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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?

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  • $\begingroup$ To be clear, when you say $S(n)$ is a "permutation group" you mean it's a group of monomial matrices? $\endgroup$
    – Qiaochu Yuan
    Commented Jun 5, 2022 at 5:07
  • $\begingroup$ Yes. It is the group of all $n\times n$ permutation matrices. Each element is a monomial matrix where the non-zero entry in each row and column is 1. There are $n!$ elements in the group $S(n)$. $\endgroup$
    – Jonas Anderson
    Commented Jun 5, 2022 at 12:48

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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).

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  • $\begingroup$ Thanks Peter! I'm a physicist and apologize in advance if I'm missing something obvious, but I do not fully understand your construction. I understand that there's always the standard irrep of $S_{n+1}$ ($n\times n$ complex unitary matrices). There's clearly a subgroup isomorphic to $S_{n}$ there. Can I always choose the elements in $S_{n}$ to be represented by the natural (or defining) permutation representation of $S_{n}$ ($n\times n$ monomial matrices)? $\endgroup$
    – Jonas Anderson
    Commented Jun 7, 2022 at 2:54
  • $\begingroup$ This being true I get that there's a larger (finite) group in $U(n)$ given by the elements of $S_{n+1}$. Can you talk me through the proof that this additional generator is monomial for $n\ge 4$? I know it's beyond the scope of my question, but if you could outline how to calculate the `extra' generator in a small example I'd be very grateful. $\endgroup$
    – Jonas Anderson
    Commented Jun 7, 2022 at 2:57
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    $\begingroup$ Inside C^{n+1} with the usual action of S_{n+1} and basis e_1,e_2,..., define vectors b_i by b_i=a(e_1+...+e_n)+be_{n+1}+ce_i where a,b,c are constants such that na+b+c=0 and these vectors b_i are orthogonal (which is another equation to solve for a,b,c, and has solutions). These b_i span the standard irrep of S_{n+1} and if you compute the action with respect to this basis, you'll get the usual permuataion representation for S_n and can work out the extra generator by working out how it acts on this basis. $\endgroup$
    – Peter McNamara
    Commented Jun 7, 2022 at 5:56

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