Use the isomorphism theorem to determine the group $GL_2(\mathbb{R})/SL_2(\mathbb{R})$. Use the isomorphism theorem to determine the group $GL_2(\mathbb{R})/SL_2(\mathbb{R})$. Here $GL_2(\mathbb{R})$ is the group of $2\times 2$ matrices with determinant not equal to $0$, and $SL_2(\mathbb{R})$ is the group of $2\times 2$ matrices with determinant $1$. In the first part of the problem, I proved that $SL_2(\mathbb{R})$ is a normal subgroup of $GL_2(\mathbb{R})$. Now it wants me to use the isomorphism theorem. I tried using
$$|GL_2(\mathbb{R})/SL_2(\mathbb{R})|=|GL_2(\mathbb{R})|/|SL_2(\mathbb{R})|,$$
but since both groups have infinite order, I don't think I can use this here.

Define  $f:GL_2(\mathbb{R})\to \mathbb{R}^*$ such that

$$\underbrace{f(M)=\det(M)}_{\text{Onto}}$$

Let  $M_1, M_2 \in GL_2(\mathbb{R}^*) $ then
$f(M_1 M_2)=\det(M_1 M_2)=\det(M_1)\det(M_2)=f(M_1)(M_2)$
Hence $f$ is homomorphism for $\ker(f)=\{M: M\in GL_2(\mathbb{R})\text{ such that }\det(M)=1\}=SL_2(\mathbb{R})$
From first theorem in isomorphism
$GL_2(\mathbb{R})/SL_2(\mathbb{R})\approx \mathbb{R}^*$
