Subgroup separability problem set session(non-elementary)

22 Jun

Update #1, five hours later: Zach Himes posted a great comment fixing my iffy part, and I also video chatted with Nick Cahill about it and we came up with an alternate solution which is still maybe iffy.  Adding both below.  Nick also added a comment about the second exercise, using group actions.  What do you think?

I’ve read Sageev’s PCMI lecture notes several times by this point; it’s the basis of my and many other people’s work (in particular, my friend Kasia has an impressive number of publications related to this stuff).  And every single time I get stumped on the same exercise near the end, so I thought I’d try to write up a solution, crowd-source it among my blog readers, and figure out something correct.  For reference, these are exercise 4.27 and 4.28 in his notes, but I’ll lay out the problems so you don’t need to look them up if you don’t want to.  Please comment with corrections/ideas!

A thing that mathematicians care about is the structure of a group.  We say that a particular subgroup H<G is separable if for any group element that’s not in H, we can find a finite index subgroup K that contains H but doesn’t contain g.  Intuitively, we can separate from H using a finite index subgroup.  Here’s the cartoon:

output_37OpDk

If H is separable, then given any g not in it, we can find a finite index subgroup that separates H from g.

The first exercise is to show that this definition is implied by another one: that for any group element that’s not in H, we can find a homomorphism f: G\to F where F is a finite group, so that the image of under the map doesn’t contain f(g).

So let’s say we start with such a homomorphism, and our goal is to find a finite index subgroup that contains but not g.  Since we’ve got a homomorphism, let’s use it and try K:=f^{-1}(f(H)).  Since f(g)\not\in f(H), we know this definition of excludes g, as desired.  Then we need to show that K is finite index in G and we’ll be done.

What about the first isomorphism theorem?  We have a map G\to F, and we know f(H)<F, and is a proper subgroup since f(g) isn’t in f(H).  This next bit is iffy and I could use help!  

  1. (Original) Then we have a map G\to F/f(H) induced by the map f, and the kernel of this map is K.  By the first isomorphism theorem, the index of in is the size of the image of this map.  Since F/f(H) is finite, the image of the map is finite.  So has finite index in G, as desired.  [What’s iffy here?  You can’t take quotients with random subgroups, just with normal subgroups, and I don’t see why f(H) would be normal in F unless there’s something I don’t know about finite groups.]
  2. (based on Zach Himes’ comment) By the first isomorphism theorem, ker has finite index in F.  We know ker is contained in K, since 1 is contained in f(H) [since 1 is contained in H, and f(1)=1, where 1 indicates the identity elements of G and F].  It’s a fact that if \ker f \leq K \leq G, then [G: \ker f] = [G:K][K: \ker f].  Since the left hand side is finite, the right hand side is also finite, which means that K has finite index in G, as desired.
  3. (Based on conversation with Nick Cahill) We can think of F/f(H) as a set which is not necessarily a group, and say that G acts on this set by (g, x) \mapsto f(g)x.  Then K=Stab(1):=Stab(f(H)).  By the orbit-stabilizer theorem, [G:K] = |Orb(1)|.  Since F is finite, the size of the orbit must be finite, so K has finite index in G, as desired.

The second exercise has to do with the profinite topology.  Basic open sets in the profinite topology of a group are finite index subgroups and their cosets.  For instance, in the integers, 2\mathbb{Z}, 2\mathbb{Z}+1 are both open sets in the profinite topology.  Being closed in the profinite topology is equivalent to being a separable subgroup (this is the second exercise).

So we have to do both directions.  First, assume we have a separable subgroup H.  We want to show that the complement of is open in the profinite topology.  Choose a in the complement of H.  By separability, there exists a finite index subgroup that contains and not g.  Then there’s a coset tK of which contains g.  This coset is a basic open set, so is contained in a basic open set and the complement of is open.

Next, assume that is closed in the profinite topology, so we want to show that is separable.  Again, choose some in the complement of H. Since the complement of is open, is contained in a coset of a finite index subgroup, so that is not in this coset.  Let’s call this coset K, and call its finite index n.  We can form a map f: G\to S_n, to the symmetric group on letters, which tells us which coset each group element gets mapped to.  Then is in the kernel of this map, since is contained in K, but is not in the kernel of f since it is not in that coset.  In fact no element of H is in the kernel.  So we’ve made a homomorphism to a finite group so that and have disjoint images, which we said implies separability by the previous exercise.

