Question

Consider natural numbers $m,n,k > 1$. There are finite groups $G$ containing elements $x,y$ such that $o(x) = m, o(y) = n$ and $o(xy) = k$. After embedding these groups in $S_\mathbb{N}$ we drive: Given $m,n,k >1$ there are $f,g \in S_\mathbb{N}$ such that $o(f) = m, o(g) = n$ and $o(fg)=k$. This argument shows the existence of $f,g$. But how to construct such permutations $f,g$ in $S_\mathbb{N}$ in terms of $m$, $n$ and $k$ ?

Answer 1

It is not hard in practice, at least for reasonably small values of $m,n,k$, to find permutations with the required property, and I did once convince myself that I could find a systematic way of doing it, but it was very messy and hard to describe.

Here is one way to do it, which will not produce permutations of optimally small degree, but which is at least easy to describe and even to do on a computer using GAP or Magma if you want to.

We will produce our permutations as elements of ${\rm PSL}(2,q)$ for suitable odd $q$. We do this by finding elements $X,Y \in {\rm SL}(2,q)$ of orders $2m,2n$ with product of order $2k$, and use their images in ${\rm PSL}(2,q)$. We make use of the fact that, for elements of ${\rm SL}(2,q)$ with order greater than 2 and dividing $q^2-1$, their order is detmined by their trace.

It is easiest to do if you choose $q$ such that $q-1$ is divisible by all of $2l,2m,2k$. We can assume that $m,n,k$ are all greater than 1. Let $ \lambda,\mu $ be primitive $ 2m^{\mathrm{th}} $, $ 2n^{\mathrm{th}} $ roots of unity in the field ${\mathbb F}_q$ of order $q$. Let

$ A = \left( \begin{array}{cc} \lambda & 0 \\ 1 & \lambda^{-1} \end{array} \right) $ and $ B = \left( \begin{array}{cc} \mu & \alpha \\ 0 & \mu^{-1} \end{array} \right) $, for some $ \alpha $. So $ |A| = 2m $ and $ |B| = 2n $. Now \[ AB = \left( \begin{array}{cc} \lambda\mu & \lambda\alpha \\ \mu & \alpha + \lambda^{-1}\mu^{-1} \end{array} \right) \] and $ {\rm Tr}(AB) = \lambda \mu + \alpha + \lambda^{-1} \mu^{-1} $. Let $ t $ be the trace of an element of $ S $ of order $ 2k $, and choose $ \alpha = t - \lambda\mu - \lambda^{-1}\mu^{-1} $. Then $ |AB| = 2k $.

As an example, I tried $m,n,k = 10,12,15$ with $q=121$, and took $\lambda=w^6$, $\mu=w^5$, $\alpha=w^4+w^{-4}-w^{11}-w^{-11}$, with $w$ a primitive field element. Then, using Magma to project onto ${\rm PSL}(2,121)$, I got the following permutations $x,y \in S_{122}$ with $(|x|,(y|,|xy|) = (10,12,15)$.

$x=$ (2, 105, 31, 119, 44, 29, 107, 98, 58, 120)(3, 36, 24, 18, 48, 12, 54, 30, 42, 60)(4, 19, 72, 87, 38, 117, 33, 102, 41, 66)(5, 73, 79, 104, 49, 56, 45, 118, 112, 69)(7, 57, 91, 67, 106, 109, 101, 50, 46, 114)(8, 103, 64, 81, 74, 21, 43, 40, 35, 77)(9, 32, 89, 34, 84, 65, 61, 47, 116, 80)(10, 26, 82, 17, 27, 94, 100, 70, 76, 99)(11, 37, 93, 83, 39, 52, 108, 14, 86, 68)(13, 75, 121, 22, 78, 23, 110, 97, 62, 113)(15, 85, 63, 92, 55, 122, 115, 95, 59, 16)(20, 25, 96, 28, 71, 88, 90, 111, 51, 53)

$y=$ (1, 95, 18, 77, 29, 118, 22, 71, 39, 80, 38, 67)(2, 99, 65, 119, 41, 102, 89, 51, 75, 113, 70, 53)(4, 103, 74, 86, 106, 37, 10, 7, 88, 97, 66, 90)(5, 115, 109, 47, 112, 61, 31, 78, 57, 98, 121, 15)(6, 87, 91, 26, 63, 60, 43, 120, 73, 13, 20, 68)(8, 35, 12, 76, 114, 46, 104, 94, 32, 45, 48, 34)(9, 122, 72, 59, 23, 62, 101, 30, 50, 64, 42, 33)(11, 107, 108, 40, 69, 117, 21, 49, 27, 58, 17, 19)(14, 56, 36, 54, 116, 25, 85, 79, 84, 81, 55, 83)(16, 92, 93, 110, 100, 44, 52, 24, 82, 111, 96, 105).

Added later: I should perhaps add that by starting with elements $x,y \in S_8$ with orders 10,12, and replacing $y$ by random conjugates, I rapidly found a solution in $S_8$: $x=(1,2,3,4,5)(6,7)$, $y=(1,2,4,7)(3,6,8)$, so I would guess that some kind of reasonably intelligent random algorithm is the best way to solve this in practice.