Left Cosets and Right Cosets. Recall that $GL_2(\mathbb{R})$ is the group of all invertible 2x2 matrices with real entries.
Let:
$G = (\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \in GL_2(\mathbb{R}) : ac \neq 0$)
and
H = ($\begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}: x\in\mathbb{R}$)
H is a subgroup of G. 
Show that every left coset of H in G is equal to the right coset of H in G.
First, I thought of assuming that G is abelian. But clearly that failed for me because G isn't abelian. So the other thing I thought of trying was to show $ghg^{-1}\in H$ where $g\in G$ and $h\in H$. Now with this idea, I'm stuck in showing that $ghg^{-1}\in H$. Can anyone help me out?
 A: Evaluate the multiplication $\displaystyle\begin{pmatrix}a & b \\ 0 & c\end{pmatrix}\begin{pmatrix} 1 & x \\ 0 & 1\end{pmatrix}$ explicitly. What happens as $x$ varies over $\Bbb R$?
And then what if the matrices are reversed - what do the products look like then as $x$ varies?
A: If $g = \begin{pmatrix} a & b \\
0 & c \end{pmatrix} \in G$, then
$$
g^{-1} = \begin{pmatrix}
1/a & -b/ca \\
0 & 1/c
\end{pmatrix}
$$
so if $h = \begin{pmatrix} 1 & x \\
0 & 1 \end{pmatrix} \in H$, then
$$
g^{-1}hg = \begin{pmatrix}
1 & -cx/a \\
0 & 1
\end{pmatrix} \in H
$$
Thus, if $g\in G, h \in H$, there is an $h' \in H$ such that
$$
hg = gh'
$$
Thus,
$$
Hg \subset gH
$$
and the other containment holds similarly, so $gH = Hg$.
A: The map $\varphi\colon G\to\mathbb{R}^*\times\mathbb{R}^*$ (where $\mathbb{R}^*$ denotes the multiplicative group of nonzero real numbers) defined by
$$
\varphi\colon\begin{pmatrix}a & b \\ 0 & c\end{pmatrix}
\mapsto (a,c)
$$
is a homomorphism because
$$
\begin{pmatrix}a_1 & b_1 \\ 0 & c_1\end{pmatrix}
\begin{pmatrix}a_2 & b_2 \\ 0 & c_2\end{pmatrix}
=
\begin{pmatrix}a_1a_2 & a_1b_2+b_1c_2 \\ 0 & c_1c_2\end{pmatrix}
$$
Since $H=\ker\varphi$, $H$ is normal in $G$.
