Let $n \in \mathbb{N}$. Is it true that for any $a, b, c \in \mathbb{N}$ satisfying $1 < a, b, c \leq n-2$ the symmetric group ${\rm S}_n$ has elements of order $a$ and $b$ whose product has order $c$?
The assertion is true at least for $n \leq 10$, see here.
Update on Sep 2, 2015: On Aug 10, 2015 Joachim König has posted a preprint to the arXiv which gives a positive answer to the question. Assuming that this preprint is correct, this completely answers the question -- and thus also Problem 18.49 in the Kourovka Notebook.
Update on Jun 18, 2014: The assertion is true at least for $n \leq 50$, see here (4MB text file).
The list of examples in GAP-readable format can be found here.
Added on Dec 11, 2013: This question will appear as Problem 18.49 in:
Kourovka Notebook: Unsolved Problems in Group Theory. Editors V. D. Mazurov, E. I. Khukhro. 18th Edition, Novosibirsk 2014.
Added on Nov 24, 2013: Is there really not enough known about, say, the class multiplication coefficients of ${\rm S}_n$ to answer this question?
Text of the question as of Feb 12, 2013:
This question is a follow-up on Order of elements . Derek Holt's answer to that question is nice, but it seems that the degree of the permutations it gives is a lot larger than necessary.
So, given natural numbers $m, n, k > 1$, what is the smallest $d$ such that the symmetric group of degree $d$ has elements of order $m$ and $n$ whose product has order $k$? - Clearly if the largest of the numbers $m$, $n$, $k$ is prime, then $d$ must be at least $\max(m,n,k)$, and there are some cases where $d$ actually must be larger. However a quick computation suggests that $d = \max(m,n,k) + 2$ might work always. - But does this or a similar bound hold?
EDIT: Smallest-degree examples for all $m, n, k \leq 8, m \leq n$ can be found here.