Finite solvable subgroup in $GL_2(\mathbb{C})$ which is not contained in Borel subgroup

Question

What is the example of finite solvable subgroup in $GL_2(\mathbb{C})$ which is not contained in any Borel subgroup?

Answer 1

You have the standard embedding of $Q_8\to GL_2(\mathbb{C})$ given by the action of the standard basis on $\mathbb{H}=\mathbb{C}^2$. More expliticly, these are given by the Pauli matrices. These cannot be put into an upper triangular form. This is exactly what you want, since all Borel groups are conjugate, and the cannonical one is the upper triangular matrices, so we are asking if we can put all the Quaterions in diagonal form. I recall the presentation $\langle l, n_i; i\le 3|n_i^2=n_1n_2n_3=l, l^2=1\rangle$. The condition $l^2=1$, and $l\neq 1$, implies that $l=-I$ or $l_{11}=-l_{22}$, and $l_{ii}^2=1$. Regardless, this means that either $l_{11}=-1$ or $l_{22}=-1$, denoted $h$. The relations $n_i^2=l$ them implies $n_{i, h,h}^2=-1$. But $n_{1, h,h}n_{2, h,h}n_{3, h,h}=(n_1n_2n_3)_{h,h}=l_{h,h}=-1$. This is impossible, since $n_{i,h,h}=\pm i$. There is probably a better way to do this. Oh, and this is solvable by Burnside lemma or $0\to \mathbb{Z}/(2)\to Q_{8}\to \mathbb{Z}/(4)\to 0$.