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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 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.

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An even stronger conjecture would be that it is the smallest $d$ such that there exist permutations $x,y,z∈\in S_d$ of orders $m,n,k$ such that ${\rm sgn}(x){\rm sgn}(y)={\rm sgn}(z)$ (where ${\rm sgn}(x)$ is $1$ and $−1$ for even and odd permutations). Can you find a counterexample to that? –  Derek Holt Jan 4 '13 at 22:53
@Derek: I am not sure I understand your question right - depending on $m$, $n$ and $k$, the degree $d$ may of course be smaller than $\max(m,n,k)$. I performed a brief computation and found that for $d = 3, 4, 5, 6, 7, 8 and 10$ your setting permits precisely the same triples (m,n,k), while for $d = 9$, in addition (2,5,20), (2,20,5), (5,5,10), (5,5,12), (5,10,5), (5,12,5) and (5,20,2) occur (this means e.g. that there are permutations $x$, $y$ and $z$ of orders 5, 5 and 12 with sign +1 in $S_9$, but not such that $xy=z$, etc.). –  Stefan Kohl Jan 5 '13 at 0:28
It looks as though my conjecture was over-optimisitic! –  Derek Holt Jan 5 '13 at 10:10

1 Answer 1

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?

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