I am having trouble with the following property of elliptic curve:
Let $\Gamma$ be a lattice in $\mathbb{C}$ with two generators. Show that the group $\mathbb{C}/\Gamma$ is isomorphic to $S^1\oplus S^1$.
I suspect the approach would be the same as showing $\mathbb{R}/\mathbb{Z}$ is isomorphic to $S^1$, but that the best I could try.
Any help will be appreciated.
Best Answer
Let $a,b$ a $\Bbb Z$-basis of $\Gamma$, now consider the $\Bbb R$-linear automorphism $A:\Bbb C \to \Bbb C$ given by $A(1)=a,A(i)=b$, $A$ has the property that it maps the lattice $\Lambda:=\Bbb Z\oplus \Bbb Z i$ to $\Gamma$ by construction. Thus $A$ induces an isomorphism $\Bbb C/\Gamma \cong \Bbb C/\Lambda$.
Now the map $f:\Bbb R/\Bbb Z\oplus \Bbb R/\Bbb Z \to \Bbb C/\Lambda,(x,y) \mapsto \overline{x+iy}$ is an isomorphism, so we get $\Bbb C/\Lambda \cong S^1 \oplus S^1$ as $\Bbb R/\Bbb Z \cong S^1$.
Another way to finish the argument would be to note that generally if $G_1,G_2$ are groups with normal subgroups $H_1$ and $H_2$ respectively, then $(G_1 \times G_2) / (H_1\times H_2) \cong (G_1/H_1) \times (G_2/H_2)$