$\mathbb{Z}^2$ as a subgroup of Aut($\mathbb{D}$) Let $\mathbb{D}$ denotes the unit disc in the complex plane. Let Aut($\mathbb{D}$) denote the automorphism group of the unit disc. I have the following question:
1). Does $\mathbb{Z}^2$ sits in Aut($\mathbb{D}$) as a subgroup?
2). If yes or no, is there a simple way to see that?
 A: 1) Yes, it does.
2) It's probably geometrically easier to see that $\operatorname{Aut}(\mathbb{H})$, the automorphism group of the upper half-plane contains a subgroup isomorphic to $\mathbb{Z}^2$. It is easy to see that $\operatorname{Aut}(\mathbb{H})$ contains subgroups isomorphic to $\mathbb{R}$, for example the translations $z \mapsto z + a,\; a\in \mathbb{R}$. Now take any two real numbers $\alpha,\beta$ that are linearly independent over $\mathbb{Q}$, say $\alpha = 1,\, \beta = \pi$, and consider the subgroup
$$T(\alpha,\beta) = \{z \mapsto z + m\alpha + n\beta : (m,n) \in \mathbb{Z}^2\}.$$
Transport that to $\operatorname{Aut}(\mathbb{D})$ by conjugation with a biholomorphic mapping between the unit disk and the upper half-plane.
The construction generalises, $\operatorname{Aut}(\mathbb{D})$ contains isomorphic copies of $\mathbb{Z}^n$ for all $n\in\mathbb{N}$.
A: With the answer of Fischer:we choose $\alpha$ and $\beta$ two irrational $\mathbb{Q}$ independent elements such that $k\alpha+l\beta$ not rational with $k,l$ in $\mathbb{Z}$,for example $\sqrt2,\sqrt3$.set,$$T(\alpha,\beta)=\{\mathcal{R}_{m\alpha + n\beta}| \mathcal{R}_{m\alpha + n\beta}\space{is}\space{a}\space{rotation}\space{with}\space{(m\alpha + n\beta)2\pi},(m,n) \in \mathbb{Z}^2\}\cong\mathbb{Z}^2.$$
