Let be two lattices $L$ and $L'$ given such that $L\subseteq L'$. Consider the canonical map from $\mathbb C/L$ to $\mathbb C/L'$.
Now I want to show that this map is an isomorphism (biholomorph) if and only if $L=L'$. The "if" part is obvious.
For the other direction I would like to write down the map which is as follows:
Let $L'=\mathbb Zv_1+\mathbb Zv_2$ then $L=\mathbb \alpha_1v_1\mathbb Z+\alpha_2v_2\mathbb Z$ for some $\alpha_1,\alpha_2\in \mathbb Z$. Then the above map is given by
$$z+(\mathbb Zv_1+\mathbb Zv_2)\mapsto z+(\mathbb Z\alpha_1v_1+\mathbb Z\alpha_2v_2).$$
I start I hoped that this map is not injective if $\alpha_1,\alpha_2\neq 1$, but it seems that this is not the case...
So now I would like to check if this map is holomorph by using the definition. Somehow I am not able to get any restriction of $\alpha_1$ and $\alpha_2$. Hopefully someone can help me a bit.
We denote $L'/L$ as the set of cosets of $L'$ under $L$ (looked at as abelian groups). Clearly $|L'/L|=1$ if and only if $L'=L$. We note that, denoting the map $f:\mathbb{C}/L\to \mathbb{C}/L'$, we have that $L'/L\subseteq \mathbb{C}/L$, and we have that $f(L'/L)=[L']$, a single element. Thus for $f$ to be an isomorphisms, we must have that $L'/L$ consists of a single element, so that by the above reasoning, $L'=L$ as desired.