Define the set $$ G := \left\{ \begin{pmatrix} 1 - 2n & n \\ -4n & 1 + 2n \end{pmatrix} : n \in \mathbb{Z}\right\}. $$ Show that $(G, \cdot)$ is a group using the usual matrix multiplication. Furthermore, decide whether $G \cong \mathbb{Z}$.
To show that $G$ is a group, we note that matrix multiplication is associative. The identity matrix $I_{2 \times 2}$ takes the role of the identity element in G. Since $\det(A) = 1 \neq 0$ we know that each element $A \in G$ has an inverse $A^{-1} \in G$ which can be computed by the formula $A^{-1} = \dfrac{1}{\det(A)} \cdot \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ for any $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Thus, $(G, \cdot)$ is a group.
Now, for the homomorphism part. Define the map $\phi : \mathbb{Z} \to G$ by
$$ \phi(n) = \begin{pmatrix} 1 - 2n & n \\ -4n & 1 + 2n \end{pmatrix}. $$
First, we show that $\phi$ is a homomorphism. I.e. that $\forall m, n \in \mathbb{Z}$ we have
\begin{equation} \label{eq:homomorphism_condition} \phi(n + m) = \phi(n) \cdot \phi(m) . \end{equation}
We have
\begin{equation*} \begin{split} \phi(n) \cdot \phi(m) & = \begin{pmatrix} 1 - 2n & n \\ -4n & 1 + 2n \end{pmatrix} \cdot \begin{pmatrix} 1 - 2m & m \\ -4m & 1 + 2m \end{pmatrix} \\ & = \begin{pmatrix} (1 - 2n)(1 - 2m) + n(-4m) & (1 - 2n)m + n(1 + 2m) \\ -4n(1 - 2m) + (1 + 2n)(-4m) & -4nm + (1 + 2n)(1 + 2m) \end{pmatrix} \\ & = \begin{pmatrix} 1 - 2m - 2n + 4mn - 4mn & m - 2nm + n + 2mn \\ -4n + 8nm -4m - 8mn & -4nm + 1 + 2m + 2n + 4mn \end{pmatrix} \\ & = \begin{pmatrix} 1 - 2m - 2n & m + n \\ -4n -4m & 1 + 2m + 2n \end{pmatrix} \\ & =\begin{pmatrix} 1 - 2(n + m) & m + n \\ -4(n + m) & 1 + 2(n + m) \end{pmatrix} \\ & = \phi(n + m), \end{split} \end{equation*}
which proves that $\phi$ is a homomorphism.
Next, we show that $\phi$ is bijective. To see that $\phi$ is injective, let $\phi(n) = \phi(n')$. Hence,
\begin{equation*} \begin{pmatrix} 1 - 2n & n \\ -4n & 1 + 2n \end{pmatrix} = \begin{pmatrix} 1 - 2n' & n' \\ -4n' & 1 + 2n' \end{pmatrix}, \end{equation*}
if and only if
\begin{equation*} \begin{split} 1 - 2n = 1 - 2n' \quad & \iff \quad n = n' \\ n = n' \quad & \iff \quad n = n' \\ -4n = -4n' \quad & \iff \quad n = n' \\ 1 + 2n = 1 + 2n' \quad & \iff \quad n = n'. \end{split} \end{equation*}
Thus, $\phi$ is injective.
I am stuck on showing that $\phi$ is surjective.
Hint: Prove this by induction: $$ \begin{pmatrix} 1 - 2n & n \\ -4n & 1 + 2n \end{pmatrix} = \begin{pmatrix} - 1 & 1 \\ -4 & 3 \end{pmatrix}^n $$
This follows because $\phi$ needs to be a homomorphism. Thus $$\phi(n)=\phi(n-1)\phi(1)=\phi(n-2)\phi(1)^2 = \cdots = \phi(0)\phi(1)^n = \phi(1)^n$$