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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 or monomial matrices over the field $\mathbb{C}$.)

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 $M(m,n)$.

Recently, in Are generalized symmetric groups maximal finite groups (in a certain sense)? Geoff Robinson showed for any fixed $n$ and $m\ge 7$ the group $M(m,n)$ and any additional unitary (in $U(n)$) which is not a generalized permutation generate an infinite group.

This post is to tie up some loose (at least to me) ends. I want to know if the following groups are maximally finite (in this sense) at a smaller $m$.

(1) Let $U_4 \in U(4)$ be a unitary matrix which is not in $M(m,4)$ for any $m$ (in other words, $U_4$ is not a generalized permutation matrix). Is the group generated by $M(4,4)$ and $U_4$ always infinite?

(2) Let $U_8 \in U(8)$ be a unitary matrix which is not in $M(m,8)$ for any $m$. Is the group generated by $M(2,8)$ (a signed permutation group) and $U_8$ always infinite?

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  • $\begingroup$ So I think (1) is true from studying the finite subgroups of $SU(4)$ in link. I looked for all groups with subgroups isomorphic to $M(4,4)$. I only found the 'dihedral' groups (Group XXXII) which do not have any elements outside of $M(m,4)$. Can anyone confirm this observation? $\endgroup$ Commented May 24, 2022 at 12:48
  • $\begingroup$ Can you explain more about your choice of $ m$ for $ M(m,n) $ here? Why in question (1) do you choose $ m=4 $ and in question (2) you ask about $ m=2 $? For example do you expect the answer to be any different for $ M(2,8) $ and $ M(4,8) $? $\endgroup$ Commented Jul 10, 2022 at 18:37
  • $\begingroup$ There's no deep reason. It's just the lowest $m=2^c$ ($c$ is a positive integer) for the respective $n$ which I do not know if it is maximal or not. I know that $M(2,4)$ is a subgroup of the 2-qubit Clifford group and cannot be maximal since I can always add $H$ (which is not monomial). Showing that $M(2,8)$ is maximal (in the sense of my definition above) would imply that$M(4,8)$ is maximal since $M(2,8)$ is a subgroup of $M(4,8)$, but the implication only goes one way. Hope that made sense. $\endgroup$ Commented Jul 10, 2022 at 22:09

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