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Improving a subgroup of finite index by intersecting

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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.


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 Gthis 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.


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