Topology by a collection of cosets which makes product continuous Let $G$ be any group.
Let $\mathcal{F}$ be a family of some  subgroups of $G$, along with $G$, which is closed under finite intersection.
Then all left cosets $\{xH : x \in G, H\in \mathcal{F} \}$ form a basis for a topology on $G$.
If, for every subgroup $H$ in $\mathcal{F}$, there is a normal subgroup $N$ of $G$, such that
(1) $N$ is contained in $H$ and
(2) $N$ is a member of $\mathcal{F}$
then it can be shown that the product map $G\times G\rightarrow G$ is continuous.
Question: For the product map $G\times G\rightarrow G$ to be continuous, are the conditions (1) and (2) above on family
$\mathcal{F}$ necessary?
 A: Multiplication $G\times G\to G$ is continuous iff for every basic open set $xH$ (with $x\in G$,  $H\in\mathcal F$), the pre-image $\{\,(a,b)\in G\times G\mid ab\in xH\,\}$ is open. This requires us to find $H_a,H_b\in \mathcal F$ such that $aH_abH_b\subseteq xH$. This implies $H_b\le H$, and in fact we may wlog. pick $H_b=H$ and the condition boils down to $aH_ab\subseteq xH=abH$, or $H_a\le bHb^{-1}$.
Hence it is necessary (and sufficient) to have

*

*Every conjugate of an element of $\mathcal F$ contains an element of $\mathcal F$.

And for this, it is sufficient to have

*

*For every $H\in\mathcal F$, there are $H'\in \mathcal F$ and $N$ with $H'\le N\trianglelefteq H$
This is weaker that your proposed conditions as there may be infinitely descending chains $H> H'> H''>\cdots$ with none of them normal.
Example: Consider $S_3$ with its non-normal subgroup $C_2$. Let $G=S_3\times S_3\times \ldots$ be the direct product of countably many $S_3$ and let $\mathcal F$ be all subgroups of the form $1\times\cdots\times 1\times C\times S_3\times S_3\times\cdots$.
