Group and left classes I need help with the following abstract algebra problem. It is not homework, but I need a solution.

Let $$SL_2(\mathbb{R}) = \left\{ \big( \begin{smallmatrix} a & b\\ c & d\\\end{smallmatrix} \big)\;\middle\vert\;a,b,c,d \in \mathbb{R}, ad - bc = 1 \right\},$$
$$SO_2(\mathbb{R}) = \left\{ \big( \begin{smallmatrix} a & b\\ -b & a\\\end{smallmatrix} \big)\;\middle\vert\;a,b,c,d \in \mathbb{R}, a^2 + b^2 = 1 \right\}.$$
Prove that $ SL_2(\mathbb{R})$ is a group, that $SO_2(\mathbb{R}) \leq  SL_2(\mathbb{R})$, and that $SL_2(\mathbb{R})$ can be represented as a union of nonintersecting left classes with respect to $SO_2(\mathbb{R})$ in the form:
$$SL_2(\mathbb{R}) = \bigcup_{\substack{r,p \in \mathbb{R}\\ r>0}} \begin{pmatrix}
r & 0  \\
r^{-1}p & r^{-1}  \\
  \end{pmatrix} \cdot SO_2(\mathbb{R})$$

 A: Hints
To check that $SL_2$ is a group (under matrix multiplication), you need to do the following:


*

*Show that it is closed under multiplication. Hint $\mathrm{det}(AB)=\mathrm{det}(A)\mathrm{det}(B)$.

*Find the unit element. (That would be easy, I suppose)

*Show that each element has a multiplicative inverse. Hint $\mathrm{det}(A^{-1})=\mathrm{det}(A)^{-1}$.

*Show that the multiplication is associative.


To show that $SO_2$ is a subgroup, you need to do the following:


*

*Show that it is non-empty.

*Show that for every pair $A,B\in SO_2$, the matrix $AB^{-1}$ is in $SO_2$. Hint: use what you know about the determinant.


Hopefully this outline helps to get you started. For the last problem, suppose that 
$$\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\in SL_2.$$
Multiplying this matrix with
$$\frac{1}{\sqrt{a^2+b^2}}\left(\begin{array}{cc} a & -b \\ b & a\end{array}\right)$$
(is it possible that this matrix is zero?) on the right side should give you something nice (pun intended) :)
A: You only have two things to check:


*

*If $M\in SL_2(\mathbb{R})$, then there exists real numbers $r,p$, with $r>0$, and a matrix $A\in SO_2(\mathbb{R})$ such that 
$$M=\begin{pmatrix}
r & 0  \\
r^{-1}p & r^{-1}  \\
  \end{pmatrix}A.$$
(This will show that the union convers everything.)

*If 
$$\begin{pmatrix}
r_1 & 0  \\
r_1^{-1}p_1 & r_1^{-1}  \\
  \end{pmatrix} \cdot SO_2(\mathbb{R})=\begin{pmatrix}
r_2 & 0  \\
r_2^{-1}p & r_2^{-1}  \\
  \end{pmatrix} \cdot SO_2(\mathbb{R}),$$
then $r_1=r_2$ and $p_1=p_2$. (This will show that the cosets are distinct, i.e. non-intersecting.)

A: $\def\SL\{\mathrm{SL}}\def\SO{\mathrm{SO}}$


*

*$\SL_2(\mathbb{R})$ is a group.
Note that multiplication of matrices is associative (it corresponds to composition of linear transformations); the identity matrix is the neutral element, and has determinant $1$, so it is in $\SL_2(\mathbb{R})$. The determinant of a product is the product of the determinants, so if $A,B\in\SL_2(\mathbb{R})$, then so is $AB$, since $\det(AB) = \det(A)\det(B)=1\times 1 = 1$. Finally, if $A$ has nonzero determinant, then it is nonsingular so it has a multiplicative inverse; and since $1=\det(I) = \det(AA^{-1})=\det(A)\det(A^{-1})$, it follows that $\det(A^{-1}) = \frac{1}{\det(A)}$; so if $A\in\SL_2(\mathbb{R})$, then it is invertible and  $\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{1}=1$, so $A^{-1}\in\SL_2(\mathbb{R})$ as well. Thus, $\SL_2(\mathbb{R})$ is a group.

