0
$\begingroup$

Let $G$ be a group, and $H$ a subgroup of $G$.

Is it possible to "see" from the profinite completions of $H$ and $G$ that $H$ has finite index in $G$?

Naively, does $H$ have finite index in $G$ iff the profinite completion of $H$ has finite index in the profinite completion of $G$? Certainly not...

$\endgroup$

1 Answer 1

9
$\begingroup$

If $G$ is a group and $H$ is a finite index subgroup a result of Marshall Hall says that the profinite topology on $H$ is induced from the profinite topology on $G$. Thus the profinite completion of $H$ is the closure of $H$ in the profinite completion of $G$. Moreover, the index of the closure of $H$ in the profinite completion of $G$ is $[G:H]$. All this can be found in the book of Ribes and Zalesskii.

Note that it can happen that $H$ inherits its profinite topology from that of $G$ without being finite index, for example this happens for every finitely generated subgroup of a finitely generated free group. The closure of $H$ in the profinite completion of $G$ will then be the profinite completion of $H$ and will not have finite index.

In general the inclusion of $H$ in $G$ will not induce an inclusion of profinite completions and it is possible for a subgroup of a group to be dense in the profinite topology and be infinite index. For example let $G$ be an infinite simple group. Then the profinite completion of $G$ is trivial and so the trivial subgroup is dense.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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