If $G/H$ is homeomorphic to $\mathbb R^n$ then $H$ is normal subgroup

Question

Let $G$ be a topological group and $H$ be a closed subgroup such that $G/H$ is homeomorphic (as a manifold) to the abelian group $\mathbb R^n$.

Then, is it necessary that $H$ has to be normal subgroup of $G$?

Answer 1

No,take $G=\pmatrix{a&b\cr 0&c}, a,c>0 0$, $G$ is isomorphic to $\mathbb{R}^3$, consider the subgroup $H=\pmatrix{a&0\cr 0&c}$ of $G$, $G/H$ is isomorphic to $\mathbb{R}$, but $H$ is not normal in $G$.

Answer 2

Let $G=$ the matrices in $GL_2(\Bbb{R})$ with positive determinant and let $H$ be the diagonal matrices in $G$. Then $H$ is a closed subgroup that is not normal, and $G/H$ is homeomorphic to $\Bbb{R}^2$.