Let $G$ be a topological group and $H$ be a finite subgroup of $G$. Assume further that $G$ is $T_1$ (this ensures $H$ is closed in $G$ and thus $G/H$ is again $T_1$) and $C_2$. If the coset space $G/H$ is compact, is $G$ also compact?
I believe I can show by the openness of $p:G\to G/H$ that if $G$ is furthermore Hausdorff then $G/H$ is compact then $G$ is compact. Just take any sequence $\{x_n \}$ in $G$, the image $\{p(x_n)\}$ in $G/H$ should have a converging subsequence. WLOG, assume $\{p(x_n)\}$ is convergent in $G/H$ to $\bar{y}$. $\bar{y}$ can be lifted to $y_1,\dots, y_h$ where $h$ is the order of $H$. If $G$ is furthermore Hausdorff (plus $C_2$), then we can separate them by nested neighborhood bases $\{ U_{1i} \}_i, \dots, \{U_{hi} \}_i$ (any $U_{pi}$ and $U_{qj}$ are disjoint if $p\ne q$. Each group can be mapped by $p$ to a nested neighborbood of $\bar{y}$) and each $\{x_n\}$ has to fall in one of them. By Pigeon Hole principle, we can select a converging subsequence of $\{x_n\}$ in one of those $h$ nested neighborhoods converging to some $y_i$. Please let me know if my argument is wrong.
Now instead of being Hausdorff, $G$ is just $T_1$ (and $C_2$). I feel like the argument above might not work. So is it still true in this case that $G$ is compact when $G/H$ is?