wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> Abelian Groups?
        
(Message started by: Barukh on Oct 23^rd, 2003, 2:59am)

Title: Abelian Groups?
Post by Barukh on Oct 23^rd, 2003, 2:59am

1. Prove that any group in which every element x satisfies x² = 1 is abelian.
2. What if x³ = 1 for every x?

Title: Re: Abelian Groups?
Post by towr on Oct 23^rd, 2003, 3:39am

If, like me, you're wondering what the ** Abelian groups are, go to http://mathworld.wolfram.com/AbelianGroup.html
in short:
"A group for which the elements commute (i.e., AB = BA for all elements A and B) is called an Abelian group."

Something else crucial I didn't know:
"A group G is a finite or infinite set of elements together with a binary operation which together satisfy the four fundamental properties of closure, associativity, the identity property, and the inverse property. "

Title: Re: Abelian Groups?
Post by towr on Oct 23^rd, 2003, 4:28am

I think I have 1)..
::[hide]
using random elements x and y from any group G satisfying forall x in G: xx=1

(xy)(xy) = 1
((xy)x)y = 1
((xy)x)yy = y
(xy)x = y
(xy)xx = yx
xy=yx, so for all x,y: xy=yx, and thus the group is Abelian.
[/hide]::

Title: Re: Abelian Groups?
Post by Barukh on Oct 23^rd, 2003, 5:58am

Good work, towr! Besides, I am glad you've learned some new stuff.

Title: Re: Abelian Groups?
Post by towr on Oct 23^rd, 2003, 6:05am

on 10/23/03 at 05:58:09, Barukh wrote:

Besides, I am glad you've learned some new stuff.

Yes, me too. I mean, that's what puzzles are for really, aren't they..

on the second question ::[hide] I'd guess no, but I don't know how to proof it.. Perhaps find a group for which it doesn't work?[/hide]::

Title: Re: Abelian Groups?
Post by wowbagger on Oct 23^rd, 2003, 6:08am

Funny what basic things these really clever people don't know... ;)

Not sure whether that's what you're after, Barukh, but in case 2.:
[hide]
(xy)(xy)(xy) = 1 leads to
(xy)(xy) = (yy)(xx) which means that

(xy)^-1 = y^-1x^-1 = (yy)(xx)
[/hide].

Title: Re: Abelian Groups?
Post by Barukh on Oct 23^rd, 2003, 7:59am

on 10/23/03 at 06:05:19, towr wrote:

on the second question ::

towr, your guess is right, as is the way to proof. However, it's not trivial, so here's a hint: [hide] one of the possible groups has matrices as elements[/hide]

Title: Re: Abelian Groups?
Post by Barukh on Oct 23^rd, 2003, 8:02am

on 10/23/03 at 06:08:12, wowbagger wrote:

Not sure whether that's what you're after, Barukh, but in case 2.:

wowbagger, your derivation is correct; however, how does it settle the question if the group is abelian?

Title: Re: Abelian Groups?
Post by wowbagger on Oct 23^rd, 2003, 8:11am

on 10/23/03 at 08:02:20, Barukh wrote:

how does it settle the question if the group is abelian?

Heh, it doesn't. As I said, I wasn't sure what you expected. It seems that the question from the first part carries over to the second. Anyway, it's nice to fiddle about with this stuff every once in a while, even though my "result" is probably not very interesting.
I was never really good at finding counter-examples.

Title: Re: Abelian Groups?
Post by Barukh on Oct 25^th, 2003, 7:49am

on 10/24/03 at 22:44:22, Pietro K.C. wrote:

How about ...

It's late. Is anything wrong with the above?

Pietro, I'm afraid, G is not a group: [hide]not every composition of 2 rotations is again a rotation in the group[/hide]. Or maybe, I didn't understand the definition of the group.

