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 https://mathoverflow.net/questions/167349
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 $S_m$.