Bad, bad Polymath for not posting last week. But...last weekend was our fall break, so these last two weeks have only comprised 8 school days, and I was very busy all week and weekend, so I beg your indulgence and forgiveness.

Three things happened this past week that were very pleasant:

- A student who has been working pretty hard, but getting D's* on his tests got an A– on his recent test. Of course the grade itself made me happy, but the best part was the expression on his face as he turned it in. He didn't need me to grade it to know he had done well. He was almost able to keep a straight face, but he cracked a little smile and said "that went a lot better." After he left the room I skimmed his test, and it was clear he was right.
- The principal was talking with some students (she didn't say which) about how things were going with them, and one of them mentioned having me for math, adding "I love him". I know that my classes are basically going pretty well this year, but it's nice to know that at least one student likes coming to class every day. It's also nice to know that my principal is the kind of person who will tell me things like this. Although I did fail to come up with the obvious reply: "Oh, I guess my well-placed bribe worked perfectly, then!"
- It's only once every year or two that I'm just about to get annoyed with a kid for doing something disruptive, but then it turns out to be so funny that I can't be annoyed. Like when I was in a high school choir, and we were rehearsing a requiem. The words were "
*et de profundo*" (or 'out of the depths'), and we all cracked up, and the conductor was mad at us until he saw what we were laughing at: a large, sweaty guy walking behind the conductor across the stage wheeling a Porta-Potty on a dolly behind him. He had to admit that was funny. So anyway, I was telling my algebra students my well-rehearsed opinion about classic rate problems (like "a train leaves Chicago going 60 miles per hour..."). These are the kinds of problems that so many adults cite as evidence of how stupid high school math is, because they're so useless. To the surprise of my students, I actually agree with that. So I tell them that I'll give them a real problem that's not stupid: "Design a new kind of airplane that won't crash and kill everyone on board." They become silent. "Too hard a problem?" They nod. "Well let's start with something easier...a train leaves Chicago going 60 miles per hour...." That usually gets a laugh and a recognition that you have to start with easy mathematical modeling problems before you try the hard ones. And I proceed with an example. And about halfway through the example, one of my quieter and more talented students starts folding up a piece of paper. I let it go for a minute, but it soon becomes clear he's folding a paper airplane. I get a little bit annoyed. "Are you folding a*paper airplane*?" And he comes back with "I'm designing a new kind of plane that won't crash and kill people."

I also concluded that a *huge* difference between 7th graders who "get" math and ones that don't is their grasp of how division works. Using divisibility tests (not in this kind of detail, though I teach the ones for 7 and 11), finding prime factorizations, and understanding the details of remainders (that is, the basics of modular arithmetic) are completely unnatural for some students, and those intuitions are extremely hard to teach. A student who can fly through the long division problem for 37,574÷7 and correctly get 5367-R5 (remainder of 5), can still not tell me very quickly what 5367 times 7 is afterwards. When asked to find prime factorizations of 24 and 48 on the same page, it doesn't occur to them to use the factorization for 24 and just add one more two. It's a revelation to them when they finally understand that if a number is divisible by 15 then it *must* be divisible by 3 as well. I can only conclude that this is a more abstract idea than it seems at first, and/or that grade school teachers don't really spend time on what division really means, but just on the long division algorithm.

And this is the paragraph where I'm supposed to wrap it up with a nice observation on how all this applies to the world at large. But it's not coming this week, so I'll just leave it at that.

*I am aware that plurals don't require an apostrophe, but *B+s* just looks so wrong compared to *B+'s* that I just have to use one when pluralizing letter grades. So sue me.

My feeling is that the division/factorization issue represents some sort of basic awareness of numbers as "entities", rather than just digit strings.

My own awareness of numbers is synaesthetic, they seem to have shapes which help me see how they "break apart" into their factors. Sometimes the shapes reflect the sub-factors, or the square or triangular "status" (some numbers can take more than one shape), but they're not always something I could draw or shape from clay.

Plurals: I just had the same issue with pluralizing the phrase "gentleman's C" someplace else. I think the standard rules of English don't quite cover using letters as symbols, but the real point is readability, so I'd do the same as you did.

