Let $G$ be an infinite group and let $H$ be a finite subgroup. Does there exist a finite index normal subgroup of $G$ which contains $H$?

Question

Let $G$ be an infinite group, and let $H$ be a finite subgroup. Must there exist some finite index proper normal subgroup of $G$ which contains $H$?

If this does not hold for general infinite groups $G$, will it hold whenever $G={\rm Gal}(\overline{K}:K)$, for some field $K$?

Background: This question was motivated by my previous question, regarding stabilizers and the action of profinite groups.

Answer 1

Let $G$ be the set of complex numbers which eventually map to $1$ under repeated squaring. Then $G$ is a group under complex multiplication.

Let $H\subsetneq G$ be a proper subgroup. As odd numbers are invertible modulo $2^n$, for $n\geq 1$, we have: $$e^{2\pi i \frac{2m+1}{2^n}}\in H\qquad\implies \qquad e^{2\pi i \frac{1}{2^n}}\in H.$$ Thus $H$ is completely characterised by the maximal $n\in \mathbb{N}$ such that $e^{2\pi i \frac{1}{2^n}}\in H$. Then $H$ is cyclic of order $2^n$.

In particular $H$ may be finite, but there are no finite index proper subgroups for it to be contained in.

Answer 2

Let $G$ be a group. The cosets of finite index subgroups form a basis for the so-called profinite topology on $G$, which makes it a topological group. Alternatively, it can be obtained as follows: let $G\to F_i$ be the collection of all homomorphisms from $G$ to finite groups (up to isomorphism). Consider the product homomorphism $G\to\prod F_i$, and pull back the product topology from $\prod F_i$.

For the profinite topology, the closure of $\{1\}$ equals the intersection of all finite index subgroup. In general, the closure of a subgroup is the intersection of all the finite index subgroups containing it, and in particular a subgroup is closed iff it is the intersection of the finite index subgroups containing it.

In particular, the profinite topology on $G$ is Hausdorff if and only if $G$ is residually finite. In this case, all finite subsets are closed, and hence every finite subgroup is the intersection of finite index subgroups containing it. (Similarly, in a profinite group, every closed subgroup is intersection of all open finite index subgroups containing it.)

This being said, a Galois group of a Galois extension is residually finite (actually profinite) so one can use the above.

However, asking to be contained in a proper normal finite index subgroup is too much, as the finite subgroup might generate the whole group as a normal subgroup. Just consider a finite group such as the symmetric group $S_n$ and the subgroup being generated by a single transposition. This can also happen in infinite profinite groups.

However, in specific absolute Galois groups one can have a positive answer. For instance, in a finite field, the absolute Galois group is torsion-free so the result is true (and essentially void). For the absolute Galois group of $\mathbf{Q}$, the only nontrivial finite subgroups have order 2, so they have a trivial image in every, say, quotient of odd order. Hence they are contained in all normal open subgroups of odd index. The same argument applies to every field of characteristic zero, provided it has at least one finite extension of odd degree $\ge 3$.