Prove that $G/N$ is Abelian Below is a homework problem I have gotten stuck on. I would really appreciate a hint, but please do not just give the answer away.

Let $G \subset \mathcal{M}_2(\mathbb{R})$ such that each $m \in G$ is
  upper triangular with a nonzero determinant.  Let $N = \left\{
 \bigl(\begin{smallmatrix} 1&b\\ 0&1 \end{smallmatrix} \bigr) : b \in
 \mathbb{R}\right\}$. Prove that:
• $N$ is a normal subgroup of $G$. 
• $G/N$ is abelian.

Solution: Let $m \in G$; then $\exists \alpha, \beta, \delta \in \mathbb{R} : m = \bigl(\begin{smallmatrix}
\alpha & \beta \\ 0&\delta
\end{smallmatrix} \bigr)$; $\alpha\delta \neq 0$. $m$ is invertible, hence $m^{-1} =  \frac{1}{\alpha \delta}\bigl(\begin{smallmatrix}
\delta & -\beta \\ 0&\alpha
\end{smallmatrix} \bigr)$. Let's look at $m N m^{-1}$. 
\begin{align*}
m N m^{-1} &= \left\{ \left(\begin{matrix}
\alpha & \beta \\ 0&\delta
\end{matrix} \right) \left(\begin{matrix}
1&b\\ 0&1
\end{matrix} \right) \frac{1}{\alpha \delta}\left(\begin{matrix}
\delta & -\beta \\ 0&\alpha
\end{matrix} \right) : \alpha, \beta, \delta, b \in \mathbb{R} \right\} \\
&= \left\{\left(\begin{matrix}
\alpha& \alpha b + \beta \\ 0&\delta
\end{matrix} \right) \frac{1}{\alpha \delta}\left(\begin{matrix}
\delta & -\beta \\ 0&\alpha
\end{matrix} \right) : \alpha, \beta, \delta, b \in \mathbb{R} \right\} \\
&= \left\{ \frac{1}{\alpha\delta}\left( \begin{matrix}
\alpha\delta& -\beta\alpha + \alpha^2 b + \beta\alpha \\ 0 &\alpha\delta
\end{matrix} \right) : \alpha, \beta, \delta, b \in \mathbb{R} \right\} \\
&= \left\{\left( \begin{matrix}
1& \frac{\alpha b}{\delta} \\ 0 & 1
\end{matrix} \right) : \alpha, \beta, \delta, b \in \mathbb{R} \right\} \\
\end{align*}
Since $\alpha \neq 0 $ and $ \delta \neq 0$, it follows that $\frac{\alpha b}{\delta}$ is a valid real number. Therefore $mNm^{-1} \in G$, hence $N$ is normal.
Now consider $G/N$: the group of right cosets of $N$ in $G$. Let $Nm_1$ and $Nm_2$ be two right cosets of $N$ in $G$. Then 
\begin{align*}
Nm_1Nm_2 &= Nm_1m_2\\
&= N  \left(\begin{matrix}
a & b  \\ 0& d
\end{matrix} \right)  \left(\begin{matrix}
\alpha & \beta \\ 0&\delta
\end{matrix} \right)\\
&= N  \left(\begin{matrix}
a\alpha & a\beta+ b \delta  \\ 0& d\delta
\end{matrix} \right) \\
....
\end{align*}

This is where I get stuck. I figure there must be a matrix in $n \in N$ such that, for any two upper triangular matrices $m_1$ and $m_2, m_1 n m_2 = m_2 n m_1$, but I can't seem to find it. 
 A: Consider the map $$G \to \mathbf R^\times \times \mathbf R^\times$$
given by  $$\bigl(\begin{smallmatrix} a&b\\ 0&c \end{smallmatrix} \bigr) \mapsto (a, ac).$$
It's easy to check that it is a homomorphism, and its kernel is precisely $N$. Therefore $N$ is normal in $G$, and since $G/N$ injects into the abelian group $\mathbf R^\times \times \mathbf R^\times$, it is abelian.
A: I never know whether to call cosets left or right, but $G/N$ stands for cosets of the form $gN$ with $g \in G$.  I guess people call these left cosets, even though you are multiplying by $N$ on the right.  Anyway, it does not materially affect your calculation, but I thought I would mention it.
To show that $G/N$ is abelian, you need
$$
(m_1 N) (m_2 N) = (m_2 N) (m_1 N),
$$
which is equivalent to
$$
m_1 m_2 N = m_2 m_1 N,
$$
which in turn is equivalent to $m_1^{-1} m_2^{-1} m_1 m_2 \in N$.  This expression is called the commutator of $m_1^{-1}$ and $m_2^{-1}$ (the inverses just appear because of how we wrote the initial calculation).  So you just need to consider $m_1^{-1} m_2^{-1} m_1 m_2$ and argue that it must have $1$'s on the diagonal.  
Further hint: 

 Instead of writing out the entire matrix multiplication, it is more transparent if you observe that the diagonal entries in the product of two upper triangular matrices are the products of the diagonal entries in those two upper triangular matrices.

