Minimum size of the generating set of a direct product of symmetric groups Let $m$ and $n$ be positive integers. Let $S_m$ and $S_n$ be the symmetric groups on the sets $\{1,\dots,m\}$ and $\{1,\dots,n\}$, respectively. What is the minimum size of a generating set for the direct product $S_m\times S_n$?
 A: Let $S_m$ act on $\{1,2,\ldots,m\}$ and $S_n$ on $\{1',2',\ldots,n'\}.$
[adding more detail to the following paragraph as per request]
A fact given in most elementary texts on permutation groups is that the 2-cycles
$(12),(13),(14),\ldots,(1m)$ generate all of $S_m$.
With that result known we next see that two generators $\sigma=(12)$ and $\tau=(123\cdots m)$ generate all of $S_m$. This is because we get sufficiently many transpositions by conjugating the former by powers of the latter. For example $\tau\sigma\tau^{-1}=(23)$, $\tau^2\sigma\tau^{-2}=(34)$
et cetera. Further conjugating gives $(13)=(23)(12)(23)$ and so forth.
Similarly we can use $\tau'=(23\cdots m)$ in place of $\tau$: $\tau'^k\sigma\tau'^{-k}$ gives the 2-cycles $(1j), 1<j\le m$, and having these suffices.
I think that two generators will always suffice to generate the direct product. 
If $m$ and $n$ are both odd, then it is clear that the permutations $\alpha=(123\ldots m)(1'2')$ and
$\beta=(12)(1'2'3'\ldots n')$ generate the whole thing, because $\alpha^2$ and $\beta^n$ generate $S_m$ and $\alpha^m$ and $\beta^2$ generate $S_n$. The general observation here is that if a permutation $\sigma$ is the product of two
disjoint cycles of coprime lengths, then the individual cycles belong to the subgroup generated by $\sigma$.
If either $m$ (resp. $n$) is even, then we use $(23\ldots m)(1'2')$ (resp.
$(12)(2'3'\cdots n')$) instead as the other generator. The key is that the longer
cycle is of an odd length, so the above observation applies.
Addedum: A single generator obviously won't do :-)
