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 monomial matrix (also known as generalized symmetric matrices). In other words, $U_n$ is not a product of a permutation and a diagonal matrix.
A permutation group, $S(n)$, is maximal if the addition of any non-monomial matrix, $U_n$, generates a group $G=\langle S(n), U_n \rangle$ where the order of $G$ is infinite. 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?
Best Answer
The standard representation of Sn+1 is faithful and n-dimensional. We may also assume it preserves a Hermitian inner product. When restricted to a standard copy of Sn, it becomes isomorphic to the permutation representation. Use this isomorphism to choose a basis, and hence realise Sn+1 as living between your S(n) and U(n), hence providing a negative answer to the last question. (Sn+1 is not generated by a monomial matrix as there is no homomorphism from Sn+1 to Sn for n at least 4).