I want to show that a certain Lie subgroup (i.e. generated by the exponential of elements in some Lie subalgebra) of a Lie group is closed. My knowledge of the subject of Lie groups is rudimentary, and I would really appreciate any pointers towards relevant results.


My question
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**Is the following statement true?**



> 
> Let $G$ be a connected finite-dimensional real Lie group, and $H$ a Lie subgroup (i.e. generated by $\exp(\mathfrak{h})$ for some sub-lie algebra $\mathfrak{h}$ of the lie algebra $\mathfrak{g}$ of $G$). If the center of $H$ (i.e. $Z(H) = \lbrace x \in H: xh = hx\text{ for all }h \in H\rbrace$) is a closed subset of $G$, then $H$ is a closed subset of $G$.
> 
> 
> 


If true I would love a reference. If for some reason it is true only for semi-simple $G$ with finite center, that would be great as well.


Some context
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[The closed subgroup theorem](https://en.wikipedia.org/wiki/Closed-subgroup_theorem) says that any connected subgroup of $G$ which is closed is in fact a Lie group. On the wikipedia page there are [some criteria](https://en.wikipedia.org/wiki/Closed-subgroup_theorem#Conditions_for_being_closed) which allow one to deduce that the group associated to some lie algebra $\mathfrak{h} \subset \mathfrak{g}$ is closed, but they don't seem to apply in my case of interest.


I found an [encyclopedia of math article](https://encyclopediaofmath.org/wiki/Lie_group,_solvable) where the claim is made that any connected subgroup of a simply connected solvable real Lie group is closed. This is false as noted in the answers to [this stackoverflow question](https://math.stackexchange.com/questions/3237589/subgroups-of-solvable-lie-groups-are-closed). However for Lie subgroups it seems to be true. The article gives [a paper of Malcev](https://mathscinet.ams.org/mathscinet-getitem?mr=13165) as a reference, there is an [an errata for the paper](https://mathscinet.ams.org/mathscinet-getitem?mr=18651), and also some errors are pointed out by Chevalley in the mathscinet review (where he implies that the particular result that interests me was not new at the time but, alas, does not give a reference). I can use this to solve some, but not all, of the cases that interest me.


Finally, I found [some papers](https://mathscinet.ams.org/mathscinet-getitem?mr=56613) by Morikuni Goto which seem likely to contain results I could piece together to either get my statement, or something similarly useful.


He gives conditions on the Lie algebra of a connected Lie group $H$ that guarantee that every isomorphic embedding has a closed image. In particular if $Z(H)$ is compact then any continuous isomorphic image of $H$ in a Lie group is closed.


However, the subgroups $H$ that interest me have non-compact center, which happens to be closed in the larger group (this means some isomorphisms could send $H$ to a non-closed subgroup inside some other group).



Is a Lie subgroup whose center is closed, a closed subgroup itself?
No. Consider a group of the form $G=V\rtimes K$ where $V$ is a Euclidean group and $K$ is a compact 2-torus and $K$ acting faithfully on $D$, with no nonzero invariant vector. (For instance $G=(\mathbf{R}^2\rtimes\operatorname{SO}(2))^2$.) Let $D$ be a dense line in $K$: then $V\rtimes D$ is dense and non-closed in $G$, and has a trivial center, hence its center is closed in $G$.


(However, I think that the statement is true when the subgroup is semisimple: in a connected Lie group, an immersed Lie subgroup that is semisimple with the assumption that its center is closed in $G$, has to be closed — I'm not sure of a reference for this.)


