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I have some general questions around the profinite topology on a group $G$.

On the page http://groupprops.subwiki.org/wiki/Profinite_topology one can read, that

The profinite topology on a group is a topology on the underlying set of the group defined in the following equivalent ways:

1. It has as a basis of open subsets all left cosets of subgroups of finite index.


2. It has as a basis of open subsets all right cosets of subgroups of finite index.


3. It has as a basis of open subsets all cosets of normal subgroups of finite index.

Under the profinite topology, any group becomes a topological group.

1. I always hear, that there is ONE profinite topology on a group. But every set of subgroups of finite index, which is closed under taking intersections gives us a "profinite" topology for a group. But then maybe some elements of the above basiselements aren't open sets in this topology. So are these topologies also called profinite, although they aren't the above defined one?




2. If we have a profinite group. How is the connection between the "profinite topology" and the definition of a profinite group.
   If we got a "profinite" topology on $G$. What about the profinite completion $G'$. How can I find a basis of open subsets in $G'$. How is the connection between a basis of $G$ and a basis of $G'$?

Thanks for help.

A good reference for both questions is Chapter 3 of Profinite groups by Ribes and Zalesskii.

Question 1

It is reasonable to say that there are many different "profinite topologies" on a group $G$. Section 3.1 of the above reference does the following. Let $G$ be a group and let $mathcal{N}$ be a collection of finite-index normal subgroups of $G$, which is filtered from below, i.e., for any $N_1,N_2inmathcal{N}$ there is some $Ninmathcal{N}$ such that $Nleq N_1cap N_2$. Then one defines a topology on $G$ whose base is the collection of cosets of all elements of $mathcal{N}$. Such a topology is called "a profinite topology" on $G$, and I'll call it the $mathcal{N}$-profinite topology on $G$.

The special case when $mathcal{N}$ is all (normal) finite-index subgroups of $G$ is usually referred to as "the" profinite topology on $G$. This terminology is justified by the following equivalent perspective.

Let $mathcal{C}$ be a collection of finite groups, which is closed under isomorphism, quotients, and finite subdirect products (Ribes and Zalesskii call this a formation of finite groups). Now let $G$ be any group, and let $mathcal{N}$ be as before. If $mathcal{N}subseteqmathcal{C}$ then the $mathcal{N}$-profinite topology on $G$ is called "a pro-$mathcal{C}$ topology on$G$". On the other hand, if $mathcal{N}$ is precisely the collection of finite-index normal subgroups $N$ of $G$ such that $G/N$ is in $mathcal{C}$, then the resulting topology is called "the pro-$mathcal{C}$ topology".

If $mathcal{C}$ is the class of all finite groups, then the pro-$mathcal{C}$ topology is the the profinite topology. Other well-studied examples are when $mathcal{C}$ is the class of finite $p$-groups, for some fixed prime $p$. In this case, the pro-$mathcal{C}$ topology is also called the pro-$p$ topology.

Question 2

Section 3.2 of the above reference deals with this.
Given a group $G$ and a collection $mathcal{N}$ of finite-index normal subgroups filtered from below, one can construct the profinite completion of $G$ with respect to $mathcal{N}$, which I will denote $G'_{mathcal{N}}$. In particular,
$$
G'_{mathcal{N}}=varprojlim_{Ninmathcal{N}}G/N,
$$
which is a profinite topological group under the "sub-product" topology (i.e., the subspace topology induced by the product topology on $prod_{Ninmathcal{N}}G/N$).

The special case when $mathcal{N}$ is all normal finite-index subgroups of $G$ yields the usual profinite completion $G'$.

Note that we have a canonical homomorphism $tau_{mathcal{N}}colon Gto G'_{mathcal{N}}$ which sends $g$ to $(gN)_{Ninmathcal{N}}$. The connection to profinite topologies on $G$ is as follows. The $mathcal{N}$-profinite topology on $G$ is precisely the coarsest topology on $G$ such that $tau_{mathcal{N}}$ is continuous with respect to the sub-product topology on $G'_{mathcal{N}}$. Dually, the sub-product topology on $G'_{mathcal{N}}$ is precisely the finest topology on $G'_{mathcal{N}}$ such that $tau_{mathcal{N}}$ is continuous with respect to the $mathcal{N}$-profinite topology on $G$. (See this link.) From this, one can compute relationships between bases for the two topologies, etc.

Final Comment

Despite all of this, the terminology "profinite topology on $G$" is still possibly problematic for the following reason. Given any group $G$, the profinite completion $G'$ is a profinite topological group. On the other hand, while $G$ is a topological group under the profinite topology, it is not necessarily a profinite space (i.e., compact, Hausdorff, totally disconnected). For example, the profinite topology on $mathbb{Z}$ is Hausdorff but not compact, since any countable compact Hausdorff group must be finite (see this question; the case of $mathbb{Z}$ is also dealt with more explicitly here). The profinite topology on an infinite simple group is trivial, and so not Hausdorff or totally disconnected.

Edit: To put a finer point on the previous comment, here is a quote from Profinite and residually finite groups by B. Hartley: "A topology on a group $G$ determined by a set $mathcal{N}$ of normal subgroups of finite-index [satisfying certain properties] will be called a cofinite topology. Such topologies have been called profinite topologies by a number of authors; however we prefer to reserve that terminology for the situation when the resulting group is actually profinite."