# Equations in finite subgroups of unitary groups

Let $n$ be an integer. Andreas Thom mentioned that Camille Jordan showed that there exists some $m \in {\mathbb N}$ (depending on $n$), such that for any pair of $n \times n$-unitaries $u,v \in U(n)$ which generate a finite group, one has $[u^m,v^m] = 1_n$. In particular, there are two distinct words of length $\ell=2m$ over the alphabet $\{u,v\}$ that represent the same group element.

My question is

what is known about how small such an $m_n$ or $\ell_n$ can be?

The connection with the Jordan-Schur Theorem is that if $G$ is a finite group generated by $u$ and $v$, and $H$ a normal abelian subgroup of index $m$, then $u^mH=H$ and $v^mH=H$ (since any group element raised to the order of the group is the identity) and so $u^m$ and $v^m$ belong to $H$, hence, $H$ being abelian, they commute. So if $[G:H]=m$ with $H$ abelian then we have such an $m$.

• I guess you mean, for given $n$, what is the smallest $m_n\ge 1$ with this property? (Compute or at least estimate $m_n$?)
– YCor
Nov 25, 2016 at 2:43
• – YCor
Nov 25, 2016 at 2:45
• OK, it's closely related anyway: first you consider $U(n)$ while the other question also considers $O(n)$. Second, you specify to a special case (two generators and their powers). Thom's answer to the linked question provides you a reasonable upper bound on $m_n$.
– YCor
Nov 25, 2016 at 3:22
• PS these provide upper bounds rather close to $n!$. Here the worse case could be closer to the exponent of the symmetric group (lcm of first $n$ integers), which seems to rather grow exponentially.
– YCor
Nov 25, 2016 at 4:40
• The shortest known identity for $Sym(n)$ is of length $\exp(C \log(n)^4 \log(\log(n)))$. I would expect that this is the crucial case, which dominates the estimates for the length of laws for all finite subgroups. There is no non-trivial known (at least to me) lower bound for the length of such an identity (be it for $Sym(n)$ or all finite subgroups of $U(n)$). The trivial lower bound is linear. Nov 25, 2016 at 22:10