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

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.

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.

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.

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.

• This reminds me of the result that a finite index subgroup always contains a finite index normal subgroup. Perhaps you can show that the number of distinct images of $H$ under the action of $\mathrm{Aut}(G)$ is finite (here is, presumably, where finite generation would come into play) and then take their intersection? That intersection is certainly characteristic. Commented Dec 29, 2020 at 18:55

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.

• I'm sorry, but I'm missing something: I grant that there are only finitely many isomorphism classes of $G$-sets of order $n$; but how does imply that there are only finitely many subgroups of index $n$? Why can't distinct subgroups yield isomorphic $G$-sets? Commented Dec 29, 2020 at 23:22
• @Shaun: Okay, I think this works: we may restrict to normal subgroups, by the usual result. Now suppose $H_1$ and $H_2$ are normal subgroups of finite index in $G$, and suppose $f\colon G/H_1\to G/H_2$ is an isomorphism of the $G$-sets under the usual action. Then for all $h\in H_1$, $f(H_1) = f(hH_1) = hf(H_1)$, so $H_1$ fixes $f(H_1)$ in the action on $G/H_2$. But stabilizers of a point in $G/H_2$ are conjugates of $H_2$, which by assumption are all equal to $H_2$, so $H_1=H_2$. Commented Dec 29, 2020 at 23:31
• @Arturo: a subgroup of index $n$ is the same thing as a pair consisting of a transitive $G$-set of order $n$ (there are finitely many isomorphism classes of these) and a choice of basepoint in this $G$-set (whose stabilizer is the desired subgroup), and there are $n$ choices of basepoint. This argument can be used to write down a generating function relating the number of homomorphisms $G \to S_n$ and the number of subgroups of index $n$, given here: mathoverflow.net/questions/376175/… Commented Dec 30, 2020 at 2:25
• (I should clarify what I mean by "the same thing as." I mean the groupoid of pointed transitive $G$-sets of size $n$ is equivalent to the (discrete) groupoid of subgroups of index $n$. I don't mean that distinct points have distinct stabilizers; what I do mean is that points have the same stabilizer iff there's an automorphism of the $G$-set exchanging them.) Commented Dec 30, 2020 at 7:39