Following the discussion I have with Yves Cornulier in the following question [Finiteness theorems for profinite groups](https://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 $[N,K]=N$? 



When is the semidirect product of profinite groups a profinite group?
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 $N$.


