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This question already has an answer here:

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

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marked as duplicate by Dietrich Burde, Shailesh, Luiz Cordeiro, R_D, user91500 Sep 20 '16 at 5:29

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  • $\begingroup$ 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). $\endgroup$ – user321268 Sep 19 '16 at 19:36
  • $\begingroup$ Do you know how to multiply 2x2 matrices ? $\endgroup$ – user171326 Sep 19 '16 at 21:54
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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.

All points have been shown on MSE already, e.g., see here.

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The subgroup test say that:

a nonempty subset of a group is itself a group if the inverse of any element in the subset multiplied with any other element in the subset is also in the subset.

So you have to prove that $A^{-1}X \in T$ for all $A$ and $X$ in $T$.

It is simpler to show that the inverse of an upper triangular matrix (if it exists) is upper triangular and the product of two upper triangular matrices is upper triangular. And, obviously, the identity matrix is upper triangular.

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