Let $G$ be a group. Consider the set
$\mathcal{N}=\lbrace N\text{ }|\text{ }N\subseteq G\text{ such that }N \text{ is a normal subgroup of }G\rbrace $.
Consider the set
$B=\lbrace\sigma N\text{ }|\text{ }\sigma\in G, \text{ }N\in\mathcal{N}\rbrace$
We define a topology on $G$ by declaring the set $B$ to be a base.
Given this topology, I have the following questions
- Is it true that for any $\sigma\in G$, the set $\lbrace\sigma N\text{ }|\text{ }N\in\mathcal{N}\rbrace$ is a neighborhood base of $\sigma$ ?
- The map $f:G\times G\rightarrow G$ defined by $(g,h)\mapsto gh$ is continuous ?
My idea:
Let $U$ be an open set containing $\sigma$. Then $U$ is of the form $$U=\bigcup_{i}g_iN_i$$ As $\sigma\in U$, we have $\sigma\in g_iN_i$ for some $g_i\in G$ and $N_i\in \mathcal{N}$. As $\sigma\in g_iN_i$, we have $\sigma N_i=g_iN_i\subseteq U$.
Consider any $\sigma,\tau\in G$. To prove that $f$ is continuous, we have to show that given any neighborhood $W$ of $\sigma\tau$, we have to find a neighborhood $U$ of $\sigma$ and a neighborhood $V$ of $\tau$ such that $f(U\times V)\subseteq W$ (This is just the definition of continuity). We have $f(U\times V)=UV$ (This is by the definition of $f$). Any arbitrary neighborhood W of $\sigma\tau$ contains a set of the form $(\sigma\tau) N$ for some $N\in\mathcal{N}$( for any $g\in G$ the set $\lbrace gN\text{ }|\text{ }N\in\mathcal{N}\rbrace$ is a neighborhood base of $g$). Consider the neighborhoods $U=\sigma N$ of $\sigma$ and $V=\tau N$ of $\tau$. We have $$UV=\sigma N\tau N=(\sigma\tau)N\subseteq W$$ The step $\sigma N\tau N=(\sigma\tau)N$, follows from the fact that $N$ is normal in $G$.