I am reading the article "The structure of approximate groups" by Breuillard, Green and Tao.
At some point they state the following theorem 1.6:
Let A be a K-approximate subgroup of a group G. Then there exists a subgroup G_0 of G and a finite normal subgroup H of G_0 such that
- G_0/H is nilpotent
- A can be covered by O_K(1) left translates of G
- A^4 contains H and a generating set of G_0.
The first condition says that G_0 is finite-by-nilpotent. Immediately after this theorem, the authors claim that G_0 is virtually nilpotent (nilpotent-by-finite) as well.
"Indeed, the stabilizer in G_0 of the conjugation action on H has finite index in G_0 and is a central extension of a finite index subgroup of G_0/H, and therefore is nilpotent."
The problem is that I don't understand this sentence.
I tried to make sense out of it, and this is what I found:
Denote by S the set of all subgroups of G_0, then G_0 acts on S by conjugation, i.e. g.A=gAg^{-1}. The stabilizer Stab(H) of H in G_0 under this action clearly is the normalizer of H in G_0. So Stab(H)=N_{G_0}(H). Since H is normal in G_0, this normalizer simply is G_0 itself. Then the index [G_0:N_{G_0}(H)]=1.
This seems a bit weird, I find that the finite index is always 1. Am I doing something wrong? And moreover, why is this stabilizer an central extension of a finite index subgroup of G_0/H?
Any help would be very wellcome, thank you in advance.
You are looking at the wrong action. Take the finite index subgroup $G_1<G_0$ whose action by conjugation on $H$ is by the identity map $G_1\to Aut(H)$. Then $H_1:=H\cap G_1$ is contained in the center of $G_1$; $G_1/H_1$ embeds in $G_0/H$ and, hence, nilpotent. Thus, $G_1$ is nilpotent too.