6
$\begingroup$

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?

$\endgroup$

1 Answer 1

7
$\begingroup$

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 pro-p 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 3-generated 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.

$\endgroup$
3
  • $\begingroup$ Andrei, is $T$ free pro-$q$ and not free pro-$p$? Also, $T$ is $2$-generated and not $3$-generated. $\endgroup$
    – Pablo
    Feb 1, 2015 at 12:02
  • $\begingroup$ @Pablo I have edited the answer. $\endgroup$ Feb 1, 2015 at 13:05
  • $\begingroup$ @Pablo "$k$-generated" means generated by some $k$-tuple of elements. Hence 3-generated implies 2-generated by definition, and the trivial group is 17-generated. The optimal number is encapsulated in the (generating) rank, which is by definition the least $k$ such that $G$ is $k$-generated. $\endgroup$
    – YCor
    Feb 1, 2015 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.