Improving a subgroup of finite index by intersecting

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Improving a subgroup of finite index by intersecting

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[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 indexintersection] If $G$ is a group and $H$ and $K$ are subgroups then $|G:H\cap K|\leq |G:H||G:K|$.[/lemma]
[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.
[/proof]

[add] Click here for the same proof expressed in a more elementary way. {{
<br />
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 $g(H\cap K)=g'(H\cap K).$ [/add]

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

[EXAMPLE normal]

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 $|G:H|=|G:gHg^{-1}|$ 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.

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

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 [ref Lemma #indexintersection], $K$ is of finite index in $G$.  But $K\subseteq H$ and so $g\notin K$ as required.

[/EXAMPLE]

Note that we didn't get much control on the index of $K$ in $G$---this proof gives a bound of $n^n$, where $n=|G:H|$.  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 [[A way of getting proper normal subgroups of small index|this]] Tricki article.  From that point of view, it's easy to see that $|G:K|$ divides $n!$.

The intersection idea really comes into its own if you want your subgroup to be better than normal---[cut]characteristic, for instance.|| characteristic, for instance.  A subgroup is ''characteristic'' if it is invariant under every automorphism.[/cut]

[EXAMPLE characteristic]
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 [ref Example #normal], $G$ 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 semidirectrf]If $A$ and $B$ are residually finite and $A$ is finitely generated then any [[w:Semidirect_product|semidirect product]] $G=A\rtimes B$ is residually finite.[/proposition]

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 [ref Example #normal] 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 charlemma]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 $K$.[/lemma]
[proof charlemma] Let $k=|A:K|$ 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 [ref Lemma #indexintersection] the result follows.
[/proof]
[/EXAMPLE]

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