two distinct elements of $GL_2(\mathbb{R})/SL_2(\mathbb{R})$. I am trying to understand factor groups better (If $N$ is a normal subgroup of a group $G$ then the cosets of $N$ in $G$ form a group $G/N$ under the operation $(aN)(bN)=abN$). But, I get a little confused when applying this to matrices groups, so could someone proved an example of two distinct elements of $GL_2(\mathbb{R})/SL_2(\mathbb{R})$ and how you would use coset multiplication to multiply these two cosets.
 A: The determinant gives the homomorphism $det: GL_2(R)\rightarrow R^*$ with kernel $SL_2(R)$.   So the elements $\begin{pmatrix} r& 0\\ 0 & 1\end{pmatrix}$ are distinct in the quotient group.
A: One way of doing this is to just use some representatives. For example, let $A, B \in \mathrm{GL}_2$ be the matrices
$$ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$$
I'll write $[A]$ for the image of $A$ in the quotient group $\mathrm{GL}_2 / \mathrm{SL}_2$. The class $[A]$ can be written as a coset: $[A] = A \cdot \mathrm{SL}_2$. In the case of $A$, we have $\det A = 1$, and so $A \in \mathrm{SL_2}$, so $[A] = A \cdot \mathrm{SL}_2 = \mathrm{SL}_2 = [I_2]$, where $I_2$ is the $2 \times 2$ identity matrix.
How about $[B]$? We cannot just pull $B$ inside of $\mathrm{SL}_2$, since $B$ is not an element of $\mathrm{SL}_2$. However, since $\det B = -2$, we could write
$$ B = -2 \begin{pmatrix} \frac{-1}{2} & -1 \\ \frac{-3}{2} & -2 \end{pmatrix} = -2C$$
for that matrix $C = \frac{B}{-2}$. Then,
$$[B] = B \cdot \mathrm{SL}_2 = -2 C \cdot \mathrm{SL}_2 = -2 I_2 \mathrm{SL}_2 = [-2 I_2]$$
It's not hard to see that actually, for any matrix $X \in \mathrm{GL}_2$ we have $[X] = [(\det X) I_2]$, and so now we have an easy way to multiply classes: $[X][Y] = XY \cdot \mathrm{SL}_2 = [(\det XY) I_2]$.
