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.

**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.