Let $F$ be a free profinite group, and let $A,B \leq F$ be finitely generated closed subgroups. Must $A \cap B$ be finitely generated?
1 Answer
The following example shows that the answer is no.
Let $p$ and $q$ be two different primes. First I want to construct two generated profinite group $G=A\ltimes H$ isomorphic to semi direct product of a infinitely generated free pro$p$ group $H$ and a cyclic free pro$q$ group $A=\langle a \rangle$.
It can be done, for example, in the following way. Start with a two generated free profinite group $F$ and a normal subgroup $N$ such that $F/N\cong A$. Let $H=N/N_p$ be the maximal prop quotient of $N$. Clearly $H$ is infinitely generated. Then $G=F/N_p\cong A\ltimes H$ satisfies the requiered conditions.
Now, let $T=<g_1,g_2>$ be a two generated free pro$q$ group. Define its action on $H$ in such way that $g_1$ and $g_2$ act as it does $a$. Form a semidirect product $T\ltimes H$ and put $G_i= <g_i,H>$ ($i=1,2$). It is clear that $G_i\cong G$ are two generated.
$T\ltimes H$ is a 3generated profinite group. Since its Sylow subgroups are free, $T\ltimes H$ has cohomological dimension 1, and so it is a subgroup of a free profinite group. However $G_1\cap G_2=H$ is not finitely generated.

$\begingroup$ Andrei, is $T$ free pro$q$ and not free pro$p$? Also, $T$ is $2$generated and not $3$generated. $\endgroup$– PabloFeb 1, 2015 at 12:02


$\begingroup$ @Pablo "$k$generated" means generated by some $k$tuple of elements. Hence 3generated implies 2generated by definition, and the trivial group is 17generated. The optimal number is encapsulated in the (generating) rank, which is by definition the least $k$ such that $G$ is $k$generated. $\endgroup$– YCorFeb 1, 2015 at 14:04