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Let $S(m,n)$ be the generalized symmetric group which is a wreath product of the cyclic group of order $m$, denoted here by $\mathbb{Z}_m$, and the symmetric group $S_n$. A standard unitary representation of $S(m,n)$ is given by the semi-direct product of the $n\times n$ permutation matrices and $n \times n$ diagonal matrices with $m$-th roots of unity entries. (These matrices are sometimes called generalized permutation matrices.)

For finite $m$ and $n$ all such unitary matrices form a finite subgroup of $U(n)$ of size $n! m^n$. I'll denote this group of unitary matrices as $M(m,n)$. One way to see that $M(m,n)$ is not a maximal finite subgroup of $U(n)$ is to note that $S(2m,n)$ is a finite group which has $S(m,n)$ as a subgroup. I'm interested in a slightly different question.

Let $U'\in U(n)$ be a unitary matrix which is not in $M(m,n)$ for any $m$ (in other words, $U'$ is not a generalized permutation matrix). Is the group generated by $M(m,n)$ (for some fixed $m$) and $U'$ always infinite?

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  • $\begingroup$ The question is equivalent to the question for $m=1$: is it true that for $U'$ unitary and not monomial, $\langle U',S_n\rangle$ is infinite? $\endgroup$
    – YCor
    Commented May 20, 2022 at 12:17
  • $\begingroup$ Yes. That is equivalent to the question I intended to ask. If it is finite for $m=1$, I'm interested in when/if it becomes infinite for larger $m$. $\endgroup$
    – Jonas Anderson
    Commented May 20, 2022 at 13:52
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    $\begingroup$ Let $H_2 = 2^{-1/2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$, then the group generated with $S_{2}$ is finite, as is the group generated by $H_4 = H_2 \otimes H_2$ with $S_{4}$, I think. Computing minimal polynomials of a few random products $H_{16}PH_{16}$ shows that many of these matrices have infinite order, though. $\endgroup$
    – Padraig Ó Catháin
    Commented May 20, 2022 at 14:56
  • $\begingroup$ @PadraigÓCatháin Excellent point! The matrix groups $M(1, 2), M(2, 2)$, and $M(4,2)$ are all subgroups of the 1-qubit Clifford group (which is isomorphic to the octahedral group). The Clifford group is a maximal finite subgroup of $U(2)$ which contains $H_2$. See arxiv.org/abs/math/0001038 for more details. $\endgroup$
    – Jonas Anderson
    Commented May 21, 2022 at 3:34
  • $\begingroup$ @PadraigÓCatháin Thanks again! $M(1,4)$ and $M(2,4)$ are subgroups of the 2-qubit Clifford group which is a maximal finite subgroup of $U(4)$. This Clifford group contains $H_2 \otimes H_2$ (and $H_2 \otimes I_2$ and $I_2 \otimes H_2$ for that matter). I'm especially interested in whether the above conjecture is true for $M(1, 8)$, a purely permutation group. $\endgroup$
    – Jonas Anderson
    Commented May 21, 2022 at 3:42

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I think the answer is "yes" when $m >6$. By arguments along the lines of Frobenius, Schur and Blichfeldt, if we set $G = \langle M(m,n), U^{\prime} \rangle $ and assume that $G$ is finite, then the non-scalar elements of $G$ whose eigenvalues all lies on an arc of length less than $\frac{\pi}{3}$ on the unit circle generate an Abelian normal subgroup of $G$, say $A$. When $m >6$, this Abelian group has rank $n$, and may be assumed to consist of diagonal matrices. In that case, $C_{G}(A)$ also consists of diagonal matrices (using that $A$ has rank $n$). By Clifford's theorem, we may conclude that the given representation of $A$ is now induced from a $1$-dimensional representation of a subgroup of $G$ of index $n$. In other words, $G$ now consists of monomial matrices.

I suppose, to be more precise, this argument shows that if $m > 6$, any finite unitary overgroup of $M(m,n)$ is conjugate (via a unitary matrix) to a finite group of monomial matrices ( what you call "generalized permutation matrices" are what I am calling monomial matrices).

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  • $\begingroup$ Can you give me some more details and perhaps a link? This is exactly the type of answer I was looking for, I just don't fully understand it yet. $\endgroup$
    – Jonas Anderson
    Commented May 21, 2022 at 3:45
  • $\begingroup$ The 1962 book of Curtis and Reiner has a section "On theorems of Frobenius, Schur and Burnside" (or something like that), and al the Character Theory book by Isaacs. The upshot is that in a finite subgroup of $U(n,\mathbb{C})$, any two elements which have all their eigenvalues close enough together on the unit circle commute. $\endgroup$
    – Geoff Robinson
    Commented May 21, 2022 at 10:18
  • $\begingroup$ When $m >6$, $M(m,n)$ has a natural diagonal subgroup $T$ of order $m^{n}$ , generated by $n$ elements of order $m$, each with $m-1$ eigenvalues $1$. Hence the group $A$ in the answer contains $T$, and necessarily needs $n$ generators. Then the continuation as in the answer should be clear. $\endgroup$
    – Geoff Robinson
    Commented May 21, 2022 at 10:22
  • $\begingroup$ Thanks, that makes sense. For clairification this argument works for all $n$, correct? $\endgroup$
    – Jonas Anderson
    Commented May 21, 2022 at 12:50
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    $\begingroup$ Yes, it works for all n. $\endgroup$
    – Geoff Robinson
    Commented May 21, 2022 at 15:12

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