Okay math friends, help me out so I can help out my summernar!  So far in summernar we’ve read these lectures by Sageev and some parts of the Primer on Mapping Class Groups, and I’m curious what people will bring out next.

Advertisements

5 Responses to “Subgroup separability problem set session(non-elementary)”

  1. Zach Himes June 22, 2016 at 1:20 pm #

    I think I have a proof for the first problem, but I’m not sure (I’m sorry I don’t know how to present the math more cleanly).
    Let J be the kernel of the group homomorphism f:G->F and let K be the preimage of f(H). Since J is a normal subgroup of G, K is contained in the normalizer of J. Therefore, JK is a subgroup of G (theorem 18 in section 3.3 of Dummit and Foote) that contains H. Since the index of J in G is finite (because it is equal to the size of G in F), and [G:J]=[G:JK][JK:J] (problem 11 of section 3.2 in D&F), we must have that [G:JK] is finite. Finally, consider an element g not contained in H and suppose that g=jk for some j in J and some k in K. Then f(g)=f(jk)=f(j)f(k)=f(k). But f(k) is in f(H) and by assumption f(g) is not contained in f(H).

    • yenergy June 22, 2016 at 2:38 pm #

      Testing if latex works in comments: JK=K because J is preimage of 1 and K is preimage of a subgroup. But rest of argument follows anyway! So great!

      • yenergy June 22, 2016 at 2:39 pm #

        Cool! So Zach, to do latex in comments you write $$ around what you want to TeX, and the word “latex” after the first $ sign. So $ latex f:G\to H $ ends up as f:G\to H once you take out the spaces.

  2. Nick June 22, 2016 at 3:39 pm #

    For Closed implies Separable:

    H is a subgroup of G, closed in the profinite topology, g not in H. Since G\H is open, there is a basic open set in G\H that contains g, and we’ll say it’s a coset of the finite index subgroup K, all by definition so far.

    There are finitely many cosets of K in G, and H is contained in some subset of them that doesn’t contain gK. So let’s say H is covered by {eK, aK, bK, cK} for the sake of explanation, and they’ve all actually got an element of H in them. Since cosets are always disjoint these are the only cosets that contain any elements of H.

    Now we look at G/K and look at what H can do under this action. If I look at where h in H moves one of the cosets covering H, aK for instance, then aK contains an element of H, so haK also contains an element of H. Therefore it has to be one of the {eK, aK, bK, cK}. This means that the image of H in Aut(G/K) lies in the subgroup that preserves the subset {eK, aK, bK, cK}. Since g takes K to gK, and gK is disjoint from H, it isn’t one of the {eK, aK, bK, cK}, so it doesn’t belong to the subgroup that preserves that set. Now we can use the previous exercise on im(H) in Aut(G/K)

  3. Nick June 23, 2016 at 7:37 am #

    Here’s a little more detail for my argument for that first question:

    Setup: H a subgroup of G, g not contained in H, F a finite group, and f a homomorphism from G to F such that f(g) is not contained in f(H)

    F acts on the coset space F/f(H), which is finite and has at least two elements, since f(H) and f(g)f(H) are different cosets by our assumptions. This action is really a homomorphism from F to Aut(F/f(H)), so we can compose it with our homomorphism from G to F to get a homomorphism from G to Aut(F/f(H)), i.e. an action of G on F/f(H).

    So now we have an action of G on F/f(H), and I claim stab(f(H)) is the subgroup we are looking for. H is contained in stab(f(H)), since h in H moves f(H) to f(h)f(H) = f(hH) = f(H). g is not in stab(f(H)), since it moves f(H) to f(g)f(H) which we assumed is not the same coset as f(H). It has finite index by the Orbit-Stabilizer theorem; [G: stab(f(H))] = |Orb(f(H))|, and Orb(f(H)) has at least two distinct elements (f(H) and f(g)f(H)) so this is our proper finite index supgroup.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: