The set of upper triangular matrices is a subgroup of GL2(R)

Proposition

Let

$$G=\left\{ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \;\middle|\; a,b,c\in\mathbb{R},\ ac\neq 0 \right\}.$$

Then $G$ is a subgroup of $GL_2(\mathbb{R})$.

Proof

Since

$$\det\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} = ac \neq 0$$

for every element of $G$, each such matrix is invertible; hence $G \subseteq GL_2(\mathbb{R})$. We verify the three subgroup axioms.

(1) Closure under multiplication.

Let

$$A=\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}, \qquad B=\begin{pmatrix} d & e \\ 0 & f \end{pmatrix} \in G.$$

Then

$$AB=\begin{pmatrix} ad & ae+bf \\ 0 & cf \end{pmatrix}.$$

Since $a,c,d,f\neq 0$, we have $ad\neq 0$ and $cf\neq 0$; hence $AB\in G$.

(2) Identity element.

The identity matrix

$$I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

satisfies $1\cdot 1\neq 0$, so $I\in G$.

(3) Closure under inverses.

For $A=\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}\in G$,

$$A^{-1}= \begin{pmatrix} \dfrac{1}{a} & -\dfrac{b}{ac} \\[2mm] 0 & \dfrac{1}{c} \end{pmatrix}.$$

Since $\dfrac1a\neq 0$ and $\dfrac1c\neq 0$, it follows that $A^{-1}\in G$.

Therefore $G$ is a subgroup of $GL_2(\mathbb{R})$. $\blacksquare$