Let $G$ be a profinite topological group with two closed subgroup $G\_1$ and $G\_2$. Suppose $G\_1$ is normal in $G$ and $G=G\_1G\_2$. Let $H\_i$ be an open subgroup in $G\_i$ for $i=1,2$.


**Question:** Is $ H\_1H\_2:=\{h\_1h\_2\mid h\_1\in H\_1 \text{ and } h\_2\in H\_2\}$ also open in $G$?



Openness of product of two open subgroups
Yes (even without assuming $G\_1$ normal).


Indeed, first, since $G\_1G\_2=G$, for some finite subset $F$ we have $FH\_1H\_2F=G$, i.e., finitely many left-right translates of the compact subset $H\_1H\_2$ cover $G$. Hence (Baire) $H\_1H\_2$ has nonempty interior.


Next, $H\_1\times H\_2$ left-right acts on $G$ and $H\_1H\_2$ is a single orbit (the orbit of 1). Hence by homogeneity, every point is an interior point. This means that $H\_1H\_2$ is open.


