Dimension of image of one parameter subgroup. [closed]

If $$G$$ is a Lie group then $$\eta : \mathbb{R}\to G$$ is called one parameter subgroup if it is a continuous group homomorphism.

I need to show that the images of one-parameter subgroups in a Lie group $$G$$ are precisely the connected Lie subgroups of dimension less than or equal to $$1$$.

I am not getting any idea how to start. Any hints are appreciated.

Thank you.

closed as off-topic by Moishe Kohan, Saad, Alexander Gruber♦Apr 9 at 16:17

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• The place to start would be to write carefully the definition of a Lie subgroup that you have. Then compare it with the setting of the problem. – Moishe Kohan Apr 8 at 21:41

This is not true, consider the 2-dimensional torus $$T^2$$, there exists morphisms $$\mathbb{R}\rightarrow T^2$$ whose image are dense, such an image is not a subgroup since it is not closed.
• @Saikat, the dimension $0$ case should not be hard for you. Now for dimension 1, Proposition 5.2, page 34 of Bump's Lie Groups is a very good start. You just need to know via Whitney's Embedding theorem that every Lie group as a manifold, sits inside some $\mathbb{R}^{n^2}=Gl(n, \mathbb{R})$. Now use the matrix exponential. – Laz Apr 8 at 22:06
• @TsemoAristide is right about his counterexample. Just precompose the homomorphism $\mathbb{S}^1\rightarrow \mathbb{T}^2$ give here (math.uchicago.edu/~womp/2007/lie2007.pdf) with $exp:\mathbb{R}\rightarrow \mathbb{S}^1$. We need to require closedness, which is also used in the lemma of Bump's book I cited. – Laz Apr 8 at 23:55