I have the following situation: let $\mathfrak{g}$ be an $n-$dimensional real Lie algebra of the connected Lie group $G$ and let $\mathfrak{r}(\mathfrak{g})$ be the solvable radical of $\mathfrak{g}$. If I consider $R = \langle \hbox{exp}{(\mathfrak{r}(\mathfrak{g}))}\rangle$, then is $R$ a closed subgroup of $G$?
My idea was to prove that $clR$ is the maximal solvable Lie subgroup of $G$, but I got nowhere.
Any help is appreciated. Thanks.