Question about the coset in $GL_2(\mathbb{R}) $

group-theorylinear algebrasolution-verification

Considering $G=GL_2(\mathbb{R})$ the group of invertible matrixes $2\times2$ with real entries and its following subgroup

$$H=\left\{\left(\begin{matrix} c & 0\\0&c\end{matrix}\right)|c\in\mathbb{R}_+^* \right\} .$$

In order to show that there are infinite cosets of $H$ in $G$, I take $A, B\in G$ such that they are in the same coset, so $AB^{-1}\in H$.

If $A=\left(\begin{matrix} a & b\\c&d\end{matrix}\right) $ and $b=\left(\begin{matrix} r & s\\t&u\end{matrix}\right)$, then

$$AB^{-1} =\frac{1}{\det B} \left(\begin{matrix} au-bt & – as+br\\cu-dt&-cs+dt\end{matrix}\right)\in H $$

if, and only if, $au-bt=-cs+dt$ and $cu-dt=-as+br=0$.

Another approach using the Theo Benedit's observation, if $A\in GL_2(\mathbb{R})$, a coset has the form

$$AH=\{cA; c \in \mathbb{R}^*\}. $$

Let's consider $d:= \det(A)\neq0$. If $ B\in H$, such that $\det(B) = b$ and then $ \det(AB)= \det(A)\cdot\det(B)=db \neq 0$. Then every element in the coset has non-zero determinant.

Conversely,if $M\in GL_2(\mathbb{R})$, with $\det(M)= m\neq 0$. Let $N= A^{-1}M$. Thus $M= AN$ and

$$\det(N)=\det(A^{-1}M)=\det(A^{-1})\cdot\det(M)=\frac{m}{d}\neq 0,$$

which means that $N\in H$ and $M =AN \in AH$. Hence, I proved that

$$AH=\{M\in GL_2(\mathbb{R});\det(M)= m\neq0\} $$

and since there are infinite possible choices for $ m $, we have infinite cosets.

Best Answer

first of all, lets notice that a coset of $H$ is of the form $$D\cdot H=\left\{D\cdot \begin{pmatrix} c & 0\\ 0 & c \end{pmatrix}\middle|\;c\in \mathbb{R}_+\right\}= \left\{c\cdot D\cdot \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}\middle|\;c\in \mathbb{R}_+\right\}= \left\{c\cdot D \middle|\;c\in \mathbb{R}_+\right\} $$ for some matrix $D\in GL_2(\mathbb{R})$. Now, consider the following family of matrices: $\left\{\begin{pmatrix} 1 & n\\ 0 & 1 \end{pmatrix}\,\middle|\; n\in\mathbb{N}\right\}\subset GL_2(\mathbb{R})$.

Now, lets assume that there exists $m,n\in \mathbb{N}$ s.t the matrices $A=\begin{pmatrix} 1 & n\\ 0 & 1 \end{pmatrix},\; B=\begin{pmatrix} 1 & m\\ 0 & 1 \end{pmatrix} $ are in the same coset. than there exists matrix $D\in GL_2(\mathbb{R})$ s.t $A=a\cdot D,\; B=b\cdot D$, for $\mathbb{R}\ni a,b>0$. so we can conclude that $\frac{1}{a}\cdot A=\frac{1}{b}\cdot B$ which implies $a=b$. it follows that $A=B$, so $m=n$. So, each matrix from the family is in a different coset.

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