Are Hall subgroups of symmetric groups determined up to isomorphism by their order?

Question

(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?

Answer 1

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.