Question on properly defining a function (an isomorphism actually)

Question

Let's say that we have the set $G=\{A_x |x\in \mathbb{Z}\}$, where $A_x=\begin{pmatrix} 1&0&0\\ 0&1&0\\ x&0&1 \end{pmatrix}, x\in \mathbb{Z}$.
I would like to show that $(G,\cdot)\cong (Z,+)$. This is pretty easy, but I would like to make sure that the function, $f:G\to \mathbb{Z}, f(A_x)=x, \forall A_x\in G$ is properly defined. It seems odd to me to define a function $\forall A_x \in G$, since the matrix $A_x$ is uniquely determined by choosing an $x\in \mathbb{Z}$.
I would be tempted to write $f(A_x)=x, \forall x\in \mathbb{Z}$, but this looks wrong since $f$ is defined on $G$, not $\mathbb{Z}$. Please let me know which of these is the correct way to define $f$(it is obvious that $f$ is an isomorphism, I just want to make sure that I define it rigorously).

Answer 1

Yes, it is a perfectly fine way to define it. I will redefine it in a more complex way with which, perhaps, you will feel more confortable. Define$$\begin{array}{rccc}h\colon&\mathbb Z^{3\times3}&\longrightarrow&\mathbb Z\\&\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}&\mapsto&a_{31}\end{array}$$and define$$\begin{array}{rccc}g\colon&\mathbb Z&\longrightarrow&\mathbb Z^{3\times3}\\&m&\mapsto&\begin{bmatrix}1&0&0\\0&1&0\\m&0&1\end{bmatrix}\end{array}.$$Then $f=h\circ g$.