Question

(a) Prove that $\mathbf{G}\mathbf{=}\left\{\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)|a,b,d\in ℝ\mathrm{and}\mathrm{ad}\ne 0\right\}$ is a group under matrix multiplication and that$\mathbf{N}\mathbf{=}\left\{\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right)|b\in ℝ\right\}$is a subgroup of $\mathbf{N}$.

Short answer

It is proved that$\mathrm{G}$ is a group under the matrix multiplication and$\mathrm{N}$ is a subgroup of$\mathrm{G}$ .

Step-by-step solution

Prove that G is group under matrix multiplication

Let $\mathrm{A},\mathrm{B}\in \mathrm{G}$be two matrices, consider $\mathrm{A}=\left(\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ 0& \mathrm{d}\end{array}\right)$, and $\mathrm{B}=\left(\begin{array}{cc}\mathrm{e}& \mathrm{f}\\ 0& \mathrm{h}\end{array}\right)$, where $\mathrm{ad}\ne 0$and $\mathrm{eh}\ne 0$.

This implies that:

$\begin{array}{rcl}\mathrm{AB}& =& \left(\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ 0& \mathrm{d}\end{array}\right)\left(\begin{array}{cc}\mathrm{e}& \mathrm{f}\\ 0& \mathrm{h}\end{array}\right)\\ & =& \left(\begin{array}{cc}\mathrm{ae}& \mathrm{af}+\mathrm{bh}\\ 0& \mathrm{dh}\end{array}\right)\end{array}$

It is observed that $\mathrm{aedh}=\left(\mathrm{ad}\right)\left(\mathrm{eh}\right)\ne 0$. This implies that $\mathrm{ab}\in \mathrm{G}$.

Assume that $\mathrm{I}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$, it is the identity element of $\mathrm{G}$. Consider$\mathrm{B}=\left(\begin{array}{cc}{\mathrm{a}}^{-1}& -\mathrm{b}\\ 0& {\mathrm{d}}^{-1}\end{array}\right)$ . This implies that $\mathrm{AB}=\mathrm{I}=\mathrm{BA}$.

Thus, every element of$\mathrm{G}$ has unique inverse. Therefore,$\mathrm{G}$ is a group under the matrix multiplication.

Prove that N is normal subgroup

To prove that N is a subgroup, we need to show that $\mathrm{A},\mathrm{B}\in \mathrm{N},{\mathrm{AB}}^{-1}\in \mathrm{N}$ .

Assume that $\mathrm{A},\mathrm{B}\in \mathrm{N}$as:

$\mathrm{A}=\left(\begin{array}{cc}1& \mathrm{a}\\ 0& 1\end{array}\right)$, and$\mathrm{B}=\left(\begin{array}{cc}1& \mathrm{b}\\ 0& 1\end{array}\right)$ .

Therefore,

$\begin{array}{rcl}{\mathrm{AB}}^{-1}& =& \left(\begin{array}{cc}1& \mathrm{a}\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& \mathrm{b}\\ 0& 1\end{array}\right)\\ & =& \left(\begin{array}{cc}1& \mathrm{a}-\mathrm{b}\\ 0& 1\end{array}\right)\\ & \in & \mathrm{N}\end{array}$

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