Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.11
Let $F$ be a field and let $$G = \left\{ \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \ |\ a,b,c \in F, ac \neq 0 \right\} \leq GL_2(F).$$
(1) Prove that the map $\varphi : \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \mapsto a$ is a surjective homomorphism from $G$ to $F^\times$. Describe the fibers and kernel of $\varphi$.
(2) Prove that the map $\psi : \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \mapsto (a,c)$ is a surjective homomorphism from $G$ to $F^\times \times F^\times$. Describe the fibers and kernel of $\psi$.
(3) Let $H = \left\{ \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} \ |\ b \in F \right\}$. Prove that $H$ is isomorphic to the additive group $F$.
Solution:
(1) Note that \begin{align*}&\ \varphi\left( \begin{bmatrix} a_1 & b_1 \\ 0 & c_1 \end{bmatrix} \begin{bmatrix} a_2 & b_2 \\ 0 & c_2 \end{bmatrix} \right) \\= &\ \varphi\left( \begin{bmatrix} a_1a_2 & a_1b_2 + b_1c_2 \\ 0 & c_1c_2 \end{bmatrix} \right) = a_1a_2\\
=&\ \varphi\left( \begin{bmatrix} a_1 & b_1 \\ 0 & c_1 \end{bmatrix} \right) \varphi\left( \begin{bmatrix} a_2 & b_2 \\ 0 & c_2 \end{bmatrix} \right)\end{align*} so that $\varphi$ is a homomorphism. Moreover, $\varphi$ is clearly surjective. It is also clear that the fiber over $a \in F^\times$ is $\left\{ \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \ |\ b,c \in F, c \neq 0 \right\}$. In particular, the kernel of $\varphi$ is $\left\{ \begin{bmatrix} 1 & b \\ 0 & c \end{bmatrix} \ |\ b,c \in F, c \neq 0 \right\}$.
(2) Note that the condition $ac \neq 0$ is equivalent to $a \neq 0$ and $c \neq 0$; so that if $\begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \in G$ we have $a,c$ arbitrary in $F^\times$. Thus $\psi : G \rightarrow F^\times \times F^\times$ is surjective.
Moreover, \begin{align*}&\ \psi \left( \begin{bmatrix} a_1 & b_1 \\ 0 & c_1 \end{bmatrix} \begin{bmatrix} a_2 & b_2 \\ 0 & c_2 \end{bmatrix} \right) \\=&\ \psi \left( \begin{bmatrix} a_1a_2 & a_1b_2 + b_1c_2 \\ 0 & c_1c_2 \end{bmatrix} \right) = (a_1a_2, c_1c_2)\\ =&\ \psi \left( \begin{bmatrix} a_1 & b_1 \\ 0 & c_1 \end{bmatrix} \right) \psi \left( \begin{bmatrix} a_2 & b_2 \\ 0 & c_2 \end{bmatrix} \right),\end{align*} so $\psi$ is a homomorphism.
Clearly the fiber over $(a,c)$ is $\left\{ \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \ |\ b \in F \right\}$, and the kernel of $\psi$ is $\left\{ \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} \ |\ b \in F \right\}$.
(3) Define $\theta : H \rightarrow F$ by $\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} \mapsto b$. $\theta$ is clearly bijective, and we have \begin{align*}&\ \theta \left( \begin{bmatrix} 1 & b_1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b_2 \\ 0 & 1 \end{bmatrix} \right) \\=&\ \theta \left( \begin{bmatrix} 1 & b_1 + b_2 \\ 0 & 1 \end{bmatrix} \right) = b_1 + b_2\\ =&\ \theta \left( \begin{bmatrix} 1 & b_1 \\ 0 & 1 \end{bmatrix} \right) \theta \left( \begin{bmatrix} 1 & b_2 \\ 0 & 1 \end{bmatrix} \right),\end{align*} so that $\theta$ is a homomorphism. So $H \cong F$.
