Let $G$ be f.g. with $H\le G$ s.t. $[G:H]<\infty$. Then $\exists K\le H$ with $K\sqsubseteq G$ and $[G:K]<\infty$.

Question

This is Exercise 4.29 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to Approach0, it is new to MSE.

The Details:

Definition: A subgroup $H\le G$ is characteristic in $G$, written $H\sqsubseteq G$, when, for all $\sigma\in {\rm Aut}(G)$, we have $\sigma(H)=H$ (or, equivalently, $\sigma(H)\le H$).

The Question:

Let $G$ be a finitely-generated group with $H\le G$ such that $[G:H]<\infty$. Then there exists $K\le H$ characteristic in $G$ with $[G:K]<\infty$.

Thoughts:

It is well-known that any finite index subgroup of a finitely-generated group is itself finitely-generated. Hence $H$ is finitely-generated. However, I don't recall this result in the book so far.

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It might help to note that

$$[G:K]=[G:H][H:K]$$

as cardinal numbers, once we have found a candidate for $K$. This is proven early on in the book.

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A special case is when $K=H$. All we need to prove there is that $K$ is characteristic in $G$. There's nothing to say, though, that equality is always possible.

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I have asked a question here on characteristic subgroups before: An abelian, characteristically simple group is divisible (supposedly).

I have a few years of experience with combinatorial group theory, so, given that we're talking about finitely-generated groups, I feel as if I should be able to answer this; I guess my main difficulty is the property of being characteristic.

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Please help :)

Answer 1

Given a subgroup $H$ of a group $G$ there is a unique subgroup $K$ of $H$ which is maximal with respect to the property of being characteristic in $G$: namely, it's the subgroup given by the intersection

$$K = \bigcap_{\varphi \in \text{Aut}(G)} \varphi(H)$$

of the images of $H$ under all automorphisms of $G$. So there exists a finite index characteristic subgroup iff $K$ is finite index.

As a warmup, here's an easier problem: it's a classic exercise to show that if $H$ is finite index in $G$ then there exists a subgroup $H'$ of $H$ which is normal and finite index in $G$. (We don't need $G$ to be finitely generated here.) The proof is very short; the maximal such subgroup is the intersection $\bigcap_{g \in G} gHg^{-1}$ of the conjugates of $H$, and this subgroup is exactly the kernel of the action of $G$ on the cosets $G/H$, which is finite. We get the more precise statement that if $H$ has index $n$ then $H'$ can be chosen to have index dividing $n!$, and setting $G = S_n, H = S_{n-1}$ shows that this bound is tight.

We can try to imitate this construction. Instead of just considering the action of $G$ on $G/H$ let's consider the action of $G$ on $G/\varphi(H)$ for every automorphism $\varphi$. We are done if we can show that this construction produces finitely many isomorphism classes of $G$-sets. Here is where we need the hypothesis that $G$ is finitely generated: we can actually show more, namely

Proposition: A finitely generated group $G$ has only finitely many isomorphism classes of actions on a finite set of size $n$.

Proof. There are finitely many homomorphisms $G \to S_n$. $\Box$

Translated back into subgroups this implies that $G$ has finitely many subgroups of a fixed finite index $n$, and since the subgroups $\varphi(H)$ all have the same index, in particular there are finitely many of these. So $K$ is the kernel of the action of $G$ on

$$\bigsqcup_{\varphi \in \text{Aut}(G)} G/\varphi(H)$$

which is finite, and we are done. More explicitly, if $G$ is generated by $r$ generators and $H$ has index $n$ then there are at most $n!^r$ homomorphisms $G \to S_n$, so the above set has size at most $n \cdot n!^r$, and $K$ has index dividing $(n \cdot n!^r)!$. Probably not best possible, but it works.