TITLE: When is $S_n \times S_m$ a subgroup of $S_p$?
QUESTION [27 upvotes]: I asked the following question on math.stackexchange several months ago:

Let $n,m,p>1$ be such that $S_n \times S_m \hookrightarrow S_p$. Does it imply that $p \geq n+m$?

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.

REPLY [12 votes]: There is a more general result to be had (which I presume is known to permutation group specialists). For a finite group $G,$ let $m(G)$ denote the smallest degree of a faithful permutation representation  of $G.$ Let $X_{1},X_{2},\ldots X_{t}$ be finite simple groups with possibly some of the $X_{i}$ being Abelian: then $m( X_{1} \times X_{2} \times \ldots \times X_{t}) = \sum_{i=1}^{t} m(X_{i}).$ Clearly the minimal degree of the product is at most the sum of the minimal degree of the individual components. We prove the reverse inequality by induction on $t,$ the result being clear when $t = 1.$  The proof is not so different from Derek's, but differs in places, so I give it.
Set $X$ to be the direct product of the $X_{i},$ and suppose that $X$ acts faithfully on $\Omega$ with $|\Omega| = m(X).$ Then it follows that we may suppose that $X$ acts transitively on $\Omega.$ For if $\Omega = \Omega_{1} \cup \Omega_{2}$ with each $\Omega_{i}$ being $X$-invariant, and the union being disjoint, let $N$ be the pointwise stabilizer of $\Omega_{1},$ which is proper and non-trivial. Then $N$ acts faithfully on $\Omega_{2}$ and $X/N$ acts faithfully on $\Omega_{1}.$ Since $N$ is isomorphic to a direct product of some of the $X_{i}$ and $X/N$ is isomorphic to the direct product of the remaining ones, our claim is established. If possible, let $\Gamma$ be a proper orbit (on $\Omega$) of a proper non-trivial normal subgroup $N$ of $X,$ and choose $N$ as large as possible. Let $K$ be the kernel of the action of $N$ on $\Gamma.$ Since $X$ is a direct of simple groups, all subnormal subgroups of $X$ are normal, so $K \lhd X.$ But $X$ then permutes the fixed points of $K$ on $\Omega,$ while $X$ is also transitive, so $K =1$ and $N$ acts faithfully on $\Gamma.$ 
Now all orbits of $N$ on $\Omega$ have size $|\Gamma|$, and $X/N$ acts faithfully as a permutation group on the set of such orbits by maximality of $N.$ 
Thus $|\Omega| \geq m(N) m(X/N) \geq \sum_{i} m(X_{i})$ . Hence we may suppose that each non-trivial normal subgroup of $X$ acts transitively on $\Omega.$ Then for any $i,$ we see that $\prod_{j \neq i} X_{j}$ acts regularly on $\Omega,$ since $X_{i}$ permutes the fixed points of any element of that product, while the product also acts transitively. But likewise, since $t >1$, $X_{i}$ must also act regularly. If we label to that $|X_{i}| \leq |X_{j}|$ whenever $i \leq j,$ then we see that we must have $t =2$ and $|X_{1}| = |X_{2}| = |\Omega|.$
If $X_{1}$ is Abelian, then we have a contradiction since an elementary Abelian group of order $p^{2}$ can't act faithfully on $p$ points when $p$ is prime. If $X_{1}$ is non-Abelian, we note that $m(X_{1}) \leq \frac{|X_{1}|}{3}$, since $X_{1}$ has a corefree subgroup of order greater than $2.$ Thus $m(X_{1} \times X_{2}) \leq 2 \frac{|X_{1}|}{3},$ a contradiction.
  The special result about symmetric groups can be recovered from this one, since if $Y$ is a direct product of symmetric groups, then $F^{*}(Y)$ is a direct product of simple groups,
and we get an alternating group factor of degree $k$ for each symmetric group of degree $k \geq 5,$ while we get a cyclic factor of degree $k$ for each symmetric group $S_{k}$ with $k \leq 3,$ while we get two cyclic factors of degree $2$ for each symmetric group factor $S_{4}.$ Since $S_{4}$ has minimal degree $4,$ this works as required.
(Later edit: A couple of further remarks: if $S$ is a non-Abelian finite simple group, there is indeed a faithful action of $S \times S$ on a set of $|S|$ points in which both factors act regularly. The action is on the cosets of the diagonal subgroup $\Delta(S)$ of $S \times S.$ It is also the case, with a little more thought, that any such simple group $S$ has $m(S) \leq \frac{|S|}{12},$ though the much cruder bound above was sufficient for that argument. For if $S$ has even order, then if $S$ has a Sylow $2$-subgroup of order less than $16$,$S$ has a subgroup of order $12$ or $56$ by transfer arguments, while if $S$ has odd order, $|S|$ is divisible by by $p^{3},$ again by transfer, where $p$ is the smallest prime divisor of $|S|$, as was known to Burnside (the latter is just a device to avoid invoking the full strength of the Feit-Thompson odd order theorem)).