Let $G$ be a Lie group and $H$ be a closed subgroup of $G$. Show that if $H$ and homogeneous space $\frac{G}{H}$ are connected, then $G$ is connected. Let $G$ ge a Lie group and $H$ be a closed subgroup of $G$. Show that if $H$ and homogenuous space $\frac{G}{H}$ are connected, then $G$ is connected.
Remark: For this proof I use the following theorem:

Theorem: Let $G$ be a Lie group and $M$ smooth manifold, suppose that $G$  acts transitively (smooth) on $M$. Let  $M_{\alpha}$ any connected component of $M$ and $G^{0}$ the connected component of identity in $G$, then $G^{0}$  acts transitively (smooth) on $M_{\alpha}$, furthermore, $M_{\alpha} \cong \frac{G^{0}}{G^{0}_{p}}$ for $p\in M_{\alpha}$, where $G^{0}_{p}$ isotropy group of $p$.

Considering $M=\frac{G}{H}=M_{\alpha}$ because $\frac{G}{H}$ is connected, we have that $G$ acts transitively on $G/H$ because $G/H$ is a homogeneous space, then, by the previous theorem  $G^{0}$ acts transitively on $G/H$.
Furthermore, as $H$ is connected closed subgroup of $G$, then $H\subseteq G^{0}$, therefore $H$ acts transitively on $\frac{G}{H}$. If you read the proof of this theorem will note that the action can say it is:
$$\begin{array}{rcl} G\times \frac{G}{H}&\rightarrow& \frac{G}{H} \\ (g,g^{*}H) &\rightarrow & gg^{*}H \end{array}$$
 Then, by the following theorem:

Theorem: Let $G$ be a Lie group, $M$ smooth manifold $G\times M \rightarrow M$ be a transitive and smooth  action, then $G_{p}$ (isotropy group) is closed subgroup of $G$ and $M \cong \frac{G}{G_{p}}$ for each $p\in M$.

we have $\frac{G}{H} \cong \frac{H}{H_{p}}$ where  taking $p=eH$ we have:
$$H_{p}=\left\{h\in H : heH=eH\right\}=H.$$
Therefore, $\frac{G}{H} \cong \frac{H}{H}$, then $G\cong H$, and as $H$ is connected, then $G$ is connected.
My questions: I need to know if my test has a mistake and if any of you have another proof.
 A: Your argument has a significant issue in the paragraph after the theorem you quote:  $H$ almost never acts transitively on $G/H$ by any action, and the induced action of $H$ on $G/H$ coming from the natural action of $G$ on $G/H$ is transitive iff $G = H$, so is very rare indeed.  In other words, by stating $H$ acts transitively on $G/H$, it automatically follows that $G  = H$, so $G$ is connected if $H$ is.
Here' a sketch of a proof.  Consider the projection $\pi:G\rightarrow G/H$ and note that $\pi$ is an open map.
Now, assume $H$ is connected, $G/H$ is connected, but $G$ is disconnected.
Pick nonemtpy open sets $U$ and $V$ witnessing the fact that $G$ is disconnected.  Because $\pi$ is an open map, $\pi(U)$ and $\pi(V)$ are nonempty open subsets of $G/H$.  Since $G/H$ is connected, we must have $\pi(U)\cap \pi(V)\neq \emptyset$ so pick $x\in \pi(U)\cap \pi(V)$.  Being in the intersection means that $\pi(y) = \pi(z) = x$ for some $y\in U$ and $z\in V$.  Since $\pi(y) = \pi(z)$, $y = zh$ for some $h\in H$.
Now, show that the relatively open subsetes $U\cap yH$ and $V\cap yH$ of $yH$ disconnect $yH$.  Since $yH$ is homeomorphic to $H$, it follows that $H$ is disconnected as well, contradiction.