*$\SO_2(\mathbb{R})$ is a subgroup of $\SL_2(\mathbb{R})$. 
From the definition, it is clear that $\SO_2(\mathbb{R})$ is a subset of $\SL_2(\mathbb{R})$, since the determinant of a matrix of the form
$\left(\begin{array}{rr}a&b\\-b&a\end{array}\right)$ is $a^2+b^2$, which by assumption will equal $1$. Also, the identity is an element of $\SO_2(\mathbb{R})$. 
To show it is a subgroup, you need to show that it is closed under products and inverses. For products, assume that $a^2+b^2=c^2+d^2=1$. Then:
$$\left(\begin{array}{rr}
a&b\\-b&a
\end{array}\right) \left(\begin{array}{rr}c&d\\-d&c
\end{array}\right) = \left(\begin{array}{cc}
ac-bd & ad+bc\\
-bc-ad & -bd+ac\end{array}\right),$$
which is of the desired $\left(\begin{array}{rr}x&y\\-y&x\end{array}\right)$, with $x=ac-bd$, $y=ad+bc$. And the determinant is $1$, again because the product of two matrices with determinant $1$ is of determinant $1$, so $(ac-bd)^2 + (ad+bc)^2 = 1$. Or, if you just expand, we get:
$$\begin{align*}
(ac-bd)^2 + (ad+bc)^2 &= a^2c^2 - 2abcd + b^2d^2 + a^2d^2+2abcd+b^2c^2 
\\
&= a^2c^2+b^2d^2+a^2d^2+b^2c^2 \\
&= (a^2+b^2)c^2 + (b^2+a^2)d^2 \\
&= c^2+d^2=1.\end{align*}$$
So if $A,B\in \SO_2(\mathbb{R})$, then $AB\in\SO_2(\mathbb{R})$. That is, $\SO_2(\mathbb{R})$ is a submonoid of $\SL_2(\mathbb{R})$.
For the inverse, note that the inverse of $\left(\begin{array}{rr}a&b\\-b&a\end{array}\right)$ with $a^2+b^2=1$ is $\left(\begin{array}{rr}a&(-b)\\-(-b)&a\end{array}\right)$, so if $A\in\SO_2(\mathbb{R})$, then $A^{-1}\in \SO_2(\mathbb{R})$. Thus, $\SO_2(\mathbb{R})$ is a subgroup of $\SL_2(\mathbb{R})$.

*If $r,p,s,q\in\mathbb{R}$, $r,s\gt 0$, and 
$$\left(\begin{array}{rr}
r & 0\\
r^{-1}p & r^{-1}\end{array}\right)^{-1}\left(\begin{array}{rr}
s & 0\\
s^{-1}q & s^{-1}\end{array}\right)\in SO_2(\mathbb{R})$$
then $r=s$ and $p=q$. Indeed, the product is equal to
$$\left(\begin{array}{rr}
r^{-1} & 0\\
-r^{-1}p & r\end{array}\right)\left(\begin{array}{rr}
s & 0\\
s^{-1}q & s^{-1}
\end{array}\right) = \left(\begin{array}{cc}
\frac{s}{r} & 0\\
-\frac{ps}{r} + \frac{qr}{s} & \frac{r}{s}\end{array}\right).$$
If this is in $\SO_2(\mathbb{R})$, then $\frac{r}{s}=\frac{s}{r}$, so $r^2=s^2$, hence (since both are positive) $r=s$; and then $-\frac{ps}{r}+\frac{qr}{s} = -p+q$ must equal $0$, so $p=q$.
Therefore, the lateral classes listed are pairwise distinct.

*If $\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\in\SL_2(\mathbb{R})$, then there exist $r\gt 0$, $p\in\mathbb{R}$, such that 
$$\left(\begin{array}{cc}
a&b\\c&d\end{array}\right)\in\left(\begin{array}{rr}r&0\\r^{-1}p&r^{-1}\end{array}\right)\SO_2(\mathbb{R}).$$
If $b=0$ and $a\gt 0$, then we are done: we can take $r=a$, $p=ac$; note that since $ad-bc=1$, we will necessarily have $d=\frac{1}{a}=r^{-1}$. If $b=0$ and $a\lt 0$, then 
$$\left(\begin{array}{cc}
a&0\\c&d\end{array}\right)=\left(\begin{array}{rr}-a&0\\-c&-d\end{array}\right)\left(\begin{array}{rr}-1&0\\0&-1\end{array}\right)\in \left(\begin{array}{rr}-a&0\\-c&-d\end{array}\right)\SO_2(\mathbb{R}),$$
which has the desired form with $r=-a$, $p=ac$. 
If $a=0$, then note that
$$\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\left(\begin{array}{rr}0 &1\\-1&0\end{array}\right) = \left(\begin{array}{rr}-b & a\\-d & c\end{array}\right)$$
so we are back to the previous case.
Finally, if $a\neq 0$ and $b\neq 0$, we want to find $x$ and $y$ with $x^2+y^2=1$ such that
$$\left(\begin{array}{rr}
a&b\\c&d\end{array}\right)\left(\begin{array}{rr}x&y\\-y&x\end{array}\right)$$
has a $0$ in the $(1,2)$ entry. The $(1,2)$ entry is $ay+bx$, so we need $y=-\frac{b}{a}x$. Then plugging that into $x^2+y^2 = 1$ yields that we want $x=\frac{a}{\sqrt{a^2+b^2}}$. Indeed,
$$\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\left(\begin{array}{cc}\frac{a}{\sqrt{a^2+b^2}} & \frac{-b}{\sqrt{a^2+b^2}}\\\frac{b}{\sqrt{a^2+b^2}}&\frac{a}{\sqrt{a^2+b^2}}\end{array}\right),$$
then the right hand factor is in $\SO_2(\mathbb{R})$, and the product has a $0$ on the upper right corner, which is all we need.
