Connected, one-dimensional Lie groups Show that a connected, one-dimensional Lie group $G$ is isomorphic to $\mathbb{R}$ or $S^1$. 
So far my approach has been to show a non-trivial, one-parameter subgroup of $G$ is surjective, but I have not really made much progress. 
I have only just begun studying Lie groups, so my knowledge of theoretical results is basically limited to the definition of the exponential map
$\exp_G: T_e G\to G$ and a few results regarding this.  
 A: Since $G$ is one-dimensional, its exponential map is of the form
$$\exp:\mathbb{R}\to G.$$
Moreover, since $G$ is connected and abelian, the exponential map is surjective. Thus, either $\ker\exp=\{0\}$ and $G\cong\Bbb R$, or $\ker\exp=r\mathbb{Z}$ for some $r>0$ and $\exp$ factors through an isomorphism $S^1\cong \Bbb R/r\Bbb Z\to G$.
A: One way would be to go to Lie algebras. If $G$ is one-dimensional then its Lie algebra is also one-dimensional. As such it is abelian. So all one-dimensional Lie groups have isomorphic Lie algebras. Therefore by Lie's third theorem they all have isomorphic universal covering groups.
Now it can be easily checked that $\mathbb{R}$ has itself as a universal covering group. Therefore $\mathbb{R}$ is a universal covering group for every Lie group of dimension one. In particular every Lie group of dimension one is a homomorphic image of $\mathbb{R}$. Now you can check that every (proper) closed subgroup of $\mathbb{R}$ is of the form $r\mathbb{Z}$ for some $r\in\mathbb{R}$. Thus the only quotients of $\mathbb{R}$ that could be candidates for a Lie group are $\mathbb{R}$ itself and $S^1$. This completes the proof.
