Locally profinite group is a topological group $G$ such that every open neighbourhood of the identity in $G$ contains a compact open subgroup of $G$.
Profinite group is the compact locally profinite group.
My questions are:
If G is a profinite group, how to show directly that the obvious map
$$G\rightarrow \underleftarrow{\lim }G/K$$
is a topological isomorphism? where K ranges over the open normal subgroups of G.
And after prove this topological isomorphism.
how to prove profinite group is the the limit of an inverse system of finite discrete groups?
Thank you for sharing your idea.
First of all, the open normal subgroups of $G$ form a fundamental system. Indeed, any compact open subgroup $K\subset G$ has finite index, so the intersection $\bigcap_{g\in G/K}gKg^{-1}$ is also of finite index, and these intersections form a fundamental system.
Let $f_K\colon H\to G/K$ be a compatible family of group homomorphisms, for an arbitrary topological group $H$ and compact open normal subgroup $K\subset G$. We want to show the existence of a homomorphism $f\colon H\to G$ such that $f_K=\pi_K\circ f$, where $\pi_K\colon G\to G/K$ is the projection. For $h\in H$, I claim the intersection of the $K$-cosets $f_K(h)\subset G$ is nonempty. Indeed, otherwise, $\{G\backslash f_K(h)\}_{K\subset G}$ is an open cover of $G$, and hence by compactness it has a finite subcovering, a contradiction.
Thus, $\bigcap_{K\subset G}f_K(h)$ must be nonempty, and since the $K$ form a fundamental system, the intersection is a single point. Define $f(h)$ to be this one point.