Is it true that a connected, simply connected, nilpotent $n$-dimensional Lie group $G$ is homeomorphic to $\mathbb R^n$?
EDIT:
Maybe a possible argument is the following: Since $G$ is simply connected, $G$ cannot contain any non-trivial maximal compact subgroups. By a theorem associated to Iwasawa and Malcev, all maximal compact subgroups are conjugate and thus have the same dimension. By a theorem by Hochschild(?), $G/K$ is diffeomorphic to $\mathbb R^n$, where $K$ is a(ny) maximal compact subgroup. But for $G$ simply connected, connected, nilpotent, $K$ must be trivial, whence $G$ itself is diffeomorphic to $\mathbb R^n$.
Best Answer
If a Lie group $G$ is nilpotent, its Lie algebra $\mathfrak g$ is nilpontent, and there is an ideal $\mathfrak h\subseteq\mathfrak g$ of codimension $1$. It follows that there is a normal subgroup $H\subseteq G$ whose Lie algebra is $\mathfrak h$, this subgroup is closed because $G$ is simply connected, and we have a short exact sequence of Lie groups $$1\to H\to G\to\mathbb R\to 0$$ This extension of groups is split, so that $G$ is in fact a semidirect product of $\mathbb R$ and $H$; to exhibit a splitting, let $\mathfrak a$ be a subspace of $\mathfrak g$ complementary to $\mathfrak h$: the $1$-dimensional closed Lie subgroup of $G$ tangent to it maps isomorphically to the $\mathbb R$ here, so its inverse splits the exact sequence. Since $H$ is nilpotent and of smaller dimension than $G$, by induction we can assume that $H$ is diffeomorphic to $\mathbb R^n$ for some $n$ and then $G\cong\mathbb R\rtimes H$ is diffeomorphic to $\mathbb R^{n+1}$.