Posted by: David Harmon | October 28, 2006 at 11:09 PM

The apostrophe use is fine, since you're ducking out of the English Language for a moment to describe something that it doesn't really cover, it's OK to defy convention.

I reckon that the whole "numbers as entities" thing is a bit of a problem in the way that maths is taught but I for one, couldn't think of an alternative way of teaching it to little ones.

Posted by: Paul Carpenter | October 29, 2006 at 05:36 AM

Two things. First, regarding the apostrophe, I was specifically taught that one of the proper uses of the apostrophe was pluralizing numbers and symbols. So in fact (at least according to *my* junior high grammar text) it would have been wrong *not* to include it. Secondly, regarding division. It seems to me that this is an interesting parallel to the fact that understanding the construction of quotient groups (or rings or whatever) is a sort of "litmus test" for understanding abstract algebra... Perhaps there is something fundamental going on here on a cognitive psychology level? Intuitively, I agree with David that the problem probably lies with the ability to see numbers as things-in-themselves. (And,in the abstract algebra case, with seeing groups, etc. as things-in-themselves.)

Posted by: Cat lover | October 29, 2006 at 12:10 PM

Interesting that you would bring up quotient groups in this context. I never really made that connection myself, but that probably was one of the abstractions that brought me to the conclusion that I wasn't cut out to be a mathematician. I can follow the definitions of quotient groups, but I don't think I ever got an intuition for them. That, and normal subgroups. I was never able to find an example of a normal vs. non-normal subgroup that really made me understand why we care about normal ones. A reference to a good post online somewhere about it would be appreciated.

Posted by: Polymath | October 30, 2006 at 06:48 AM

for quotient groups, begin with the integers-mod-n.

for normal subgroups ... well, they're the ones that allow the formation of quotient groups!

"D's" looks right to me, too.

Posted by: vlorbik | October 31, 2006 at 10:42 AM

oh, and. minimal counterexample.

we call of subgroup H of G *normal in G* if the collection of the "left cosets"

{xH | x \in G}

coincides with the collection of "right cosets"

{Hx | x \in G}.

when H is normal, we can make a group out of the set

of its (left) cosets via (xH)*(yH) = (xy)H.

let G denote the symmetric group on three elements:

G = {(), (12), (23), (13), (123), (132)}.

consider the subgroup

B = {(), (12)}.

the left cosets of B are then:

B itself,

(13)B = {(13), (123)} (= (123)B), and

(23)B = {(23), (132)} (= (132)B).

but these do *not* form a group in the manner we considered a moment ago (as this would entail

(13)B*(12)B = (123)B \not= B

but at the same time

(123)B*(132)B = B

-- this is a contradiction).

see how easy that was?

Posted by: vlorbik | October 31, 2006 at 11:26 AM

vlorbik,

when i read the definitions for normal subgroups, i can follow them when i take excruciating care to calculate all the cosets (like you showed). and i know that every normal subgroup generates quotient groups with special properties involving isomorphisms between the subgroup and the kernel, and...something...i can't remember.

but i've never seen a example that makes me say "oh, THAT's why a normal subgroup is so much cooler."

not that it's your responsibility to find one--i think it's just that i haven't learned to think about the abstractions well enough.

Posted by: Polymath | November 01, 2006 at 08:55 AM

An example that (perhaps) shows why normal subgroups are so much cooler.

Consider the set of n by n matrices with entries in the real numbers and determinant one. This set (which we call SL_n) forms a group with the operation of matrix multiplication. Now, SL_n has lots and lots of subgroups. For instance, we could take any element A of SL_n and look at the set A,A^2,A^3, et cetera. Or we could look at all the matrices in SL_n that have rational entries. (It's pretty clear that these form a subgroup.) And so on and so forth. We could go on defining subgroups of SL_n all day, if we felt like it. But what about normal subgroups? These there are very few of! In fact, if n is odd, there is only the trivial subgroup, and if n is even, there is only the trivial subgroup and the one generated by -I (where I is the identity matrix). (The general fact, if we allowed, say, complex numbers, is that the only normal subgroups are the ones that are generated by scalar multiples of I.) Intuitively, what is going on is that conjugation (in any matrix group) amounts to a change of basis, so in order to be normal, a subgroup must be invariant under (certain) changes of basis. Clearly, scalar multiples of the identity have this property; intuitively it makes sense that nothing else does. (As usual, proving this is another matter.)

