I have a group with five generators $\sigma_i$, and the following relations:
\begin{split} \sigma_i^2 = \varepsilon \\ |i-j| \neq 1 \implies (\sigma_i\sigma_j)^2 = \varepsilon \\ (\sigma_0\sigma_1)^4 = \varepsilon \\ (\sigma_1\sigma_2)^3 = \varepsilon \\ (\sigma_2\sigma_3)^3 = \varepsilon \\ (\sigma_3\sigma_4)^4 = \varepsilon \\ (\sigma_0\sigma_1\sigma_2\sigma_3\sigma_4\sigma_3\sigma_2\sigma_1)^n = \varepsilon \\ \end{split}
Note that without the last relation this is the Coxeter group $[4,3,3,4]$. So we can think of this as the Coxeter group $[4,3,3,4]$ with an extra relation. Equivalently this is $I_2(n)\wr S_4$.
The task is to prove:
$$ (\sigma_0\sigma_4\sigma_1\sigma_3\sigma_2\sigma_1\sigma_3)^{2n} = \varepsilon $$
or equivalently, that the group $\left\langle\sigma_0\sigma_4,\sigma_1\sigma_3,\sigma_2\right\rangle$ is isomorphic to $I_2(n)\wr S_3$.
Why do I believe this could be true?
I've use the GAP system to confirm that this is true for $n < 16$. While always finite some of these groups begin to get very large, so I think it's very likely this holds for all $n$.
What have I done so far?
Besides confirming it holds for cases, I've tried a couple of approaches.
The first was to simply hope that $\sigma_0\sigma_1\sigma_2\sigma_3\sigma_4\sigma_3\sigma_2\sigma_1 = (\sigma_0\sigma_4\sigma_1\sigma_3\sigma_2\sigma_1\sigma_3)^2$ as if I could prove that, it would solve the problem. It turns out that is not the case.
Next I hoped that $\sigma_0\sigma_1\sigma_2\sigma_3\sigma_4\sigma_3\sigma_2\sigma_1$ and $(\sigma_0\sigma_4\sigma_1\sigma_3\sigma_2\sigma_1\sigma_3)^2$ were conjugates in $[4,3,3,4]$, since conjugates have the same order in a group. GAP can solve for this and it found they were in fact not conjugates.
Jykri Lahtonen pointed out that outer automorphisms preserve order, so I could in addition checking the two values are in the same conjugacy class I could check if applying the dual automorphism to one of the values makes it the conjugate of the other. So if I could show $\sigma_4\sigma_3\sigma_2\sigma_1\sigma_0\sigma_1\sigma_2\sigma_3$ and $(\sigma_0\sigma_4\sigma_1\sigma_3\sigma_2\sigma_1\sigma_3)^2$ are conjugates this would also prove the result. However GAP tells me these are not conjugates either.
Where does this problem come from?
This problem sort of looks like random nonsense. Just something made from bashing together random generators until something stuck. However it is part of a larger problem which I will try to briefly justify here.
The Coxeter group $[4,3,3,4]$ corresponds to the symmetry of the 5-dimensional hypercubic honeycomb, and the additional relation that we give gives a subsymmetry of the 4-torus in 8-dimensional Euclidean space. The 4-torus being a quotient of 4-space.
The relation I am aiming to prove shows that this symmetry has a nice relationship to similar symmetries of the 2-torus in 4-space.
This relationship is useful to me for proving that certain types of polyhedral embeddings exist in 8-space.
