Prove $\mathbb{Z}$ is isomorphic to G

Question

Group $G$ is defined as \begin{bmatrix} {1-n}& {-n}\\ n & 1+n\\ \end{bmatrix} under matrix multiplication for all $n \in \mathbb{Z}$ .

We are asked to prove that $\mathbb{Z}$ is isomorphic to $G$.

I believe our $\phi$ is just $G$ so I proved the matrices of $n_1=n_2$ for injective.

Surjective is giving me a problem. I know I want to say there is some $n_1,n_2$ such that I want $\phi (n_1) = n_2$ but I do not know how to go about this.

Answer 1

First show that $G$ is a subgroup of $SL_n(\mathbb{Z})$. So it is really a group. Then show that is infinite cyclic, hence isomorphic to $\mathbb{Z}$. That the map you have from G to $\mathbb{Z}$ is surjective, is trivial, because for given $m\in \mathbb{Z}$, what could the preimage be?

Answer 2

Hint: I would recomment you try to multiply $\left[\begin{smallmatrix} {1-n}& {-n}\\ n & 1+n\\ \end{smallmatrix}\right]$ with $\left[\begin{smallmatrix} {1-m}& {-m}\\ m & 1+m\\ \end{smallmatrix}\right]$ and see what happens. After that, it should be obvious what $\phi$ ought to be.