I'm going through this Wikipedia section on Group Filtrations. Let $G$ be a group and $G_n$ a descending filtration of subgroups ($G_{n+1} \subset G_n$).
I've already proved that it forms a topology. So I'm trying to show that the map $f: (b,c) \mapsto bc^{-1} : G\times G \to G$ is continuous. I've got that $f^{-1} (\bigcup_{n \in \Bbb{N}} a_n G_n) = \bigcup f^{-1} (a_n G_n)$.
But $f^{-1} (a_n G_n) = $ all $(b,c)$ st. $bc^{-1} \in a_n G_n \iff 1 \in c b^{-1}a_n G_n \iff cb^{-1} a_n = 1 \iff (b,c) \in \{ (b, a_n^{-1} b) : b \in G \} = G \times $ what? I don't get how to transform that into a union of subsets of $G \times G$ of the form $d_i G_i \times e_j G_j$.
Thanks.
If $bc^{-1} \in aG_n$, then $bG_n \times cG_n \subset f^{-1}(aG_n)$. To prove this, let $x = bg \in bG_n$ and $y = ch \in cG_n$. Then:
$$a^{-1}xy^{-1} = a^{-1}bgh^{-1}c^{-1} = (a^{-1}bc)(cgh^{-1}c^{-1}) \in G_n$$
The last assertion follows from the observations that $a^{-1}bc \in G_n$ by assumption and $cgh^{-1}c^{-1} \in G_n$ since $gh^{-1} \in G_n$ by assumption and $G_n$ is normal.