Order of products of elements in symmetric groups
The question has meanwhile been answered in the positive in:
Joachim König, A note on the product of two permutations of prescribed orders. European Journal of Combinatorics 57 (2016), 50-56.
The proof makes a case distinction based on the smallest $n$ such that there are elements of given orders $a \leq b \leq c$ in the symmetric group on $n$ points whose product is the identity:
$n = c$ is sufficient,
$n = c+1$ is needed and sufficient, and
$n = c+2$ is needed.
It analyzes the various subcases, and makes use of Corollary 4.4 and Lemma 4.5 in:
A.L. Edmonds, R.S. Kulkarni, R.E. Stong: Realizability of branched coverings of surfaces. Trans. Amer. Math. Soc. 282(2) (1984), 773--790.
First let me paraphrase the question. Given integers $m,n,k$ each at least 2, set $d:=\max(m,n,k)+2$. Do there exist elements $a,b$ in the symmetric group $S_d$ such that $|a|=m$, $|b|=n$ and $|ab|=k$?
It is convenient to write $c=ab$. A simple argument shows that we can assume $m\leq n\leq k$ and $d=k+2$. [The equation $b*a=bcb^{-1}$ shows we can swap $m$ and $n$ as $|bcb^{-1}|=k$. The equation $a^{-1}c=b$ shows we can swap $n$ and $k$ as $|a^{-1}|=m$. Thus we may assume $m\leq n\leq k$.]
It is easy to prove the result for small cases such as $m=n=2$. With these simple ideas Stefan's table of data can be simplified, and extended. It seems that there may be results already in the literature. Can an expert help? What about the special case when $m,n,k$ are each powers of the same prime?