Tag Archives: algebra

## 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:

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), 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.

## Back to basics: algebra and variables (nytimes and sony)

3 Jan

I’ve had many students get tripped up over the concept of variables, which is the key idea in algebra.  Unfortunately these were calculus students, so we had to try to catch up on algebra while learning calculus (also needed to review trigonometry and geometry.  No joke, once a student asked me how to find the area of a rectangle while I was proctoring a calculus exam.  I was speechless.  I checked back later and student had figured it out.)  But it’s not just calculus students who have trouble with algebra.  I ran into this tweet last week:

Snark aside, I want to solve the problem in this article.  First, let’s go way back to basics and talk about variables.  It’s absolutely a leap to go from the concrete world of numbers to something more representational.  Instead of jumping straight into abstraction, let’s use analogies!

First, note that variable implies something that varies.  While we used to use just numbers and equations like 8+2=10, now we have expressions like 8x+2 or x*x.  The value of this expression varies depending on what value we assign to x.  This is the tricky part, because we’re used to things that stand by themselves and don’t vary depending on some other context.  OR DO WE?

Let’s take this expression:

By itself, the sentence is ambiguous.  Are you saying you are buying extremely cheap cookies in Japan or Hawaii?  Or do you feel like eating some scrumptious small baked goods?  Or am I hanging out in your kitchen helping you bake?  In this expression, “yen” could mean three different things.  And depending on what value you put in for “yen” you’ll have completely different meanings.

How can we resolve the ambiguity?  Well, we could add some context.  For instance:

1. He offered me cake, but I had a yen for cookies.
2. I couldn’t afford a real meal, but I had a yen for cookies.

By adding context we narrowed down the sentence meaning to be unambiguous.  This is like adding an = sign to an expression.  So with 8x+2, the value of the expression could’ve been 10 (if x=1), or 0 (if x=-4), or 3000 (if x=374.75), or anything really.  But if we add an = sign and have 8x+2=18, then we’ve declared that 8x+2 is unambiguously 18.  So is unambiguously $\frac{18-2}{8} = 2$.

Sometimes we still might have multiple solutions to an equation.  My sentence could be “I was in Japan, and I had a yen for cookies.”  In fact, sentences like that basically make up all the captions in this blog.

So I could have an equation like x*x=1, and we’d have two solutions: x=1 works, and so does x=-1.

That’s that for the concept!  Let’s apply some algebra to that tweet.

We have a total of $15,000,000 brought in. Some of this is from$6 digital rentals, and some is from $15 sales. And we have 2000000 transactions in total. So let’s figure out how many transactions came from rentals, and how many from sales. Then we can figure out how much money sales brought in, and how much money rentals brought in. That was a heavy paragraph, so let’s break it down. What do we want in the end? How much money came from rentals, and how much came from sales. How can we calculate how much money came from rentals? Well, we can figure out how many rentals there were, and then multiply by 6, since each rental is$6.  Similarly, we can figure out the number of sales, and multiply by 15.  So let’s figure out the number of rentals and sales, using algebra.

First, we name our variables.  Let’s use for the number of rentals occurred, and use for the number of sales.  We know that R + S = 2,000,000 because there were two million transactions.  Also, we know that 6R + 15S= 15,000,000 because that’s how much money was brought in.  To make this a little easier to read, let’s cut out the 000,000 and add it in at the end.  Our system of equations is now:

• R+S=2
• 6R+15S=15

Now let’s solve.  If we solve the first equation for R, we’ll have R=2-S.  Notice that we have a variable in our solution for R, since the value of depends on the value of S.  But we can resolve the ambiguity by plugging this equation into our second equation.  Then we have

6(2-S)+15S=15

So, distributing, we have

12-6S+15S = 15

and grouping like terms gives

9S=15-12=3

Which means that S=3/9=1/3.  Going back to that solved equation earlier, this means that R=2-S=2-1/3=5/3.

So there were 1/3 million sales and 5/3 million rentals.  Going back to our “what are we looking for” paragraph, this means that $10,000,000 came from rentals, and$5,000,000 came from sales.

Of course, you could’ve just put this into wolfram instead of breaking it down into all the pieces like we did.  This post was mostly aimed at my old math 34a students at UCSB.  Likely the people who usually read the math posts on this blog won’t be interested in this post, but maybe you can pass it on to someone else!