Exercise 1.4.10 (Group $\left\{ \begin{pmatrix} a & b\\ 0 & c \end{pmatrix} \mid a,b,c \in \mathbb{R},\ a\ne 0, \ c \ne 0 \right\}$)

Proof. Let $G=\left\{\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\mid a,b,c\in ℝ,a\ne 0,c\ne 0\right\}$.

(a)

Let $M=\left(\begin{array}{cc}\hfill a_1\hfill & \hfill b_1\hfill \\ \hfill 0\hfill & \hfill c_1\hfill \end{array}\right)$ and $N=\left(\begin{array}{cc}\hfill a_2\hfill & \hfill b_2\hfill \\ \hfill 0\hfill & \hfill c_2\hfill \end{array}\right)$ be elements of $G$, so that $a_1\ne 0,c_1\ne 0,a_2\ne 0,c_2\ne 0$. Then $$\begin{array}{llll}\hfill \mathit{MN}& =\left(\begin{array}{cc}\hfill a_1\hfill & \hfill b_1\hfill \\ \hfill 0\hfill & \hfill c_1\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a_2\hfill & \hfill b_2\hfill \\ \hfill 0\hfill & \hfill c_2\hfill \end{array}\right)& \hfill & \\ \hfill & =\left(\begin{array}{cc}\hfill a_1{a}_2\hfill & \hfill a_1{b}_2+b_1{c}_2\hfill \\ \hfill 0\hfill & \hfill c_1{c}_2\hfill \end{array}\right),& \hfill & \end{array}$$

where $a_1{a}_2\ne 0,c_1{c}_2\ne 0$. Therefore $\mathit{MN}\in G$, so $G$ is closed under matrix multiplication.

(b)

Let $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\in G$, so that $a\ne 0,c\ne 0$. Then $\det <!--\; FUNCTION\; APPLICATION\; -->(A)=\mathit{ac}\ne 0$, so $A$ is invertible, and $$A^{-1}=\frac{1}{\mathit{ac}}\left(\begin{array}{cc}\hfill c\hfill & \hfill -b\hfill \\ \hfill 0\hfill & \hfill a\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill a^{-1}\hfill & \hfill -b{a}^{-1}{c}^{-1}\hfill \\ \hfill 0\hfill & \hfill c^{-1}\hfill \end{array}\right),$$

where $a^{-1}\ne 0,c^{-1}\ne 0$. Therefore $A^{-1}\in G$, so $G$ is closed under inverses.

(c)

Let $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\in G$. Then $\det <!--\; FUNCTION\; APPLICATION\; -->(A)=\mathit{ac}\ne 0$, so $A\in {\mathrm{GL}}_2(ℝ)$. This shows that $G\subset {\mathrm{GL}}_2(ℝ)$. Moreover $I_2\in G$, so $G\ne \varnothing$. Since $G$ is closed under matrix multiplication and under inverses, $G$ is a subgroup of ${\mathrm{GL}}_2(ℝ)$.

(d)

Let $H=\left\{\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill a\hfill \end{array}\right)\mid a,b\in ℝ,a\ne 0,\right\}$.

Then $H\subset G$, and $I_2\in H$, so $H\ne \varnothing$. Moreover if $M=\left(\begin{array}{cc}\hfill a_1\hfill & \hfill b_1\hfill \\ \hfill 0\hfill & \hfill a_1\hfill \end{array}\right),N=\left(\begin{array}{cc}\hfill a_2\hfill & \hfill b_2\hfill \\ \hfill 0\hfill & \hfill a_2\hfill \end{array}\right)\in H$, then $a_1\ne 0$, $a_2\ne 0$, and

$$\mathit{MN}=\left(\begin{array}{cc}\hfill a_1{a}_2\hfill & \hfill a_1{b}_2+b_1{a}_2\hfill \\ \hfill 0\hfill & \hfill a_1{a}_2\hfill \end{array}\right),$$

where $a_1{a}_2\ne 0$, so $\mathit{MN}\in H$. Moreover, if $A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill a\hfill \end{array}\right)\in H$, then $a\ne 0$, so $A$ is invertible, and

$$A^{-1}=\left(\begin{array}{cc}\hfill a^{-1}\hfill & \hfill -b{a}^{-2}\hfill \\ \hfill 0\hfill & \hfill a^{-1}\hfill \end{array}\right),$$

where $a^{-1}\ne 0$, so $A^{-1}\in H$. This shows that $H$ is a subgroup of $G$, a fortiori a subgroup of ${\mathrm{GL}}_2(ℝ)$.