I asked the following question on [math.stackexchange](https://math.stackexchange.com/questions/448772/when-is-mathfraks-n-times-mathfraks-m-a-subgroup-of-mathfraks-p) several months ago:



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


Derek Holt gave a positive answer. However, because the problem is easy to solve for small values of $p$, I am really curious to know if there exists an "elementary" solution.



When is $S_n \times S_m$ a subgroup of $S_p$?
I think I can do this with an elementary argument, but I have to rush off somewhere soon, so I will answer quickly and hope I get it roughly right!


Assume $p < n+m$, so we can assume also that $n>p/2$. $S\_n$ must have a faithful orbit in its action on $p$ points, and if that orbit has size $k$ then $n \le k \le p < 2n$.


Let $H$ be the point stabilizer of the action of $S\_n$ on this orbit of length $k$. Then since $|S\_n:H| < 2n$ and any proper subgroup of $A\_n$ has index at least $n$ (I am assuming $n \ge 5$, since you can do small cases separately), we cannot have $H < A\_n$, so $H$ must be a maximal subgroup of $S\_n$.


But then the action of $S\_n$ on this orbit is primitive, and primitive groups have trivial centralizer. (That's because the point stabilizer in a primitive group fixes a unique point, since otherwise its fixed points would be a block, and hence the centralizer fixes that point and hence all points.)


So the other factor, $S\_m$ must fix all points in that orbit, but that's impossible, because $m > p-k$.


