## An excursion into Group Theory.

When you have been regularly exposed to bad teaching its entirely

possible to miss something that deserves a big open mouthed “wow”, pass

without even something as small as an acknowledgment. This almost

happened to me.

I have been learning abstract algebra these days out of pure boredom

and I am fascinated doing every minute of it. Especially with basic

group theory. Anyone who has learned group theory under Anna University

would consider it extremely boring and a mind-numbing subject better

left to mathematicians (one good example of such a subject is measure

theory). Full of definitions and proofs of what seem to be utterly

useless concepts, completely devoid of any examples. In other words, dry.

There was absolutely no way one would even be able to concentrate

and sleep, let alone work your way through group theory. But that never

deterred me in trying to learn it then. That without doubt, was my most

boring and utterly confused learning excursion. I was exposed to

concepts out of the blue without any idea of where or how they come

about and where they would ever be used.

In particular was the concept of co-sets. The way I was introduced

to them, it seemed as if the only purpose of defining what a co-set

entirely was to prove that order of a subgroup divides the order of the

finite group (Lagrange’s theorem). And the proof was tedious and

equally mind numbing. But when I came across the second time, the proof

was the definition of elegance itself. When I realised that it was

actually the much feared Lagrange’s theorem it really blew me off! The

biggest mental ograsm that I’ve had in sometime! This definitely

deserves a big “WOW”.

And for the benefit of my more mathematically inclined readers, here is an abridged version of the proof.

Let G be a group and H any subgroup of G. Then, lets define the

relation “a is congruent to b mod H” for any two elements a and b in G

whenever ab’ (note : b’ is the inverse of b in G) is in H. It is

obvious for anyone to see that the above relation is an equivalence

relation, and hence would divide the the group G into k distinct

equivalence classes. Now consider the set Ha, which is the set of all

elements obtained by multiplying the elements of H with a, i.e Ha = { x

| x = ha, for all h in H }. We can use the definition of Ha to get to a very interesting fact.

x = ha is true for any x in Ha, multiplying by a’ on both sides we

see that xa’ is in H. Thus Ha contains only elements which are

congruent to a mod H. In fact, Ha contains all elements in G which are

congruent to a mod H, i.e, Ha = [a]. now n(Ha) + n(Hb) +…. = o(G) if

G is finite. Since for any a in G, n(Ha) = o(H) and there are totally k

equivalence classes k * o(H) = o(G). That is the much famed Lagrange’s

theorem. One small definition from basic set-theory and it simplifies

the whole proof of an amazing theorem! Does this deserve a wow or what!

Signing Off,

Vishnu Vyas.

AnneA great fortune is a great slavery…

ChristopherOctober 31, 2006 at 1:42 am