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

Following the discussion I have with Yves Cornulier in the following question 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 $[N,K]=N$?

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could you clarify what you mean by "each element acts continuously"? Do you mean that the action of $K$ endowed with the discrete topology, on $N$, is continuous (i.e. $K$ maps to the bicontinuous automorphisms of $N$)? (Then Will Sawin's example works) Or do you mean something stronger, such as: "every topology cyclic subgroup of $G$ acts continuously"? –  YCor Jun 4 '12 at 18:07
I meant that for each fixed $k \in K$ the map $n \mapsto n^k$ is a continuous map on $N$. –  Yiftach Barnea Jun 4 '12 at 18:48
then Will's example answers your question. –  YCor Jun 4 '12 at 19:11
@Yves: there are various ways to set a topology on a non-finitely generated profinite group. Without a clarification what the topology is in Will's example I cannot understand why it is continuous. –  Yiftach Barnea Jun 4 '12 at 19:17
He considers a product of cyclic groups; he obviously uses the product topology. –  YCor Jun 4 '12 at 19:26

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 $N$.

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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.

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I just looked it up and Ellis in fact proved any separately continuous action of a locally compact groups is automatically jointly continuous. So that is the minimum extra assumption you need. –  Benjamin Steinberg Jun 4 '12 at 16:39
@Benjamin: thanks for pointing out Ribes and Zalesskii reference. I suspected this the case, but did not really think about it. I assume separately contnuous means continuous in each coordinate when you fix an element in the other coordinate. Is Ellis's theorem just what you stated in the comment? Indeed I want each element of $K$ to act continuously, otherwise I think the question won't be very interesting. –  Yiftach Barnea Jun 4 '12 at 17:19
There are two Ellis theorems that are related and relevant to your question. The first theorem says that if $G$ is a group with a locally compact topology such that inversion is continuous and multiplication is separately continuous (in the sense of your comment) then the multiplication is jointly continuous and so G is a topological group. He also proved that if a locally compact group $G$ acts on a locally compact space $X$ such that the action is separately continuous, then the action is jointly continuous. –  Benjamin Steinberg Jun 4 '12 at 17:42
I believe that if merely assume that $N$ has a basis of open normal subgroups invariant under the action of $K$ and each element of $K$ is continuous, then again the semidirect product if profinite. This condition says precisely that $K$ is an equicontinuous family for the natural uniformity on $N$ and this makes the compact-open topology equal pointwise convergence and so I think everything is ok. –  Benjamin Steinberg Jun 4 '12 at 17:49
@Benjamin: of course you are right in your second comment. The characteristic based condition implies this independantely of the action of $K$. –  Yiftach Barnea Jun 4 '12 at 18:51