*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 *g *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 *g *from H using a finite index subgroup. Here’s the cartoon:

The first exercise is to show that this definition is implied by another one: that for any group element *g *that’s not in H, we can find a homomorphism * *where F is a finite group, so that the image of *H *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 *K *that contains *H *but not *g*. Since we’ve got a homomorphism, let’s use it and try . Since , we know this definition of *K *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 , and we know , and is a proper subgroup since f(g) isn’t in f(H). **This next bit is iffy and I could use help! **

- (
*Original)*Then we have a map induced by the map*f*, and the kernel of this map is*K*. By the first isomorphism theorem, the index of*K*in*G*is the size of the image of this map. Since*F/f*(*H*) is finite, the image of the map is finite. So*K*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.] - (
*based on Zach Himes’ comment)*By the first isomorphism theorem, ker*f*has finite index in F. We know ker*f*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 , then . 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. - (
*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 . Then . By the orbit-stabilizer theorem, . 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, 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 *H *is open in the profinite topology. Choose a *g *in the complement of *H*. By separability, there exists a finite index subgroup *K *that contains *H *and not *g*. Then there’s a coset *tK *of *K *which contains *g*. This coset is a basic open set, so *g *is contained in a basic open set and the complement of *H *is open.

Next, assume that *H *is closed in the profinite topology, so we want to show that *H *is separable. Again, choose some *g *in the complement of *H*. Since the complement of *H *is open, *g *is contained in a coset of a finite index subgroup*, *so that *H *is not in this coset. Let’s call this coset *K, *and call its finite index *n. *We can form a map , to the symmetric group on *n *letters, which tells us which coset each group element gets mapped to. Then *g *is in the kernel of this map, since *g *is contained in *K*, but *H *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 *g *and *H *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.