Matrix group isomorphic to $\mathbb{Z}$ 
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 $$\phi(n)=\phi(n-1)\phi(1)=\phi(n-2)\phi(1)^2 = \cdots = \phi(0)\phi(1)^n = \phi(1)^n$$
