Elliptic curves $\mathbb C/\Gamma , \mathbb C/\Gamma'$ are isomorphic iff $\Gamma=\lambda\Gamma'.$

Question

Let, $\Gamma, \Gamma'$ be lattices of $\mathbb C$, define ellicptic curves by $\mathbb C/\Gamma , \mathbb C/\Gamma'$, then

$\mathbb C/\Gamma , \mathbb C/\Gamma'$ are isomorphic $\Leftrightarrow$ $\Gamma=\lambda\Gamma'.$

The $\Leftarrow$ part is easy, but how to prove the $\Rightarrow$ part?

(Rmk:

I am reading Serre's A Course In Arithmetic, it doesn't particularly treat elliptic curves, and just writes:

"Let us associate to a lattice $\Gamma$ of $\mathbb C$ the elliptic curve $E_\Gamma =\mathbb C/ \Gamma$. It is easy to see that two lattices $\Gamma, \Gamma'$ define isomorphic elliptic curves if and only if they are homothety."

That's all I know about elliptic curves now, so I am not sure what does "isomorphic" mean in original question. ... I thought it means group isomorphism, but I am not sure now. It is helpful that anyone clarifies what is the author talking about.)

Answer 1

Hint:

A map $\mathbb{C}/\Gamma \to \mathbb{C}/\Gamma'$ comes from a map $\phi\colon \mathbb{C} \to \mathbb{C}$ such that $\phi( z + \gamma) = \phi(z) + f(\gamma)$, for some function $f$. Hence for the derivative we have $\phi'(z+ \gamma) = \phi'(z)$, and so $\phi'(z)$ is a $\Gamma$ periodic functions and thus constant $\equiv \alpha$. Therefore $\phi(z) = \alpha z + \beta$ with the extra condition $\alpha(\Gamma) \subset \Gamma'$. We have found all the maps between tori. Now it's easy.