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?
Best Answer
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.