "Hall subgroups of the symmetric groups", J. Combinatorial Theory **1** (1966), 271--279,

which deals with the non-solvable case, show that the only Hall subgroups of $S_n$, apart from the trivial group, $S_n$ itself, and its Sylow subgroups, are

* $S_{n-1}$ (and its conjugates) when $n$ is prime
* Hall $\\{2,3\\}$-subgroups of $S_7$ and $S_8$,



and that in the second case there is only one conjugacy class of Hall subgroups.

Since the Hall $\\{2,3\\}$-subgroups of $S_7$ and $S_8$ don't have the same order as any symmetric group, that gives a positive answer to your question.