Abelian Lie group with no nonzero element $A$ such that $A+A=0$

Question

Suppose that an $n$-dimensional abelian Lie group has no nonzero element $A$ such that $A+A=0$ (using + for the group operation). Does it follow that the group is isomorphic to the vector space $\mathbb{R}^n$? This seems likely to me, since a simply connected abelian Lie group is a vector space, and the examples that are not simply connected have the topology of a direct product of a torus with a vector space. Is it necessary to assume that the group is connected?

Answer 1

Yes, let $X$ such an Abelian Lie group, the universal covering map $p:\hat X=R^n\rightarrow X$ is a surjective morphism of Lie group. Suppose that $p$ is not injective, you have $p(B)=0$, $B\neq 0$. Since $p$ is a covering, there exists a neighborhood $0\in U\subset \hat X$ such that the restriction of $p$ to $U$ is injective. There exists an integer $n$ such that ${B\over 2^n}\in U$, this implies that $p({B\over 2^n})\neq 0$. You can take $m<n$ such that $p({B\over 2^m})\neq 0$ and $p({B\over 2^{m-1}})=0$ and write $A=p({B\over 2^m})$. $A\neq 0$ and $2A=0$.

We conclude that if $p$ is not injective, there exists $A\in X, A\neq 0, 2A=0$. So if there does not exist an element $A$ such that $2A=0$, $p$ is injective and is henceforth an isomorphism.