# Existence of a solution for an equation in a permutation group

Here is a concrete example, but I'm looking for methods in general :

Let $S_{13}$ be the permutation group.
Let $i : S_2 \times S_3 \times S_4 \times S_4 \to S_{13}$ be the canonical injection.
Let $H=i(S_2 \times S_3 \times S_4 \times S_4)$ be the subgroup of $S_{13}$.

Let $\sigma \in S_{13}$ be the following product of transposition : $$\sigma = (1,6)(2,10)(4,7)(5,11)(9,12)$$

Let the equation : $$\sigma p_1 \sigma p_2 \sigma p_3 \sigma = p_4 \sigma p_5$$ with the variables $p_i \in H$.

Question : Is there a solution to this equation ? which one ?

Remark : Equivalently, is there $h_i \in H$ such that $\sigma h_1 \sigma h_2 \sigma h_3 \sigma h_4 \sigma \in H$ ?

• Hello. Could you provide some motivation for the question? – Julien Dec 13 '13 at 1:02
• This equation is a particular case of one of the pentagonal equations of the second fusion ring here. In general, $\sigma$ is a permutation matrix in $M_{13}(\mathbb{C})$ and the variables $p_i$ are invertible matrices in $M_{2}(\mathbb{C}) \oplus M_{3}(\mathbb{C}) \oplus M_{4}(\mathbb{C}) \oplus M_{4}(\mathbb{C})$. If the general equation has no solution, the fusion ring isn't categorifiable. In fact I hope there is a solution, and here I'm looking for a solution with permutations. – Sebastien Palcoux Dec 13 '13 at 1:44
• Ok, thanks. Maybe you should add the "finite-groups" tag while you're at it. – Julien Dec 13 '13 at 1:56

Here is a solution:

$S_{13}$ acts on $\{1,2,3,4,5,6,7,8,9,10,11,12,13 \}$.
The invariant subsets of $H$ are $\{ 1,2 \}$, $\{ 3,4,5 \}$, $\{ 6,7,8,9 \}$, $\{ 10,11,12,13 \}$

Take $h_1=h_2=e$, so we have to find $h_3, h_4 \in H$ such that $\sigma h_3 \sigma h_4 \sigma \in H$
(with $\sigma = (1,6)(2,10)(4,7)(5,11)(9,12)$)

Next take:
$h_3=(1,2)(6,8,7,9)(10,13,11,12) \in H$
$h_4 =(4,5)(6,9)(7,8)(10,12)(11,13) \in H$

Then $\sigma h_3 \sigma h_4 \sigma = (1,2)(6,8,7,9)(10,13,11,12)=h_3 \in H$

Remark: I have found this solution by using diagrams, a bit like solving a generalized Sudoku...
I will try to explain this method later.