Denote $G=\mathrm{Aff}(R)^{\circ}$ the identity component of the affine group of $\mathbb{R}$, and $\varphi$ any injective homomorphism from $\mathbb{Z}^2 \to G$.
My question is how can I prove that $\varphi(\mathbb{Z}^2)$ cannot be discrete in $G$? do you know any other example of $G$ where we can prove this? " $\varphi ( \mathbb{Z}^2)$ is not discrete in $G$"
Unsure if it is truely research level. Anyways, the Abeliean subgroups of $\text{Aff}(\mathbb{R})^{o}$ are either the embedded torus $G_{m}(\mathbb{R})^{o} = GL_{1}(\mathbb{R})^{o} = \mathbb{R}_{>0}$ (you specified the identity component) or a subgroup isomorphic to the unipotent group $\mathbb{G}_{a}(\mathbb{R})$. For the purposes of the question, these two groups are isomorphic (they are not as alg. groups, even though you used this tag). So you end up with an injective homomorphism $\phi:\mathbb{Z}^{2}\to \mathbb{G}_{a}(\mathbb{R})$. Obviously, by specifying the values over basis elements, you get a map of the form $\phi(z_1 ,z_2)=\alpha\cdot z_1 + \beta\cdot z_2$ for some $\alpha,\beta\neq 0$. If $\alpha,\beta$ are commensurable, you don't have injectivity. Hence they are not commensurable. As known, in this case, corollary to Kroncker's lemma will show you that $\phi(\mathbb{Z}^2)$ will contain elements arbitrary close to the identity, in particular, you will want to take rational approximations to $-\alpha/\beta$.