Showing that a factor group of upper triangular matrices is abelian This is my first time studying abstract algebra, so my apologies if I misunderstand some concepts/notation.
In Abstract Algebra: Theory and Applications am I currently studying normal subgroups and factor groups. Asked is the following:

Question: Let $(T,\times)$ be a group with $T \subseteq \mathbb M_2(\mathbb R)$ the set of invertible upper triangular matrices. Let $U \subseteq T$, such that for $A \in U$, it holds that $u_{11} = u_{22} = 1$ (diagonal is 1). Show that $T/U$ is abelian.   

I have managed to proved the tips that are provided. That is, I have shown that $(U, \times)$ is a subgroup of $(T, \times)$, I have shown that $U$ is abelian and I have shown that $U$ is normal in $T$ (that is, $XU = \{ XA : A \in U \} = \{ AX : A \in U \} = UX$ for all $X \in T$. I can't manage the last part though. Anyone suggestions? 
 A: Consider $\phi : T\to (\mathbb{R}\setminus\{0\})^2$ defined by $$\phi\begin{pmatrix}a&b\\0&c\end{pmatrix}=(a,c).$$
You can check that $\phi$ is a group homomorphism. Then it's clear that the kernel is $U$. Hence $T/U$ is isomorphic to the image of $T$ in $(\mathbb{R}\setminus\{0\})^2$, which is abelian. Thus $T/U$ is abelian.
A: Hint: it is sufficient to show that $U$ contains the commutator subgroup $[T, T]$ of $T$, i.e. that for any $X, Y \in T$, we have $XYX^{-1}Y^{-1} \in U$. This is not so hard: letting 
$$X = \left( \begin{matrix} a & b \\ 0 & c \end{matrix} \right), Y = \left( \begin{matrix} d & e \\ 0 & f \end{matrix} \right)$$
you can actually compute $X^{-1}$ and $Y^{-1}$, and it is easy to see what happens to the diagonal entries.
A: First we can check here that $U$ is always a normal subgroup of $T$, also for $n\times n$-matrices.
A question regarding normal subgroup of group of upper-triangular matrices
Then we consider the quotient group. The cosets are represented by diagonal matrices. Obviously they commute. So $T/U$ is abelian. For a reference, see here:
Identifying cosets of normal subgroups in $UT_2(\mathbb{R})$.
A: First, some intuition.  Since the matrices in $U$ are of the form $\begin{pmatrix} 1 & u_{12}\\ 0 & 1 \end{pmatrix}$, we should expect that modding out by $U$ "kills" the upper right entry of each matrix, leaving us with diagonal matrices, which are commutative.
More rigorously, given $A = \begin{pmatrix} t_{11} & t_{12}\\ 0 & t_{22} \end{pmatrix} \in T$, then $\det(A) = t_{11} t_{22} \neq 0$ so $t_{11} \neq 0$.  Then
$$
\begin{pmatrix} t_{11} & t_{12}\\ 0 & t_{22} \end{pmatrix} = \begin{pmatrix} t_{11} & 0\\ 0 & t_{22} \end{pmatrix} \overbrace{\begin{pmatrix} 1 & t_{12}/t_{11}\\ 0 & 1 \end{pmatrix}}^{\in U}
$$
so
$$
\begin{pmatrix} t_{11} & t_{12}\\ 0 & t_{22} \end{pmatrix} \equiv \begin{pmatrix} t_{11} & 0\\ 0 & t_{22} \end{pmatrix} \pmod{U} \, .
$$
Thus every element of the quotient $T/U$ can be represented by a diagonal matrix.  Since diagonal matrices commute, this shows that $T/U$ is abelian.
