Let $G$ be a group with a filtration: $$G=G_0\supset G_1\supset G_2\supset\cdots$$ where $G_i$ are normal in $G$.
I want to show that the associated topology on $G$, whose basic open sets are translates of $G_i$, makes $G$ into a topological group. For starters I would like to show that $\mu:G\times G\to G$ is continuous, when $G\times G$ is equipped with the product topology. Simpler (I imagine) I would like to first simply find all elements of $G\times G$ that are in the preimage of a single basic open. So $\mu(x,y)=xy$, and I am considering $\mu^{-1}(gG_i)$.
I am not sure how to be systematic about finding every such element. The preimage is: $$\{(x,y)\in G\times G\mid xy=gh, h\in G_i\},$$ where I can hence see that the preimage contains $\{g\}\times G_i$ and $\{e\}\times gG_i$ and $G\times \{g\}$ works since $y=g$ gives us $g^{-1}xg\in G_i$ whenever $x\in G_i$ by normality. Seems like I'm just guessing around which isn't doing me any good.
Let $\mu(x,y)=xy$ for some arbitrary $x,y\in G$. Take a neighbourhood $xyG_i$ of $xy$. If $x'\in xG_i$ and $y'\in yG_i$, then $x'=xg_1$ and $y'=yg_2$ for some $g_1,g_2\in G_i$. Then, $x'y'=xg_1yg_2$. Note $G_i$ is normal in $G$, so $y^{-1}g_1y=g_1'$ for some $g_1'\in G_i$, and we have $x'y'=xyg_1'g_2\in xyG_i$.
That proves that $\mu(xG_i\times yG_i)\subseteq xyG_i$, and as $xG_i\times yG_i$ is one of the base neighbourhoods of $(x,y)$ in $G\times G$ with the product topology, this is enough to claim continuity in $(x,y)$.