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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:

$SL_2(\mathbb{R}) = \bigcup_{\substack{r,p \in \mathbb{R}\\ r>0}} \begin{pmatrix} r & 0 \\ r^{-1}p & r^{-1} \\ \end{pmatrix} \cdot SO_2(\mathbb{R})$

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    As with your question about the center of a group, I'm fascinated by the fact that this particular style of question in group theory "needs" to be solved but is not homework. Where are you getting this question from, and why do you "need" to solve it, rather than trying to understand what it says and just using it for later purposes (whatever they may be)?2012-06-07

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

  1. $\SL_2(\mathbb{R})$ 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.

  2. $\SO_2(\mathbb{R})$ 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 $\SL_2(\mathbb{R})$.

  3. 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.

  4. 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.

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Hints

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

  1. Show that it is closed under multiplication. Hint $\mathrm{det}(AB)=\mathrm{det}(A)\mathrm{det}(B)$.
  2. Find the unit element. (That would be easy, I suppose)
  3. Show that each element has a multiplicative inverse. Hint $\mathrm{det}(A^{-1})=\mathrm{det}(A)^{-1}$.
  4. Show that the multiplication is associative.

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

  1. Show that it is non-empty.
  2. 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) :)

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You only have two things to check:

  1. 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.)
  2. 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.)