To us, here and now, it appears thus.

rants, ramblings and other things that make life worth living…

An excursion into Group Theory.

with one comment

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.


Written by vishnuvyas

March 26, 2006 at 10:27 am

One Response

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  1. Anne

    A great fortune is a great slavery…


    October 31, 2006 at 1:42 am

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