Question

(b) Use Theorem 8.11 to show that$\mathbf{N}$ is normal in $\mathbf{G}$.

Short answer

It is proved that$\mathrm{N}$ is normal in $\mathrm{G}$.

Step-by-step solution

Assume

To prove that $\mathrm{N}$is a normal subgroup, we need to show that $\mathrm{A}\in \mathrm{G},{\mathrm{ANA}}^{-1}\subset \mathrm{N}$. Assume$\mathrm{A}=\left(\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ 0& \mathrm{d}\end{array}\right)$ with$\mathrm{ad}\ne 0$ , and$\mathrm{B}\in \mathrm{N}$ is in the form $\mathrm{B}=\left(\begin{array}{cc}1& \mathrm{c}\\ 0& 1\end{array}\right)$.

Prove that N is normal subgroup

Determine the value of${\mathrm{ABA}}^{-1}$ as:

$\begin{array}{rcl}{\mathrm{ABA}}^{-1}& =& \left(\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ 0& \mathrm{d}\end{array}\right)\left(\begin{array}{cc}1& \mathrm{c}\\ 0& 1\end{array}\right)\left(\begin{array}{cc}{\mathrm{a}}^{-1}& -\mathrm{b}\\ 0& {\mathrm{d}}^{-1}\end{array}\right)\\ & =& \left(\begin{array}{cc}\mathrm{a}& \mathrm{ac}+\mathrm{b}\\ 0& \mathrm{d}\end{array}\right)\left(\begin{array}{cc}{\mathrm{a}}^{-1}& -\mathrm{b}\\ 0& {\mathrm{d}}^{-1}\end{array}\right)\\ & =& \left(\begin{array}{cc}1& -\mathrm{ab}+{\mathrm{acd}}^{-1}+{\mathrm{bd}}^{-1}\\ 0& 1\end{array}\right)\end{array}$

This implies that ${\mathrm{ABA}}^{-1}\in \mathrm{N}$

Hence, it is proved that$\mathrm{N}$is a normal subgroup.