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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 $G$?

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  • $\begingroup$ Is $\ H_i\ $ assumed to be open in $\ G\ $ or in $\ G_i?$ $\endgroup$
    – Wlod AA
    Commented Sep 26, 2022 at 13:32
  • $\begingroup$ Anyway, as it looks now, if $\ G\ $ is not discrete, and if $\ |G_1|=|G_2|=1\ $ then $\ H_1H_1\ $ is not open in $\ G$. $\endgroup$
    – Wlod AA
    Commented Sep 26, 2022 at 13:49
  • $\begingroup$ And, if the openness is relative to $\ G\ $ then the answer is trivial YES. $\endgroup$
    – Wlod AA
    Commented Sep 26, 2022 at 13:53
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    $\begingroup$ @WlodAA If $G_1=G_2=\{1\}$, then $G=G_1G_2=\{1\}$ and $G$ is discrete. $\endgroup$
    – kabenyuk
    Commented Sep 26, 2022 at 17:04
  • $\begingroup$ @kabenyuk, thank you. (I've overlooked an assumption). $\endgroup$
    – Wlod AA
    Commented Sep 26, 2022 at 19:13

1 Answer 1

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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.

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  • $\begingroup$ Could you please state the general theorem used in the last part of your answer. $\endgroup$
    – MSMalekan
    Commented Oct 28, 2022 at 18:43

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