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?

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?
