# If $H_1 \subset H_2 \subset G$ and $G/H_2,\ H_2/H_1$ are compact then $G/H_1$ is compact.

I'm trying to solve the following exercise of the book "Grupos de Lie - Luiz A. B. San Martin (exercise 18, page 55)":

Exercise: Let $$G$$ be a topological group (if necessary, $$G$$ is Hausdorff) and $$H_1 \subset H_2\subset G$$ closed subgroups of $$G$$. Show that if $$G/H_2$$ and $$H_2/H_1$$ are compact, then $$G/H_1$$ is compact.

It is easy to see that the function \begin{align*} \pi: G/H_1 &\to G/H_2 \\ g H_1 & \mapsto g H_2, \end{align*} is a continuous and open function, nevertheless $$\pi$$ also satisfies $$g_1 \cdot \pi(g_2) = \pi(g_1\cdot g_2)$$, $$\forall \ g_1 \in G,\$$ ($$g_1 \cdot (g_2 H) = (g_1 \cdot g_2) H$$).

Although I am aware of the following theorem:

Theorem: Let $$G$$ be a topological group, and $$H$$ a closed subgroup of $$G$$, if $$H$$ and $$G/H$$ are compact then $$G$$ is compact.

I can't apply it to solve my problem, once neither $$G/H_1$$ nor $$H_2/H_1$$ are topological groups.

So, I tried to adapt the proof of the theorem cited above for my case, and it is necessary to show that the function $$\pi$$ is a closed function, which I was not able to conclude.

Can anyone help me?

• Why wouldn't $G / H_1$ and $G / H_2$ be topological groups? Apr 23, 2019 at 21:45
• $H_1$ and $H_2$ are not normal subgroups. Apr 23, 2019 at 21:46
• Ah yes, I was focusing on the topological bit. In your definition of a topological group, are they always Hausdorff? Apr 23, 2019 at 22:28
• Unfortunately not. Apr 23, 2019 at 22:29
• I'm now suspecting the statement is actually false. (without the normality assumptions) Apr 26, 2019 at 15:43

Lemma. The map $$\pi$$ is closed.
Proof. Let $$F_1\subset G/H_1$$ be any closed set and $$x_2\in G/H_2\setminus \pi(F_1)$$ be any point. We shall construct a neighborhood of $$x_2$$ disjoint from $$\pi(F_1)$$. For each $$i=1,2$$ let $$q_i:G\to G/H_i$$ be the quotient maps. We have $$q_2^{-1}(x_2)=xH_2$$ for some point $$x\in G$$. Put $$F=q_1^{-1}(F_1)$$ and remark that $$F$$ is closed in $$G$$ and $$F=FH_1$$. Since $$x_2\not\in \pi(F_1)$$, the sets $$F$$ and $$xH_2$$ are disjoint. Let $$y\in xH_2$$ be any point. Since the set $$F$$ is closed, $$y\not\in F$$, and $$G$$ is a Hausdorff topological group, there exists an open neighborhood $$O_y=O_y^{-1}$$ of the identity of $$G$$ such that $$O_y^2y\cap F=\varnothing$$. Since $$F=FH_1$$, $$O_y^2yH_1\cap F=\varnothing$$ and so $$q_1(O_y^2yH_1)\cap q_1(F)=\varnothing$$. Remark that $$O_{y1}=q_1(O_yyH_1)$$ is an open neighborhood of a point $$yH_1\in G/H_1$$. Since a map $$tH_1\mapsto xtH_1$$ for each $$t\in H$$ is a homeomorphism of a space $$G/H_1$$, the set $$xH_2/H_1$$ is homeomorphic to $$H_2/H_1$$, so it is compact. Since $$\{O_{y1}:y\in xH_2\}$$ is an open cover of the set $$H_2/H_1$$, there exists a finite subset $$Y$$ of $$xH_2$$ such that $$xH_2\subset\bigcup \{O_{y1}:y\in Y\}$$. Put $$O=\bigcap\{Oy: y\in Y\}$$. We claim that $$F\cap OxH_2=\varnothing$$. Indeed, suppose to the contrary that there exists a point $$z\in F\cap OxH_2$$. Since $$xH_2\subset\bigcup \{O_{y1}:y\in Y\}$$, there exists a point $$y\in Y$$ such that $$z\in O=O_yyH_1\subset O_y^2H_1$$, a contradiction, because $$O_y^2yH_1\cap F=\varnothing$$. So a set $$q_2(OxH_2)$$ is a neighborhood of $$x_2$$ and $$q_2(OxH_2)\cap \pi(F_1)= q_2(OxH_2)\cap q_2(FH_2)=\varnothing$$. $$\square$$
Since $$H_1$$ is closed in $$G$$, the space $$G/H_1$$ is Hausdorff and we see that the map $$\pi$$ is perfect. So the space $$G/H_1$$ is compact by Theorem 3.7.2 from “General topology” by Ryszard Engelking (Heldermann Verlag, Berlin, 1989)).
• What is $F_2$ in the 11th line? Apr 30, 2019 at 6:02
• @YuiToCheng Thanks, it is $F_1$, corrected. Apr 30, 2019 at 6:04
• You deduce $q_1(O_y^2yH_1)\cap q_1(F)=\varnothing$ from $O_y^2yH_1\cap F=\varnothing$ (20th line), but $f(A)\cap f(B) \subset f(A\cap B)$ holds only if $f$ is injective. Apr 30, 2019 at 9:01