Let $G = GL(2, \mathbb R)$, and let $H$ be the subgroup
$H =\left\{ \begin{bmatrix}
1 & x\\
0 & 1
\end{bmatrix}\mid x\in\mathbb R \right\}$Describe the left and right cosets of $H$ in $G$.
Note: If $C = gH$ is a left coset, and you claim that $C = D$ where you describe $D$ as the set of matrices {\begin{bmatrix}
a & b\\
c & d
\end{bmatrix} }
satisfying specific conditions on $a, b, c, d$, then make sure to
show both $C ⊆ D$ and $D ⊆ C$.
Left coset is $g.H$ $=$ {$g.h| g ∈ G$ and $h ∈ H$}
Right coset $H.g$ $=$ {$H.g| h' ∈ H$ and $g ∈ G$}
but how to work on this problem starting from these definitions? any help please?
Best Answer
You want to classify both left and right cosets. Let's do this for the left ones, and leave the right ones as an exercise for you (the idea is the same).
We take some $g\in GL(2,\mathbb R).$ Then for all $ h\in H$: $$g=\begin{bmatrix} a & b\\ c & d\end{bmatrix} $$ $$gh=\begin{bmatrix} a & b\\ c & d\end{bmatrix} \begin{bmatrix} 1 & x\\ 0 & 1\end{bmatrix}=\begin{bmatrix} a & ax+b\\ c & cx+d\end{bmatrix} $$
Since $h$ is an arbitrary element of $H$, each member of $gH$ must be of this form. Moreover, each matrix of this form is in $gH$, since if $x \in \mathbb R$ and the matrix $m$ is of the form: $$m=\begin{bmatrix} a & ax+b\\ c & cx+d\end{bmatrix} $$ Then we can multiply $m$ by the matrix: $$h=\begin{bmatrix} 1 & x^{-1}\\ 0 & 1\end{bmatrix} $$ Which is clearly a member of $H$ (unless $x=0$, but this is the trivial case), and get: $$hm= \begin{bmatrix} 1 & x^{-1}\\ 0 & 1\end{bmatrix} \begin{bmatrix} a & ax+b\\ c & cx+d\end{bmatrix} =\begin{bmatrix} a & b\\ c & d\end{bmatrix}=g$$ And so: $$hm=g\Rightarrow m=gh^{-1}\Rightarrow m\in gH$$ (Since $h^{-1}\in H$)
We have thus classified $gH$ for all $g\in G$. As I said, the exercises for the right cosets is basically the same.