Here's a counter-example I'm familiar with: [hide]Take a group G of 3x3 matrices of the form:
1 a b0 1 c0 0 1where a, b, c are numbers 0, 1, 2, and the operation is ordinary matrix multiplication modulo 3. Simple manipulations reveal that G is not commutative, and every matrix in G raised to power 3 gives the identity matrix.

G consists of 27 elements (in group theory: has order 27). I have also seen another counter-example with the group of order 27. I am curious: is there a counter-example of order less than 27? [/hide]

Title: Re: Abelian Groups?
Post by Pietro K.C. on Oct 27^th, 2003, 6:01pm

Right you are. Duh.

I meant to define G as all possible compositions of such rotations (like defining a vector space by giving a basis and taking all linear combinations of its elements), but alas, I only "showed" x³ = 1 for the "basis" elements.

Before it got late and I posted all that nonsense, I tried to conjure up a counter-example involving matrices, which are non-commutative par excellence, but the calculations started to get a little too messy.

I thought your final question was very interesting. For all elements, x³ = 1, and the smallest non-abelian group you know of has order 3³... maybe there's something there?

So here's a follow-up: find, with proof, the non-abelian group of smallest order that satisfies [forall]x[in]G : x³ = 1.

Title: Re: Abelian Groups?
Post by Icarus on Nov 10^th, 2003, 5:10pm

on 10/23/03 at 07:59:40, Barukh wrote:

However, it's not trivial, so here's a hint: one of the possible groups has matrices as elements.

I know the question has been answered, but I would like to point out that this is not much of a hint!

All finite groups can be expressed as a group of matrices, as can all Lie groups (groups with "manifold" structure - i.e. locally they look like [bbr]ⁿ in a differentiable fashion, where n is finite).

Title: Re: Abelian Groups?
Post by towr on Nov 11^th, 2003, 12:42am

It may not be much of a hint to you, but it is to some of us. It implies there's a relatively small set of small matrices which do the trick. That helps limit the search space of all possible groups a lot
I was allready trying matrices though, but those were 2x2 matrices which didn't work out..

Title: Re: Abelian Groups?
Post by Barukh on Nov 11^th, 2003, 2:18am

towr, your last reply was a relief for me :D

Title: Re: Abelian Groups?
Post by ecoist on Dec 28^th, 2006, 9:26pm

Barukh asked if there is an example of order less than 27. The answer is no. Since every element is of order 1 or 3, the group is a finite 3-group. All p-groups of order p², where p is a prime, are abelian.

Another way to look at a counterexample is to note that GL(2,Z₃) contains a matrix of order three. This matrix and Z₃² form a semi-direct product of Z₃xZ₃ by an automorphism of order 3.

Title: Re: Abelian Groups?
Post by Eigenray on Dec 29^th, 2006, 12:59am

Suppose that G is a finite group. If more than 3/4 of the elements x satisfy x²=1, show that G is abelian.

Title: Re: Abelian Groups?
Post by Barukh on Dec 30^th, 2006, 5:24am

Let I http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subset.gif G be the set of elements satisfying x² = 1. Take s http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif I, and consider the set I(s) = I http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cap.gif sI. Since |sI| = |I| > 3/4 |G|, |I(s)| > |G|/2.

Now, for every r http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif I(s), we have r = st, t http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif I. Thus, r = s^-1t^-1. But r = r^-1 = t^-1s^-1 = ts, therefore s commutes with t. Because r was chosen arbitrarily, there are at least |I(s)| > |G|/2 elements commuting with s. Therefore, C(s) – the centralizer of s - is the whole G.

As s was also chosen arbitrarily, we get in the same way, C(I) = G. It means for every g http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif G, at least |I| > |G|/2 elements commute with g. Therefore, C(G) = G.

Title: Re: Abelian Groups?
Post by ecoist on Dec 30^th, 2006, 11:14am

Nice, Barukh! It appears that the three quarters condition is best possible because the direct product of an elementary abelian 2-group with the dihedral group of order 8 has exactly three quarters of its elements satisfying the equation x²=1.