Order of $A/2A$ for $A$ an Abelian variety

Question

Let $A$ be an Abelian variety over $\mathbb R$ of dimension $g$. Then the size of $A(\mathbb R)/2A(\mathbb R)$ is $(\# A(\mathbb R)[2])/2^g$. I'm wondering how one might go about proving such a result. If we deal with the Jacobian of a hyperelliptic curve, then I guess you could start talking about the number of components of the curve etc. Any textbooks sources would be cool too (I looked in Lang's book and couln't see anything relevant).

Answer 1

This is something simple about commutative compact real lie groups of dimension $g$. I think these are all of the form $G = (S^1)^g \times A$ where $A$ is a finite abelian group (didn't carefully check this, but I'm not worried). Hence $G/2G$ is isomorphic to $A/2A$. On the other hand $G[2] = (\mathbf{Z}/2\mathbf{Z})^g \times A[2]$. And for a finite abelian group $A$ the groups $A/2A$ and $A[2]$ have the same number of elements.