0
$\begingroup$

My task is the following:

Given the permutations $\sigma =(1 \space 2 \space 3 \space 4)$ and $\tau = (1 \space 3)$ and the permutation group $G=(S_4, \circ)$, find $|<\sigma, \tau>|$.

Is there a fast way to calculate this? If $\sigma$ and $\tau$ were disjoint, $|<\sigma, \tau>| = |<\sigma>|*|<\tau>|$ I think, but as both permutations contain $1$ and $3$, I couldn't find any faster way to calculate this, than brute forcing all elements of $<\sigma, \tau>$.

As this was an exam task, I'm convinced there is a faster way tho, but I can't find anything in the lecture script.

Thanks in advance for your answers :)

  • 0
    Write the numbers 1, 2, 3, 4 clockwise on a square and think about what the two permutations do to the numbers on the square. – 2017-02-21
  • 0
    In this particular case, it's 90 degreee rotation and reflection through the diagonal thought 2 and 4, so you should get $D_4$, right? While in this particular case the geometric interpretation works nice and well, in another exam it was $\sigma=(1 \space 3)$ and $\tau=(1 \space 5)(4 \space 6)$. There's probably a geometric interpretation here as well, but not as easy as $D_4$ I suppose... – 2017-02-21
  • 0
    $D_4$ is correct. In this new example, I can't see a geometric interpretation... – 2017-02-21

1 Answers 1

1

This might be one way to look at it. Let $G$ be a graph on $n$ vertices, and on each vertex $v_i$ place a pebble, $p_i$. Let each $\sigma\in S_n$ place edges in a cycle on $G$ connecting all $i\in[n]$ such that $\sigma(i)\ne i$. Interpret a permutation as moving these pebbles along the vertices in a single direction.

For example, given $\pi = (1 2)$ and $\sigma = (14)(35)$, we have a graph that has an edge between $1$ and $2$, and then a matching contain edges $1,4$ and $3,5$. This means that the pebbles on $3$ and $5$ repeatedly swap, while any permutation of the pebbles on the path $1,2,4$ are possible. The position of the pebbles on $3,5$ are dependent on the parity of the permutation on the path. Without working it out by hand, I would guess $|\langle\sigma, \pi\rangle| = 12$

  • 0
    That raises the question whether it is true, that if two cycles in $\pi$ and $\sigma$ share a vertex, all permutations among the vertices in those two cycles are achievable. So i.e. if $\pi=(123456)$ and $\sigma=(9874)$, can all permutations of $S_9$ be achieved, just by combining $\pi$ and $\sigma$, because they share $4$? If that's the case, $|<\pi, \sigma>|$ is actually prettty easy to compute. – 2017-02-21
  • 1
    In general, I do not believe this is the case -- your permutations $\pi$ and $\sigma$ must be generators for $S_n$, given $n$ non-fixed points. It is not true that any two permutations generate the symmetric group. The reason this works in particular is because each permutation is a singular edge on a path, which happens to an induced subgraph. – 2017-02-21