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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.

EDIT: There is now a more elementary solution to this and also to the more general problem where there can be more than two direct factors on http://mathoverflow.net/questions/167349

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nice question, but what that $U_p$ means? Is that $A_p$? Thanks –  Babak 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
    
@seirios: I think you can use the classification of maximal subgroups of $\mathfrak{A}_p$ to see it. Intransitive and imprimitive should work by induction. The primitive tend to be fairly small, but I'm not sure how to handle this case yet. –  Jack Schmidt Jul 21 '13 at 13:53
    
So for $p=11$ for instance, the only maximal subgroup that is not handle by induction are the Mathieu groups of degree 11, but its order is only divisible by 5 once, and not by 7, so $\mathfrak{S}_{m} \times\mathfrak{S}_{n}$ cannot be contained in that subgroup if $m \geq 7$ or if $5 \leq m < 7$ (since $n \geq 5$ too) or if $m < 5$ (since $n \geq 7$). Obviously this isn't enough to handle general $p$, but it works for $p=11,12,13,14$, etc. individually. –  Jack Schmidt Jul 21 '13 at 14:03

1 Answer 1

up vote 14 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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