# Let $G = GL_2\mathbb (R)$. Show that $T$ is a subgroup of $G$ [duplicate]

Let $G = GL_2 (\mathbb {R}$).

Show that $T$ = {$\begin{bmatrix} a & b \\ 0 & d \\ \end{bmatrix}$ | ad $\neq 0$ } is a subgroup of $G$

My attempt:

det$(TT^{-1})$ = det$(T)$ det($T^{-1})$ = det($T$) $1$/det($T$) = $ad$ $(1/ad)$= $1$. We are done by subgroup test

## marked as duplicate by Dietrich Burde, Shailesh, Luiz Cordeiro, R_D, user91500Sep 20 '16 at 5:29

• your attempy is incorrect. the determinant of the product $TT^{-1}$ is always 1, regardless the matrices (of course I'm refereing to invertible matrices). – user321268 Sep 19 '16 at 19:36
• Do you know how to multiply 2x2 matrices ? – user171326 Sep 19 '16 at 21:54

You need to show that with $T$ and $S$ also $TS$ and $S^{-1}$ is again in $G$. So far you only showed that $\det(I_n)=1$, which is clear anyway. In other words, why is $TS$ and $S^{-1}$ also upper-triangular, and nonsingular.
So you have to prove that $A^{-1}X \in T$ for all $A$ and $X$ in $T$.