isomorphism between torus I am working on Lectures On Riemann Surfaces by Forster. I am having trouble figuring out the following question.
1.5
a) Let $\Gamma,\Gamma'\subset\mathbb{C}$  be two lattices. Suppose $\alpha\in\mathbb{C}^*$ such that $\alpha\Gamma\subset\Gamma'$. Show that the map $\mathbb{C}\rightarrow\mathbb{C}$, $z\mapsto\alpha z$ induces a holomorphic map $\mathbb{C}/\Gamma\rightarrow\mathbb{C}/\Gamma'$, which is biholomorphic if and only if $\alpha\Gamma=\Gamma'$.
b) Show that every torus $X=\mathbb{C}/\Gamma$ is isomorphic to a torus of the form $X(\tau):=\mathbb{C}/(\mathbb{Z}+\mathbb{Z}\tau)$, where $\tau\in\mathbb{C}$ satifies $\Im(\tau)>0$.
c) Suppose $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\\$$\in SL(2,\mathbb{C})$ and $\Im(\tau)>0$. Let $\tau':=\frac{a\tau+b}{c\tau+d}$. Show that the tori $X(\tau)$ and $X(\tau')$ are isomorphic.
The only part that I have figured out is b). I used part a) and set $\alpha$ to be some rotation and dilation to manipulate the basis of $\Gamma$.
Unfortunately, I have no idea for a) and c).
 A: a) Define the map $\mathbb{C}\to\mathbb{C}$ by $z\mapsto\alpha z$. Well, we can compose this with the projection $\mathbb{C}\to\mathbb{C}/\Gamma'$ to get a map $\mathbb{C}\to\mathbb{C}/\Gamma'$. Then, this map will factor through the quotient $\mathbb{C}/\Gamma$ if and only if $\Gamma$ is in the kernel of this map. But, this is equivalent to $\alpha\Gamma\subseteq\Gamma'$. Now, since this is a holomorphic map of compact Riemann surfaces it's automatically surjective (although this was clear by inspection). So you then only care about injectivity. But, the factored map $\mathbb{C}/\Gamma\to\mathbb{C}\Gamma'$ is injective if and only if 
$$\ker(\mathbb{C}\to\mathbb{C}/\Gamma')=\Gamma$$
Well, we already know that $\alpha\Gamma\subseteq\Gamma'$. And, if we assume that $\Gamma\supseteq\ker(\mathbb{C}\to\mathbb{C}/\Gamma')=\frac{1}{\alpha}\Gamma'$ we get the other inclusion. 
b) You said you did this.
c) You will eventually show that the only maps $X(\tau)\to X(\tau')$ are those constructed in a), so you know this isomorphism will have to be of that form. So, assume that you'f found $\alpha$ such that $\alpha(\mathbb{Z}+\mathbb{Z}\tau)=\mathbb{Z}+\mathbb{Z}\tau'$. This, means that $\alpha,\alpha\tau$ need to generate $\mathbb{Z}+\mathbb{Z}\tau'$. Well, for $\alpha,\alpha\tau\in\mathbb{Z}+\mathbb{Z}\tau'$ there needs to exist $a,b,c,d\in\mathbb{Z}$ such that $\alpha=c+d\tau'$ and $\alpha\tau=a+b\tau'$. So, upon dividing you get
$$\tau=\frac{a+b\tau'}{c+d\tau'}$$
but, for this to actually generate $\mathbb{Z}+\mathbb{Z}\tau$ you need the change of basis matrix $\begin{pmatrix}a & b\\ c & d\end{pmatrix}$ to lie in $\text{GL}_2(\mathbb{Z})$. In particular, it must have determinant $\pm 1$. But, use the fact that $\tau,\tau'\in\mathfrak{h}$ to conclude that the determinant must be $1$. 
Working backwards gives you the other direction.
