Connectedness of a normal subgroup of a linear Lie group. Let $G\leq GL(n,\mathbb{R})$ be a connected linear Lie group and let $N$ be a normal subgroup of $G$ such that their Lie algebras are isomorphic:
$$\mathfrak{g}\cong\mathfrak{n}$$
Does it follow that $N$ is connected? I feel like it should be, although I don't have much intuition for Lie groups.
 A: Edit: As pointed out by Mirjam below, there is a straighforward proof, which does not need to assume that $N$ be closed in $G$. The exponential map is a local homeomorphism $0\longmapsto 1$ between the Lie algebras and the groups. If the former are equal, it follows that $N$ is open in $G$, hence $G\setminus N=\bigsqcup_i g_iN$ (where $g_i$'s are a system of representative of $G/N$) is closed in $G$. So $N$ is open/closed and nonempty in $G$. By connectedness, $N=G$.
Quotient argument: If you assume $N$ to be a closed normal subgroup in $G$, then $G/N$ is a Lie group with Lie algebra naturally isomorphic to the quotient of the Lie algebra of $G$ by the one of $N$. See here. Now if the Lie algebras of $N$ and $G$ are the same, the Lie algebra of $G/N$ is trivial. And if $G$ is connected, so is $G/N$. We deduce from this that $G/N$ is trivial, i.e. $N=G$.
Non-connected case: This is no longer true if $G$ is disconnected. For instance, consider
$$
G=\mathbb{R}\times\mathbb{Z}_2
\qquad N=\mathbb{R}\times\{0\}.$$
