Order of products of elements in symmetric groups 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.
 A: 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.
A: 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?
