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Inspired by another question, I wondered when $\mathfrak{S}_n \times \mathfrak{S}_m$ is isomorphic to a subgroup of $\mathfrak{S}_p$. Eliminating the obvious cases, the question becomes:

Let $n,m,p>1$ be such that $\mathfrak{S}_n \times \mathfrak{S}_m \hookrightarrow \mathfrak{S}_p$. Does it imply that $p \geq n+m$?

I was able to prove that the statement is true for $p \leq 10$ using David Ward's argument and the following easy results:

Claim 1: $\mathfrak{S}_n$ and $\mathfrak{A}_n$ are indecomposable.

Claim 2: If $\mathfrak{S}_n \times \mathfrak{S}_m \hookrightarrow \mathfrak{S}_p$ then $\mathfrak{S}_n \times \mathfrak{S}_m \hookrightarrow \mathfrak{A}_p$.

EDIT: Derek Holt gave a simple solution below, using however a difficult result. Therefore, an elementary solution would be appreciated.

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I'm not sure why would a french student use gothic german letters to denote a group that almost all the world, including, I think, France, denote by the latin letter S, but it may cause confusion. –  DonAntonio Jul 21 '13 at 13:44
nice question, but what that $U_p$ means? Is that $A_p$? Thanks –  B. S. Jul 21 '13 at 13:44
@BabakS, that is a gothig-german $\,A\,$ , not a $\,U\,$ . –  DonAntonio Jul 21 '13 at 13:47
ams.org/mathscinet-getitem?mr=2265507 is a related question of when there is a maximal subgroup of this form. The $p$ are much larger than $n+m$. –  Jack Schmidt Jul 21 '13 at 13:48
@DonAntonio: I think the symmetric group was always denoted by $\mathfrak{S}_n$ in all French text I read. Can it be misleading? If so I can change the notation. –  Seirios Jul 21 '13 at 13:50
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1 Answer

up vote 10 down vote accepted

Assume that $m+n>p$. Then at least one of $m,n$ - say $n$ - satisfies $n>p/2$. For $n \ge 7$, the only faithful transitive action of $S_n$ of degree less than $2n$ is the natural one. (See, for example,

Liebeck, Martin W.; Praeger, Cheryl E.; Saxl, Jan. A classification of the maximal subgroups of the finite alternating and symmetric groups. J. Algebra 111 (1987), no. 2, 365–383.)

So if $m+n \ge 13$, then there is a set of $n$ points on which $S_n$ acts naturally. Then the centralizer of $S_n$ in $S_p$ must fix every point in this orbit of $S_n$, so it has order at most $(n-p)!$, and hence it cannot contain $S_m$.

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I don't really understand why the only faithful transitive action of $S_n$ of dregree less than $2n$ is the natural one. Is it a consequence of O'Nan-Scott theorem? –  Seirios Jul 21 '13 at 16:11
Well it's a nontrivial result! I gave a reference to a paper by Liebeck, Praeger and Saxl, which does indeed use the O'Nan Scott Theorem. There is also a 1980 paper of Praeger and Saxl, which says that a primitive permutation group of degree $n$ that does not contain $A_n$ has order at most $4^n$, which was proved before the classification of simple groups, so I I would guess that your question can be answered without using CFSG. There have been stronger results proved using the classification. –  Derek Holt Jul 21 '13 at 16:34
Thank you. I will not accept your answer yet expecting a possible elementary solution. +1. –  Seirios Jul 23 '13 at 8:07
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