When is the semidirect product of profinite groups a profinite group? - MathOverflow

When is the semidirect product of profinite groups a profinite group?

2012-06-04T16:02:48Z

Following the discussion I have with Yves Cornulier in the following question http://mathoverflow.net/questions/98696/finiteness-theorems-for-profinite-groups, I would like to ask the following: Suppose $K$ and $N$ are two profinite groups and $K$ acts on $N$. Suppose further that each element of $K$ acts continuously on $N$. We can form the semidirect product $G=N \rtimes K$. If $N$ is characteristc based, that is $N$ has a base for its topology at the identity made of open characteristic subgroups, then $G$ has a structure of a profinite group such that the induce topology on $N$ and $K$ as subgroups is their original topology. In particular, this is the case if $N$ is finitely generated.

Could you give an example where $G$ does not have such a structure? More specifically, could you give an example where $K$ and $N$ are pro-$p$ groups, but $[N,K]=N$?

Answer by Will Sawin for When is the semidirect product of profinite groups a profinite group?

2012-06-04T16:28:43Z

Let $K$ be the $l$-adic numbers $\mathbb Z_l$. Let $N$ be the product of uncountably many copies of $\mathbb Z/p$. These are both profinite groups. $K$ acts on $N$ through a simple transitive action on the set of copies of $\mathbb Z/p$. This action is continuous for each element of $K$, but the total action is not a continuous map from $K \times N$ to $N$.

If $G$ were a profinite group, then the commutator action of $K$ on $N$, which can be written in terms of group operation, would have to be continuous.

If we set $l=p$ then $K$ and $N$ are pro-$p$ groups and, indeed, $[N,K]=N$. In fact the commutators of elements of $N$ with any single non-identity element of $K$ generate $N$.

Answer by Benjamin Steinberg for When is the semidirect product of profinite groups a profinite group?

2012-06-04T16:31:01Z

This was going to be a comment but became too long.

If the action mapping $K\times N\to N$ is continuous, then the semidirect product is profinite. You can find this in the book of Ribes and Zalesskii. Do you really want just that each element of $K$ acts continuously on $N$? I believe that if you just ask that $K\times N\to N$ be separately continuous (so each element of $K$ acts continuously on $N$ and also if you fix $n\in N$, then the map $k\mapsto kn$ is continuous from $K$ to $N$), then you will have that $N\rtimes K$ is semitopological and compact (that is, left and right translations are each continuous). Then by Ellis's theorem, you will get joint continuity for free and so $N\rtimes K$ will be profinite. If you do not ask the action map to be separately continuous, you may have troubles although I don't have an example off the top of head.

I think for profinite semigroups it would be easy to create a counterexample.

Added: The semidirect product $N\rtimes K$ is profinite iff $N$ has a basis of open normal subgroups which are $K$-invariant. Indeed, if $N\rtimes K$ is profinite, then the action of $K$ on $N$ is equivalent to conjugation, which is a jointly continuous action $K\times N\to N$. The corresponding map $K\to Aut(N)$ has compact image in the compact-open topology (since $K$ is compact and the map is continuous) and so the image of $K$ is equicontinuous with respect to the uniformity of $N$ (which has as a fundamental system the open normal subgroups) by Arzela-Ascoli. This equicontinuity is equivalent to $N$ having a fundamental system of $K$-invariant open subgroups.

Conversely, such a fundamental system of neighborhoods of 1 exist, then the action, the $K$ is an equicontinuous family of automorphisms of $N$ and so $K\to Aut(N)$ is continuous in the compact-open topology and hence the action $K\times N\to N$ is continuous. This gives that the semidirect product is a profinite group by a well known result that can be found in Ribes and Zalesskii.