Let $S(n)$ be the (unitary) matrix group of $n\times n$ permutation matrices. This is clearly a finite group of order $n!$. It is well known that we can add diagonal unitary matrices with any finite root of unity entries to this group and the new group is also finite. For any finite group of permutations and diagonal matrices we can always add diagonal matrices with higher root of unity entries to the group and the group will remain finite. There is, therefore, no maximal finite group in this case. Here we are interested in a slightly different notion of maximal finite group (see below).
Let $U_n\in U(n)$ be a unitary matrix which is not a (unitary) monomial matrix (sometimes calledalso known as generalized symmetric matrices). In other words, $U_n$ is not a product of a permutation and a diagonal matrix.
IfA permutation group, $S(n)$, is maximal if the addition of any non-monomial matrix, $U_n$ as, generates a generator withgroup $S(n)$$G=\langle S(n), U_n \rangle$ where the order of ($\langle S(n), U_n \rangle$) always generates an$G$ is infinite order group. When this is the case, I will refer to the group $S(n)$ as a maximal finite group.
With this definition of a maximal finite group I'm interested in the following questions:
Can we prove the existence of maximal groups $S(n)$ for some $n$?
Can we prove that there exists an integer $k$ such that for all $n>k$ and any non-monomial matrix $U_n$, that $\langle S(n), U_n \rangle$ necessarily generates an infinite order group?
I can show that $k>4$ see the $H\otimes H$ example in the comments to this question: Are generalized symmetric groups maximal finite groups (in a certain sense)?