Tricki

## Improving a subgroup of finite index by intersecting

### Quick description

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

### Prerequisites

Elementary group theory.

### General discussion

Suppose you have a subgroup of finite index in , 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 only has finitely many images under this operation, and each of those images also has finite index in , then their intersection will be invariant. Furthermore, it will have finite index by the following elementary lemma.

Lemma 1 If is a group and and are subgroups then .

Proof. Consider the actions by left-translation of on the left-coset spaces and . Now consider the diagonal action of on the Cartesian product . The stabilizer of is precisely , 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 is at most the index of times the index of then a natural approach is to find an injection from the set of left cosets of to the set of pairs where is a left coset of and is a left coset of Given a left coset how can we associate with it a coset of and a coset of ? Well, and so we have a well-defined map We would like it to be an injection. But if and then and which implies that and therefore

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 and there exists a subgroup of finite index in with then there is a homomorphism from to a finite group with .

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

which is clearly normal in . It's easy to see that for any , so this is an intersection of subgroups of finite index, but it seems that if 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 by an element , the result only depends on the coset .

Indeed, if then , so in fact

which is a finite intersection of finite-index subgroups of and so, by Lemma 1, is of finite index in . But and so as required.

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

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 is residually finite if, for every , there is a homomorphism from to a finite group such that . By Example 1, is residually finite if and only if for every there is a finite-index subgroup of that doesn't contain . We will prove the following.

Proposition 2 If and are residually finite and is finitely generated then any semidirect product is residually finite.

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

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

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