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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.

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

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    $\begingroup$ 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. $\endgroup$ – Moishe Kohan Apr 8 at 21:41
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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.

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    $\begingroup$ As far as I know, they are not group homomorphisms, however. $\endgroup$ – Randall Apr 8 at 19:01
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    $\begingroup$ If they were, the image would be a subgroup. $\endgroup$ – Randall Apr 8 at 19:02
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    $\begingroup$ Frequently, Lie subgroups are not required to be closed, see en.wikipedia.org/wiki/Lie_group#Lie_subgroup $\endgroup$ – Moishe Kohan Apr 8 at 19:11
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    $\begingroup$ @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. $\endgroup$ – Laz Apr 8 at 22:06
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    $\begingroup$ @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. $\endgroup$ – Laz Apr 8 at 23:55

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