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Definitions

Let $G$ be a finite subgroup of $U(n)$ and let $\mathcal{D} \subset U(n)$ denote the group of $n\times n$ diagonal, unitary matrices. We'll say that a matrix $T$ is diagonalizable over $G$ if there exists $g_1, g_2 \in G$ such that $g_1 T g_2 \in \mathcal{D}$. Otherwise, $T$ is not diagonalizable over $G$.

Let $M=PD$ be an $n\times n$ unitary, monomial matrix expressed (as is typical) as a product of a permutation matrix, $P$, and a unitary diagonal matrix $D$.

Question

If $P$ is not diagonalizable over $G$, can we prove that $M=PD$ is also not diagonalizable over $G$?

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    $\begingroup$ That's not a question, it's a command. Does that mean you already know it's true? $\endgroup$ Commented Sep 16 at 12:13
  • $\begingroup$ Sorry I'll rephrase it. $\endgroup$ Commented Sep 16 at 14:02
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    $\begingroup$ @JonasAnderson When $D$ is in the normalizer of $G$ in $U(n)$ then for any $d\in D$ we have $$ g_1 pd g_2 = g_1p (dg_2d^{-1})d = g_1 p \widetilde{g_2} d. $$ So if $p$ is not diagonalizable by $G$, then $pd$ is also not diagonalizable by $G$. $\endgroup$ Commented Sep 21 at 22:56
  • $\begingroup$ Good catch. Fixed it $\endgroup$ Commented Sep 22 at 11:39
  • $\begingroup$ @vittucek Thanks. I agree. But, I'm also interested in cases cases where $D$ is not in the normalizer. $\endgroup$ Commented Sep 22 at 11:58

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