Improving a subgroup of finite index by intersecting

Quick description

Given a group $G$ and a subgroup $H$ of finite index, you can find a new subgroup of finite index with better properties by taking the intersection of $H$ with other subgroups of finite index. This gives a method of constructing finite-index subgroups that are normal or characteristic in $G$ .

Prerequisites

Elementary group theory.

General discussion

Suppose you have a subgroup $H$ of finite index in $G$ , and you're looking for a finite-index subgroup that is invariant under a certain operation—for instance, you might be looking for a subgroup that's preserved by conjugation, in other words a normal subgroup. If $H$ only has finitely many images under this operation, and each of those images also has finite index in $G$ , then their intersection will be invariant. Furthermore, it will have finite index by the following elementary lemma.

Lemma 1 If $G$ is a group and $H$ and $K$ are subgroups then $K|$ .

Proof. Consider the actions by left-translation of $G$ on the left-coset spaces $G/H$ and $G/K$ . Now consider the diagonal action of $G$ on the Cartesian product $G/H\times G/K$ . The stabilizer of $(H,K)$ is precisely $H\cap K$ , and the lemma now follows from the Orbit-Stabilizer Theorem.□

Click here for the same proof expressed in a more elementary way.
The index of a subgroup counts the number of left cosets of that subgroup, so if we want to prove that the index of $H\cap K$ is at most the index of $H$ times the index of $K,$ then a natural approach is to find an injection from the set of left cosets of $H\cap K$ to the set of pairs $(A,B),$ where $A$ is a left coset of $H$ and $B$ is a left coset of $K.$ Given a left coset $g(H\cap K),$ how can we associate with it a coset of $H$ and a coset of $K$ ? Well, $g(H\cap K)H=gH,$ and $g(H\cap K)K=gK,$ so we have a well-defined map $g(H\cap K)\mapsto(gH,gK).$ We would like it to be an injection. But if $gH=g'H$ and $gK=g'K,$ then $g^{-1}g'\in H$ and $g^{-1}g'\in K,$ which implies that $g^{-1}g'\in H\cap K,$ and therefore $g(H\cap K)=g'(H\cap K).$

As a warm up, our first example illustrates how to use this idea to construct normal subgroups.

Example 1

We prove the following fact: if $g\in G$ and there exists a subgroup $H$ of finite index in $G$ with $g\notin H$ then there is a homomorphism $\phi$ from $G$ to a finite group with $\phi(g)\neq 1$ .

If $H$ were normal then we would be done. In other words, if $H$ were invariant under conjugation then we would be done. So it's natural to consider the subgroup

$K=\bigcap_{g\in G}gHg^{-1}$

which is clearly normal in $G$ . It's easy to see that $gHg^{-1}|$ for any $g\in G$ , so this is an intersection of subgroups of finite index, but it seems that if $G$ is infinite then we have intersected infinitely many subgroups. Or have we? There's a useful rule to bear in mind here.

When conjugating a subgroup $H$ by an element $g$ , the result only depends on the coset $gH$ .

Indeed, if $kH=gH$ then $kHk^{-1}=gHg^{-1}$ , so in fact

$K=\bigcap_{gH\in G/H}gHg^{-1}$

which is a finite intersection of finite-index subgroups of $G$ and so, by Lemma 1, $K$ is of finite index in $G$ . But $K\subseteq H$ and so $g\notin K$ as required.

Note that we didn't get much control on the index of $K$ in $G$ —this proof gives a bound of $n^n$ , where $H|$ . One can do a little better than this. It's an easy exercise to see that this construction of $K$ gives the same result as the normal subgroup constructed in this Tricki article. From that point of view, it's easy to see that $K|$ divides $n!$ .

The intersection idea really comes into its own if you want your subgroup to be better than normal—characteristic, for instance. ( characteristic, for instance. A subgroup is characteristic if it is invariant under every automorphism.)

Example 2

A group $G$ is residually finite if, for every $g\in G\smallsetminus 1$ , there is a homomorphism $\phi$ from $G$ to a finite group such that $\phi(g)\neq 1$ . By Example 1, $G$ is residually finite if and only if for every $g\in G\smallsetminus 1$ there is a finite-index subgroup of $G$ that doesn't contain $g$ . We will prove the following.

Proposition 2 If $A$ and $B$ are residually finite and $A$ is finitely generated then any semidirect product $G=A\rtimes B$ is residually finite.

If $g\in G\smallsetminus A$ then the image of $g$ under the quotient map $G\to B$ is non-trivial, so because $B$ is residually finite there is a map to a finite group that doesn't kill $g$ . Therefore, we may assume that $g\in A\smallsetminus 1$ . Now, because $A$ is residually finite, there is a finite-index subgroup $K$ of $A$ such that $g\notin K$ . We would like to extend $K$ to a finite-index subgroup of $G$ . If the action of $B$ on $A$ happens to preserve $K$ , then $K\rtimes B$ is naturally a finite-index subgroup of $G$ with the required properties. So we would be done if $K$ were characteristic in $A$ . Of course, it may not be, but Example 1 leads us to believe that we might be able to improve $K$ by intersecting. Indeed, the proof of the proposition is now completed by the following lemma.

Lemma 3 Let $A$ be a finitely generated group and let $K$ be a finite-index subgroup. Then $A$ has a characteristic subgroup $L$ of finite index that is contained in $K$ .

Proof of Lemma 3. Let $K|$ and let $L$ be the intersection of every subgroup of $A$ of index $k$ . By construction, $L$ is characteristic. Because $A$ is finitely generated, it has only finitely many subgroups of index $k$ (this follows from the corresponding fact for finitely generated free groups) and so, by Lemma 1 the result follows.□