Show that $T/U$ is abelian Let $T$ be the group of nonsingular upper triangular 2 x 2 matrices with entries in $\mathbb{R}$; that is, matrices of the form 
$$\begin{bmatrix}
    a & b \\
    0 & c 
\end{bmatrix}$$
where $a, b, c\in\mathbb{R}$ and $ac\neq 0$. Let $U$ consist of matrices of the form
$$\begin{bmatrix}
    1 & x \\
    0 & 1 
\end{bmatrix}$$
where $x\in\mathbb{R}$.
Prove that $T/U$ is abelian.
My attempt:
I've already proved that $U$ is normal in $T$.
Let $t_1=\begin{bmatrix}
    a & b \\
    0 & c 
\end{bmatrix}$ and $t_2=\begin{bmatrix}
    d & e \\
    0 & f 
\end{bmatrix}$
$U=\left\{\begin{bmatrix}
    1 & x \\
    0 & 1 
\end{bmatrix}: x\in\mathbb{R}\right\}$
$(t_1U)(t_2U)=t_1t_2U=\left\{\begin{bmatrix}
    ad & adx+ae+bf \\
    0 & cf 
\end{bmatrix}: x\in\mathbb{R}\right\}$
$(t_2U)(t_1U)=t_2t_1U=\left\{\begin{bmatrix}
    ad & adx+bd+ce \\
    0 & cf 
\end{bmatrix}: x\in\mathbb{R}\right\}$
How do I prove that $(t_1U)(t_2U)=(t_2U)(t_1U)$?
Thanks in advance.
 A: To prove that $T/U$ is abelian, you need to show that (why?) for any $t_1,t_2\in T$,
$$
t_1t_2t_1^{-1}t_2^{-1} \in U
$$
If $t_1$ and $t_2$ are as in your question, then
\begin{equation*}
\begin{split}
t_1t_2t_1^{-1}t_2^{-1} &= \begin{pmatrix} a &  b \\ 0 & c\end{pmatrix}\begin{pmatrix} d &  e \\ 0 & f\end{pmatrix}\begin{pmatrix} a^{-1} &  -b/ac \\ 0 & c^{-1}\end{pmatrix}\begin{pmatrix} d^{-1} &  -e/df \\ 0 & f^{-1}\end{pmatrix} \\
&= \begin{pmatrix} ad &  be+cf \\ 0 & cf\end{pmatrix}\begin{pmatrix} 1/ad &  -e/adf-b/acf \\ 0 & 1/cf \end{pmatrix} \\
&= \begin{pmatrix} 1 &  z \\ 0 & 1\end{pmatrix}
\end{split} 
\end{equation*}
for some $z\in \mathbb{R}$ as required.
A: You need to show that $\forall X \in U$, $\exists Y\in U$ such that
\begin{equation}
t_{1}t_{2}X=t_{2}t_{1}Y
\end{equation}
Try computing $Y$ explicitly in terms of the entries of $t_1, t_2, X$
A: Rather than going for a direct computation, you could ask yourself what $T/U$ is like. Using your notation:
Let $t=\begin{bmatrix}
    a & b \\
    0 & c 
\end{bmatrix}$, $U=\left\{\begin{bmatrix}
    1 & x \\
    0 & 1 
\end{bmatrix}: x\in\mathbb{R}\right\}$, then
$$tU =\left\{\begin{bmatrix}
    a & ax+b \\
    0 & c 
\end{bmatrix}: x\in\mathbb{R}\right\}$$
Since $a \neq 0$, each element $tU$ (which is actually an equivalence class) has a unique representative that is diagonal, being the representative with $x = -b/a$.
Therefore, $T/U$ is isomorphic to the group of nonsingular diagonal 2 x 2 matrices with entries in $\mathbb{R}$, and it is easy to see this group is abelian. 
