I tried some numerical examples but have no idea how to prove it in general.
Could some one give me an idea how to prove it?
2026-03-25 17:42:02.1774460522
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Prove that the group of diagonal matrices is not a normal subgroup of the group $\mathrm{GL}_n\,(\mathbb K)$ if $n\ge 2$ and $|K|\ge 3$
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Because$$\begin{pmatrix}1&1&0&0&\ldots&0\\1&-1&0&0&\ldots&0\\0&0&1&0&\ldots&0\\0&0&0&1&\ldots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\ldots&1\end{pmatrix}.\begin{pmatrix}-1&0&0&0&\ldots&0\\0&1&0&0&\ldots&0\\0&0&1&0&\ldots&0\\0&0&0&1&\ldots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\ldots&1\end{pmatrix}.\begin{pmatrix}1&1&0&0&\ldots&0\\1&-1&0&0&\ldots&0\\0&0&1&0&\ldots&0\\0&0&0&1&\ldots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\ldots&1\end{pmatrix}^{-1}$$is equal to$$\begin{pmatrix}0&1&0&0&\ldots&0\\1&0&0&0&\ldots&0\\0&0&1&0&\ldots&0\\0&0&0&1&\ldots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\ldots&1\end{pmatrix},$$which is not diagonal. Note that if $\operatorname{char}K=2$ then this would fail because then the matrix$$\begin{pmatrix}1&1&0&0&\ldots&0\\1&-1&0&0&\ldots&0\\0&0&1&0&\ldots&0\\0&0&0&1&\ldots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\ldots&1\end{pmatrix}$$wouldn't be invertible.
Because $|K|>2$ there exists two distinct non-zero elements in $K$. Say, $\alpha$ and $\beta$. Conjugate the diagonal matrix $D=diag(\alpha,\beta)$ with $$P=\left(\begin{array}{cc}1&1\\ 0&1\end{array}\right).$$ You get that $$ PDP^{-1}=\left(\begin{array}{cc}\alpha&-\alpha+\beta\\0&\beta\end{array}\right). $$ This is not diagonal because $\alpha\neq \beta$.
For examples with larger matrices just add ones along the diagonal and zeros elsewhere.