Re: Problems from Herstein's topics in algebra

From: Troubled (mkajumap_at_hotmail.com)
Date: 06/21/04 

Date: Mon, 21 Jun 2004 02:28:44 GMT

"Ken Pledger" <Ken.Pledger@mcs.vuw.ac.nz> wrote in message
news:Ken.Pledger-4D5D03.12510221062004@bats.mcs.vuw.ac.nz...
> In article <YLpBc.3242$%47.38@news01.roc.ny>,
> "Troubled" <mkajumap@hotmail.com> wrote:
>
> > On page 32 Dr. Herstein goes on to say let G be the set of 2x2 matrices
of
> > the form (a b -b a) (the 1st row is the 1st 2 numbers and the 2nd row
the
> > last 2 numbers) with the operation being standard matrix multiplication.
...
> > He then asks if the multiplication if G remind you of anything. Well, I
> > don't see anything special (other than it being closed). Can someone
please
> > give me a leading hint?
>
>
> Look closely at the entries in a typical product
>
> ( a b) ( c d)
> (-b a) (-d c). You _have_ seen those expressions before.
>
>
> >
> > #2) If G is a group of order 5 I need to show that it is abelian. I did
this
> > for a group order 3 and 4 but am having difficulty with this one. I know
> > that it is cyclic (and hence abelian) by LaGrange's theorem but the
problem
> > came up before subgroups and Lagrange thm. I basically need to know what
> > elements of G to start off with. Of course I can 1st prove Lagrange's
thm
> > and then everything will easily follow ....
>
>
> and that's certainly the neatest way to do it. However, a very
> elementary proof is possible, although much trickier for order 5 than
> for orders up to 4. This question came up in sci.math in March, and I
> gave quite a careful explanation in a thread called "Proving
> associativity". If you can't find that, post here again.
>
> Ken Pledger.

Ok, I read and understood your proof--THANKS.
While waiting for a response I DID NOT just wait but I kept with the problem
and I have a question with something else that came up. I realized that G
could have either 0, 2 or 4 elements of order 2. If G has 4 elements of
order 2 than it is abelian ( common problem is to show that if x^2 = e for
all x in G then G is abelian). My question is how to show that exactly 2
elements of order 2 can't happen (as my Caley table shows). Of course this
result is an immediate consequence of your proof but I am looking for a more
direct proof.