For another example, take the case of the symmetric group on a set X. In this case, conjugation is precisely equivalent to a "renaming" of the elements of X. So in order to be normal, a subgroup must be equivalent to itself under all relabelings of the underlying set. In the case X = {1,2,3} discussed above, the subgroup {(123),(132),()} is normal because we can clearly relabel X however we like and not change the subgroup. The subgroup {(12),()} is not normal because if we swap 1 and 3, it becomes a different subgroup, {(32),()}. Aside: it can be shown that ,as long as |X| > 4, the symmetric group on X has only one normal subgroup!

Two final notes. First, I think a key to understanding quotient groups (at least for me) is making the leap from the analogy

group elements == numbers

to the analogy

groups == numbers.

Second, it helps, when trying to understand the concept of a normal subgroup, to keep in mind that an alternative definition of "normal" is that a normal subgroup is the kernel of some homomorphism. This is a much more natural definintion than the one about being fixed under conjugation, and it also highlights the analogy between normal subgroups in group theory and ideals in ring theory. Anyway, I've gone on long enough now... hope some of it helps :-)

Posted by: Cat lover | November 01, 2006 at 11:19 PM

For some reason A-Level maths includes practically no group theory, thus I understand roughly none of the comments on it. Do they teach group theory in US high schools or do you lot just know it from uni or something?

Posted by: Paul Carpenter | November 02, 2006 at 08:28 AM

paul carpenter: right, from uni. i first saw this stuff senior year.

cat lover: very nice exposition! conjugation as relabeling ... why didn't *i* mention that?

polymath: i still maintain that Z/nZ is the killer example you seek. the point is that we put a group structure on the cosets: the very *elements* of, say, Z/2Z are

[0]:={...,-2, 0, 2, 4,...}

(the identity element) and

[1]:= {...,-3,-1,1,3,...}

(the other coset of [0] in Z). that stuff you don't remember is that the kernel (the normal subgroup) becomes the identity element in a new group etcetera. actually, i'm just guessing ... this might have been clear to you already ... far be it from me to ask you to give this matter any more thought than you feel it deserves ....

Posted by: vlorbik | November 03, 2006 at 01:00 PM

Perhaps part of the reason that normalness is such a non-obvious property is that most "obvious" groups (the ones we understand the best) are all abelian, so there are no non-normal subgroups.

Posted by: JBL | November 18, 2006 at 06:51 PM

Normal subgroups are closed under a sort of "conjugacy":

gh(g^-1) is in H for all g in G

(for all h in H, where H is a normal subgroup of G).

(In fact, thinking about it, this is precisely why G/H can be formed).

This general idea - that you can do APPLE (g) to something, then do ORANGE (h), then do APPLE^-1, and it's the same effect as if you just did an ORANGE - occurs over and over again in many diverse areas of maths.

e.g. using Fourier transforms, you can transform a problem to phase space, do a simplified calculation, then transform back - and it works just the same as if you'd done the (more complicated) calculation in the problem space.

another e.g. - the construct PA(P^-1), which occurs over and over again in linear algebra.

In a very general way, I guess this useful type of structure is part of the underlying motivation for considering normal subgroups.

Posted by: SL | January 11, 2007 at 04:05 PM

SL

thanks for that explanation. that's just the kind of thing i was looking for.

i can see now how imposing a group structure on a situation (and finding normal subgroups) could help in certain kinds of calculations or analysis, even though i still don't quite have an intuition for what they are.

much appreciated.

Posted by: Polymath | January 13, 2007 at 08:53 PM

I'm appropriating the "plane that won't crash and kill people" schtick. Thank you.

Posted by: Jonathan | February 07, 2007 at 06:39 AM