Let $GL(2,\mathbb{R})$ denote the set of all non-singular $2×2$ matrices with real entries.Show that $SL(2,\mathbb{R})$ is a normal subgroup of

Question

Let $GL(2,\mathbb {R})$ denote the set of all non-singular $2×2$ matrices with real entries.Show that $$SL(2,\mathbb{R})=\{\begin{bmatrix} a & b \\ c & d \end{bmatrix}\in GL(2,\mathbb {R}),ad-bc=1\}$$ is a normal subgroup of $GL(2,\mathbb {R})$ .

In this problem, it is easy to see that $SL(2,\mathbb{R})$ is a subgroup of $GL(2,\mathbb {R})$ . Now, in order to show that $SL(2,\mathbb{R})$ as a normal subgroup we must show that $\forall A\in GL(2,\mathbb{R})$, $ASA^{-1}\in SL(2,\mathbb{R})$, $\forall S\in SL(2,\mathbb{R})$. Now, if we try to do the calculation of $ASA^{-1}$ , by considering $S=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $A=\begin{bmatrix} p & q \\ r & s \end{bmatrix}$ , it becomes an absurdly long calculation. Is there any other way to do this ....I am not quite getting this...There may be a lot of posts concerning the same topic but I can't seem to find it either...

Answer 1

Since we just need $MSM^{-1} \in SL(2,\mathbb{R})$ for $S \in SL(2,\mathbb{R})$. we can use the determinant property: $det(MSM^{-1}) = det(S) = 1$. Hence $MSM^{-1} \in SL(2,\mathbb{R})$.

We are just using the property $det(AB) = det(A) det(B)$.