Let $G$ be a topological group and $H\leq G$ be a subgroup.
If $H,\frac{G}{H}$ are compact, then $G$ is compact.
There is a step in the book I do not understand.
Denote $O(G)$ the collection of all open sets of $G$ and this forms a lattice. Let $J$ be an $O(G)$ ideal s.t. $\cup_{x\in J}x=G$. It suffices to show $G\in J$.
Definition of lattice: $X$ is a topological space and $S$ is a collection of set s.t. $\emptyset\in S,X\in S$ and $S$ is closed under intersection and union. Then $S$ is called lattice. In this case $O(G)$ is a lattice.
Definition of ideal $I$ of a lattice $O(G)$: $I$ is an ideal if $A,B\in I$ then $A\cup B\in I$ and if $A\in I, B\in O(G)$ and $B\subset A$, then $B\in I$.
For any $x\in G, xH$ is compact and $xH$ is covered by some element of $J$. So $xH\subset J$ with $J'\in J$. Now pick open neighborhood $V$ of identity element $e$ s.t $VxH\subset J'$. This gives $VxH\in J$ and this is union of cosets of $H$. Let $I=\{A\in O(\frac{G}{H})\vert \pi^{-1}A\cap V\in J\}$ where $\pi:G\to\frac{G}{H}$. It is clear that $I$ is an non-empty ideal of $O(\frac{G}{H})$. Now $\pi(VxH)\ni\pi(x)$. So $\pi(x)\in\cup_{x\in I}I$. Since $x$ is arbitrary, we have all $\pi(G)\subset\cup_{x\in I}I$. However $\pi(G)=\frac{G}{H}$ is compact. Hence $\pi(G)\in I$ by $I$ being ideal.
$\textbf{Q:}$ Then the book says "i.e. $G\in J$ and hence $G$ is compact". Why so?
Ref: Introduction to Topological Groups by Higgins.